# Using regularization to generate synthetic controls and conformal prediction for significance tests

When viewing past synthetic control results, one of things that has struck me is that the matching of the pre-trends is really good — almost too good in many cases (appears to be fitting to noise, although you may argue that is a feature in terms of matching exogenous shocks). For example, if you end up having a pre-treatment series of 10 years, and you have a potential donor pool the size of 30, you could technically pick 10 of them at random, fit a linear regression predicting the 10 observations in the treated unit, based on 10 covariates of the donor pool outcomes over the same pre time period, and get perfect predictions (ignoring the typical constraints one places on the coefficients).

So how do we solve that problem? One solution is to use regularized regression results (e.g. ridge regression, lasso), when the number of predictors is greater than the number of observations. So I can cast the matching procedure into a regression problem to generate the weights. Those regression procedures are typically used for forecasting, but don’t have well defined standard errors, and so subsequently are typically only used for point forecasts. One way to make inferences though is to generate the synthetic weights (here using lasso regression), and then use conformal prediction intervals to do our hypothesis testing of counterfactual trends.

Here I walk through an example using state panel crime data in R, full code and data can be downloaded here.

# A Synthetic Control Example

So first, these are the packages we need to replicate the results. `conformalInference` is not on CRAN yet, so use devtools to install it.

``````#library(devtools)
#install_github(repo="ryantibs/conformal", subdir="conformalInference")
library(conformalInference)
library(glmnet)
library(Synth)``````

Then I have prepped a nice state panel dataset of crime rates and counts from 1960 through 2014. I set a hypothetical treatment start year in 2005 just so I have a nice 10 years post data for illustration. That is a pretty good length pre-panel though, and a good number of potential donors.

``````MyDir <- "C:\\Users\\axw161530\\Desktop\\SynthIdeas"
setwd(MyDir)

TreatYear <- 2005

summary(LongData)``````

Next I prep my data, currently it is in long panel format, but I need it in wide format to fit the regression equations I want. I am just matching on violent crime rates here. I take out NY, as it is missing a few years of data. (This dataset also includes DC.) Then I split it up into my pre intervention and post intervention set.

``````#Changing the data to wide for just the violent offenses
wide <- LongData[,c('State','Year','Violent.Crime.rate')]
names(wide)[3] <- 'VCR'
wide <- reshape(wide, idvar="Year", timevar="State", direction="wide")
summary(wide)
#Take out NY because of NAs
wide <- wide[,c(1:33,35:52)]

wide_pre <- as.matrix(wide[wide\$Year < TreatYear,])
wide_post <- as.matrix(wide[wide\$Year >= TreatYear,])``````

Now onto the good stuff, we can estimate our lasso regression using the pre-data to get our weights. This constrains the coefficients to be positive and below 1. But does not have the constraint they sum to 1. I just choose Alabama as an example treated unit — I intentionally chose a state and year that should not have any effects for illustration and to check the coverage of my technique vs more traditional analyses.

You can see in my notes this is different than traditional synth in that it has an intercept as well. I was surprised, but the predictions in sample were really bad without the intercept no matter how I sliced it.

``````res <- glmnet(x=wide_pre[,3:51],y=wide_pre[,2],family="gaussian",
lower.limits=0,upper.limits=1,intercept=TRUE,standardize=FALSE,
alpha=1) #need the intercept, predictions suck otherwise``````

Even though this does not constrain the coefficients to sum to 1, it ends up with weights really close to that ideal anyway (sum of the non-intercept coefficients is just over 1.01). When I use crossvalidation it does not choose weights that sum to unity, but in sample the above code and the `cv.glmnet` are really similar in terms of predictions.

``````co_ridge <- as.matrix(coef(res))
fin <- co_ridge[,"s99"]
active <- fin[fin > 0] #Does not include intercept``````

If you print `active` we then have for our state weights (and the intercept is pretty tiny, -22). So not quite sure why eliminating the intercept was causing such problems in this example. So North Carolina just sneaks in, but otherwise the synthetic control is a mix of Arkansas, California, Kentucky, and Texas. The intercept is just a level shift, so we are still matching curves otherwise, so that does not bother me very much.

``````VCR.AR 0.2078156362
VCR.CA 0.1201658279
VCR.IL 0.1543015666
VCR.KY 0.2483613907
VCR.NC 0.0002896238
VCR.TX 0.2818272850``````

If we look at our predictions for the pre-time period, Alabama had the typical crime path, with a big raise going into the early 90’s and then a fall afterward (black line), and our in-sample predictions from the lasso regression are decent.

``````pre_pred <- predict(res,newx=wide_pre[,3:51],s=min(res\$lambda)) #for not cv results

plot(wide_pre[,1],wide_pre[,2],type='l',xlab='',ylab='Violent Crime Rate per 100,000')
points(wide_pre[,1],pre_pred,bg='red',pch=21) #Not too shabby
legend(1960,800,legend=c("Observed Albama","Predicted"),col=c("black","black"), pt.bg=c("black","red"), lty=c(1,NA), pch=c(NA,21))``````

Now to evaluate post intervention, we are going to generate conformal prediction intervals using a jackknife approach. Basically doing all the jazz of above, but leaving one pre year out at a time, and trying to predict Alabama’s violent crime rate for that left out year. Repeat that same process for all prior years, and we can get a calculation of the standard error of our prediction. Then apply that standard error to future years, so we can tell if the observed trend is different than the counterfactual we estimated (given the counterfactual has errors). I generate both 90% prediction intervals, as well as 99% prediction intervals.

``````train_fun <- function(x, y, out=NULL){
return( glmnet(x,y,alpha=1,standardize=FALSE,intercept=TRUE,nlambda=100,
lower.limits=0,upper.limits=1,family="gaussian")
)
}

pred_fun = function(out, newx) {
return(predict(out, newx, s=min(out\$lambda)))
}

limits_10 <- conformal.pred.jack(x=wide_pre[,3:51],y=wide_pre[,2],x0=wide_post[,3:51],
train.fun=train_fun,predict.fun=pred_fun,alpha=0.10,
verbose=TRUE)

limits_01 <- conformal.pred.jack(x=wide_pre[,3:51],y=wide_pre[,2],x0=wide_post[,3:51],
train.fun=train_fun,predict.fun=pred_fun,alpha=0.01,
verbose=TRUE)

plot(wide_post[,1],wide_post[,2],type='l',ylim=c(150,650),xlab='',ylab='Violent Crime Rate per 100,000')
points(wide_post[,1],post_pred,bg='red',pch=21)
lines(wide_post[,1],limits_10\$lo,col='grey')
lines(wide_post[,1],limits_10\$up,col='grey')
lines(wide_post[,1],limits_01\$lo,col='grey',lwd=3)
lines(wide_post[,1],limits_01\$up,col='grey',lwd=3)
legend("topright",legend=c("Observed Albama","Predicted","90% Pred. Int.","99% Pred. Int."),cex=0.7,
col=c("black","black","grey","grey"), pt.bg="red", lty=c(1,NA,1,1), pch=c(NA,21,NA,NA), lwd=c(1,1,1,3))``````

Then at the end of the above code snippet I made a plot. Black line is observed for Alabama from 05-14. Red dots are the estimated counterfactual based on the pre-weights. The lighter grey lines are then the prediction intervals. So we can see it is just outside the 90% intervals 3 times in the later years (would only expect 1 time), but all easily within the 99% intervals.

Note these are prediction intervals, not confidence intervals. Thinking about it I honestly don’t know whether we want prediction or confidence intervals in this circumstance, but prediction will be wider.

So this approach just matches on the pre-treated same outcome observations. To match on additional covariates, you can add them in as rows into the pre-treatment dataset (although you would want to normalize the values to a similar mean and standard deviation as the pre-treated outcome series).

You may also add in other covariates, like functions of time (although this changes the nature of the identification). So for example say you incorporate a linear and quadratic trend in time, and lasso only chooses those two time factors and no control areas. You are doing something more akin to interrupted time series analysis at that point (the counterfactual is simply based on your estimate of the pre-trend). Which I think is OK sometimes, but is quite different than using control areas to hopefully capture random shocks.

# Comparing to Traditional Synth results

To see whether my error intervals are similar to the placebo approach, I used the old school synth R package. It isn’t 100% comparable, as it makes you match on at least one covariate, so here I choose to also match on the average logged population over the pre-treatment period.

``````#NY is missing years
LongData_MinNY <- LongData[as.character(LongData\$State) != "NY",c("State","Year","Violent.Crime.rate","Population")]
LongData_MinNY\$StateNum <- as.numeric(LongData_MinNY\$State)
LongData_MinNY\$State <- as.character(LongData_MinNY\$State)
LongData_MinNY\$LogPop <- log(LongData_MinNY\$Population)

state_nums <- unique(LongData_MinNY\$StateNum)

dataprep.out <- dataprep(foo = LongData_MinNY,
dependent = "Violent.Crime.rate",
predictors = c("LogPop"),
unit.variable = "StateNum",
unit.names.variable = "State",
time.variable = "Year",
treatment.identifier = 2,
controls.identifier = state_nums[!state_nums %in% 2],
time.optimize.ssr = 1960:(TreatYear-1),
time.predictors.prior = 1960:(TreatYear-1),
time.plot = 1960:2014
)

synth_res <- synth(dataprep.out)
synth_tables <- synth.tab(dataprep.res = dataprep.out, synth.res = synth_res)
synth_tables\$tab.w #a bunch of little weights across the board
path.plot(synth.res = synth_res, dataprep.res = dataprep.out, tr.intake=TreatYear,Xlab='',Ylab='Violent Crime Rate per 100,000',
Legend=c("Alabama","Synthetic Control"), Legend.position=c("topleft"))``````

Looking at the weights, it is a bunch of little ones for many different states. Looking at the plot, it doesn’t appear to be any better fit than the lasso approach.

And then I just do the typical approach and use placebo checks to do inference. I loop over my 49 placebos (-1 state for NY, but +1 state because this list includes DC).

``````#Dataframes to stuff the placebos check results into
Predicted <- data.frame(dataprep.out\$Y0plot %*% synth_res\$solution.w)
names(Predicted) <- "TreatPred"

Pred_MinTreat <- data.frame(TreatPred = Predicted\$TreatPred - LongData_MinNY[LongData_MinNY\$StateNum == 2,"Violent.Crime.rate"])

#Now I just need to loop over the other states and collect their results for the placebo tests

placebos <- state_nums[!state_nums %in% 2]
for (i in placebos){
dataprep.plac <- dataprep(foo = LongData_MinNY,
dependent = "Violent.Crime.rate",
predictors = c("LogPop"),
unit.variable = "StateNum",
unit.names.variable = "State",
time.variable = "Year",
treatment.identifier = i,
controls.identifier = state_nums[!state_nums %in% i],
time.optimize.ssr = 1960:(TreatYear-1),
time.predictors.prior = 1960:(TreatYear-1),
time.plot = 1960:2014
)
synth_resP <- synth(dataprep.plac)
synth_tablesP <- synth.tab(dataprep.res = dataprep.plac, synth.res = synth_resP)
nm <- paste0("S.",i)
Predicted[,nm] <- dataprep.plac\$Y0plot %*% synth_resP\$solution.w
Pred_MinTreat[,nm] <- Predicted[,nm] - LongData_MinNY[LongData_MinNY\$StateNum == i,"Violent.Crime.rate"]
}``````

If you look at the synth estimates for Alabama (grey circles), they are almost exactly the same as the lasso predictions (red circles), even though the weights are very different.

``````PredRecent <- Predicted[1960:2014 >= TreatYear,]
DiffRecent <- Pred_MinTreat[1960:2014 >= TreatYear,]

plot(wide_post[,1],wide_post[,2],type='l',ylim=c(100,700),xlab='',ylab='Violent Crime Rate per 100,000')
points(wide_post[,1],post_pred,bg='red',pch=21)
lines(wide_post[,1],limits_10\$lo,col='grey')
lines(wide_post[,1],limits_10\$up,col='grey')
lines(wide_post[,1],limits_01\$lo,col='grey',lwd=3)
lines(wide_post[,1],limits_01\$up,col='grey',lwd=3)
points(wide_post[,1],PredRecent\$TreatPred,bg='grey',pch=21)
legend("topright",legend=c("Observed Albama","Lasso Pred.","90% Pred. Int.","99% Pred. Int.","Synth Pred."),cex=0.6,
col=c("black","black","grey","grey"), pt.bg=c(NA,"red",NA,NA,"grey"), lty=c(1,NA,1,1,NA), pch=c(NA,21,NA,NA,21), lwd=c(1,1,1,3,1))``````

But when we look at variation in our placebo results (thin, purple lines), they are much wider than our conformal prediction intervals.

``````plot(wide_post[,1],wide_post[,2]-post_pred,type='l',ylim=c(-500,500),xlab='',ylab='Observed - Predicted (Violent Crime Rates)')
points(wide_post[,1],post_pred-post_pred,bg='red',pch=21)
lines(wide_post[,1],limits_01\$lo-post_pred,col='grey',lwd=3)
lines(wide_post[,1],limits_01\$up-post_pred,col='grey',lwd=3)

for (i in 2:ncol(PredRecent)){
lines(wide_post[,1],DiffRecent[,i],col='#9400D340',lwd=0.5)
}

legend(x=2005.5,y=-700,legend=c("Observed Albama","Lasso Pred.","99% Pred. Int.","Placebos"),
col=c("black","black","grey",'#9400D3'), pt.bg=c(NA,"red",NA,NA), lty=c(1,NA,1,1),
pch=c(NA,21,NA,NA), lwd=c(1,1,3,0.5), xpd=TRUE, horiz=TRUE, cex = 0.45)``````

So I was hoping they would be the same (conformal would cover the placebo at the expected rate), but alas they are not. So I’m not sure if my conformal intervals are too small, or the placebo checks are extra noisy. I can’t prove it, but I suspect the placebo checks are somewhat noisy, mainly because there will always be some intervention that is idiosyncratic to specific donors over long periods of time that makes them no longer good counterfactuals. This seems especially true if you consider predictions further out from the treatment year. Although I find the logic of the placebo checks pretty convincing, so I am somewhat torn.

Since we have in this example 49 donors, the two-tailed p-value for being outside the placebos would be 2/(49+1)=0.04. Here we would need an intervention that either increased violent crime rates by plus/minus 400 per 100,000, pretty much an impossible standard given a baseline of only 400 crimes per 100,000 as of 2004. The 99% conformal intervals are still pretty wide, with an increase/decrease of about 150 violent crimes per 100,000 needed to be a significant change. The two lines way outside 400 happen to be Alaska and Wyoming, not DC, so maybe a tiny population state results in higher volatility problem. But besides them there are a bunch of placebo states around plus/minus 300 as well.

So caveat emptor if you want to use this idea in your own work, I don’t know if my suggestion is good or bad. Here it suggests its more diagnostic (smaller intervals) than the placebo checks, and isn’t limited by the number of potential donors in setting the alpha level for your tests (e.g. if you only have 10 potential donors your placebo checks are only 90% intervals).

Since this is just one example, there are a few things I would need to know before recommending it more generally. One is that it may not work with smaller pre time series and/or a smaller donor pool. (Not sure of any better way of checking than via a ton of different simulations.)

# More general notes

Doing some more lit review while preparing this post, I appear to be like 15th in line to suggest this approach (so don’t take it as novel). In terms of using the lasso to estimate the synth weights, it seems Susan Athey and colleagues proposed something similar in addition to using other machine learning techniques. Also see Amjad et al. 2018 in the Journal of Machine Learning, and this workshop by Alex Hollingsworth and Coady Wing. I am not even the first one to think to use conformal prediction intervals apparently, see this working paper (Chernozhukov, Wuthrich, and Zhu, 2019) posted just a few weeks prior.

There is another R package, gsynth, that appears to solve the problem of p > n via a variable reduction technique (Xu, 2017). Xu also discusses how incorporating more information is really making different identification assumptions. So again just getting good predictions/minimizing the in-sample mean square error is not necessarily the right approach to get correct causal inferences.

Just a blog post, so again can’t say if this is an improvement over other work offhand. This is just illustrative that the bounds for the conformal prediction may be smaller than the typical permutation based approach. Casting it as a regression problem I intuitively grok more, and think opens up more possibilities. For example, you may want to use binomial logistic models instead of linear for the fitting process (so takes into account more volatility for smaller population states).

# Making a hexbin map in ggplot

In a recent working paper I made a hexbin map all in R. (Gio did most of the hard work of data munging and modeling though!) Figured I would detail the process here for some notes. Hexagon binning is purportedly better than regular squares (to avoid artifacts of runs in discretized data). But the reason I use them in this circumstance is mostly just an aesthetic preference.

Two tricky parts to this: 1) making the north arrow and scale bar, and 2) figuring out the dimensions to make regular hexagons. As an illustration I use the shooting victim data from Philly (see the working paper for all the details) full data and code to replicate here. I will walk through a bit of it though.

## Data Prep

First to start out, I just use these three libraries, and set the working directory to where my data is.

``````library(ggplot2)
library(rgdal)
library(proj4)
setwd('C:\\Users\\axw161530\\Dropbox\\Documents\\BLOG\\HexagonMap_ggplot\\Analysis')``````

Now I read in the Philly shooting data, and then an outline of the city that is projected. Note I read in the shapefile data using `rgdal`, which imports the projection info. I need that to be able to convert the latitude/longitude spherical coordinates in the shooting data to a local projection. (Unless you are making a webmap, you pretty much always want to use some type of local projection, and not spherical coordinates.)

``````#Read in the shooting data
#Get rid of missing
shoot <- shoot[!is.na(shoot\$lng),c('lng','lat')]
#Project the Shooting data
phill_pj <- proj4string(PhilBound)
XYMeters <- proj4::project(as.matrix(shoot[,c('lng','lat')]), proj=phill_pj)
shoot\$x <- XYMeters[,1]
shoot\$y <- XYMeters[,2]``````

## Making a Basemap

It is a bit of work to make a nice basemap in R and ggplot, but once that upfront work is done then it is really easy to make more maps. To start, the `GISTools` package has a set of functions to get a north arrow and scale bar, but I have had trouble with them. The `ggsn` package imports the north arrow as a bitmap instead of vector, and I also had a difficult time with its scale bar function. (I have not figured out the `cartography` package either, I can’t keep up with all the mapping stuff in R!) So long story short, this is my solution to adding a north arrow and scale bar, but I admit better solutions probably exist.

So basically I just build my own polygons and labels to add into the map where I want. Code is motivated based on the functions in `GISTools`.

``````#creating north arrow and scale bar, motivation from GISTools package
arrow_data <- function(xb, yb, len) {
s <- len
arrow.x = c(0,0.5,1,0.5,0) - 0.5
arrow.y = c(0,1.7  ,0,0.5,0)
adata <- data.frame(aX = xb + arrow.x * s, aY = yb + arrow.y * s)
}

scale_data <- function(llx,lly,len,height){
box1 <- data.frame(x = c(llx,llx+len,llx+len,llx,llx),
y = c(lly,lly,lly+height,lly+height,lly))
box2 <- data.frame(x = c(llx-len,llx,llx,llx-len,llx-len),
y = c(lly,lly,lly+height,lly+height,lly))
return(list(box1,box2))
}

x_cent <- 830000
len_bar <- 3000
offset_scaleNum <- 64300
arrow <- arrow_data(xb=x_cent,yb=67300,len=2500)
scale_bxs <- scale_data(llx=x_cent,lly=65000,len=len_bar,height=750)

lab_data <- data.frame(x=c(x_cent, x_cent-len_bar, x_cent, x_cent+len_bar, x_cent),
y=c( 72300, offset_scaleNum, offset_scaleNum, offset_scaleNum, 66500),
lab=c("N","0","3","6","Kilometers"))``````

This is about the best I have been able to automate the creation of the north arrow and scale bar polygons, while still having flexibility where to place the labels. But now we have all of the ingredients necessary to make our basemap. Make sure to use `coord_fixed()` for maps! Also for background maps I typically like making the outline thicker, and then have borders for smaller polygons lighter and thinner to create a hierarchy. (If you don’t want the background map to have any color, use `fill=NA`.)

``````base_map <- ggplot() +
geom_polygon(data=PhilBound,size=1.5,color='black', fill='darkgrey', aes(x=long,y=lat)) +
geom_polygon(data=arrow, fill='black', aes(x=aX, y=aY)) +
geom_polygon(data=scale_bxs[[1]], fill='grey', color='black', aes(x=x, y = y)) +
geom_polygon(data=scale_bxs[[2]], fill='white', color='black', aes(x=x, y = y)) +
geom_text(data=lab_data, size=4, aes(x=x,y=y,label=lab)) +
coord_fixed() + theme_void()

#Check it out
base_map``````

This is what it looks like on my windows machine in RStudio — it ends up looking alittle different when I export the figure straight to PNG though. Will get to that in a minute.

## Making a hexagon map

Now you have your basemap you can superimpose whatever other data you want. Here I wanted to visualize the spatial distribution of shootings in Philly. One option is a kernel density map. I tend to like aggregated count maps though better for an overview, since I don’t care so much for drilling down and identifying very specific hot spots. And the counts are easier to understand than densities.

In `geom_hex` you can supply a vertical and horizontal parameter to control the size of the hexagon — supplying the same for each does not create a regular hexagon though. The way the hexagon is oriented in `geom_hex` the vertical parameter is vertex to vertex, whereas the horizontal parameter is side to side.

Here are three helper functions. First, `wd_hex` gives you a horizontal width length given the vertical parameter. So if you wanted your hexagon to be vertex to vertex to be 1000 meters (so a side is 500 meters), `wd_hex(1000)` returns just over 866. Second, if for your map you wanted to convert the numbers to densities per unit area, you can use `hex_area` to figure out the size of your hexagon. Going again with our 1000 meters vertex to vertex hexagon, we have a total of `hex_area(1000/2)` is just under 650,000 square meters (or about 0.65 square kilometers).

For maps though, I think it makes the most sense to set the hexagon to a particular area. So `hex_dim` does that. If you want to set your hexagons to a square kilometer, given our projected data is in meters, we would then just do `hex_dim(1000^2)`, which with rounding gives us vert/horz measures of about (1241,1075) to supply to `geom_hex`.

``````#ggplot geom_hex you need to supply height and width
#if you want a regular hexagon though, these
#are not equal given the default way geom_hex draws them
#https://www.varsitytutors.com/high_school_math-help/how-to-find-the-area-of-a-hexagon

#get width given height
wd_hex <- function(height){
tri_side <- height/2
sma_side <- height/4
width <- 2*sqrt(tri_side^2 - sma_side^2)
return(width)
}

#now to figure out the area if you want
#side is simply height/2 in geom_hex
hex_area <- function(side){
area <- 6 * (  (sqrt(3)*side^2)/4 )
return(area)
}

#So if you want your hexagon to have a regular area need the inverse function
#Gives height and width if you want a specific area
hex_dim <- function(area){
num <- 4*area
den <- 6*sqrt(3)
vert <- 2*sqrt(num/den)
horz <- wd_hex(height)
return(c(vert,horz))
}

my_dims <- hex_dim(1000^2)   #making it a square kilometer
sqrt(hex_area(my_dims[1]/2)) #check to make sure it is square km
#my_dims also checks out with https://hexagoncalculator.apphb.com/``````

Now onto the good stuff. I tend to think discrete bins make nicer looking maps than continuous fills. So through some trial/error you can figure out the best way to make those via `cut`. Also I make the outlines for the hexagons thin and white, and make the hexagons semi-transparent. So you can see the outline for the city. I like how by default areas with no shootings are not given any hexagon.

``````lev_cnt <- seq(0,225,25)
shoot_count <- base_map +
geom_hex(data=shoot, color='white', alpha=0.85, size=0.1, binwidth=my_dims,
aes(x=x,y=y,fill=cut(..count..,lev_cnt))) +
scale_fill_brewer(name="Count Shootings", palette="OrRd")``````

We have come so far, now to automate exporting the figure to a PNG file. I’ve had trouble getting journals recently to not bungle vector figures that I forward them, so I am just like going with high res PNG to avoid that hassle. If you render the figure and use the GUI to export to PNG, it won’t be as high resolution, so you can often easily see aliasing pixels (e.g. the pixels in the North Arrow for the earlier base map image).

``````png('Philly_ShootCount.png', height=5, width=5, units="in", res=1000, type="cairo")
shoot_count
dev.off()``````

Note the font size/location in the exported PNG are often not quite exactly as they are when rendered in the RGUI window or RStudio on my windows machine. So make sure to check the PNG file.

# Weighted buffers in R

Had a request not so recently about implementing weighted buffer counts. The idea behind a weighted buffer is that instead of say counting the number of crimes that happen within 1,000 meters of a school, you want to give events that are closer to the school more weight.

There are two reasons you might want to do this for crime analysis:

• You want to measure the amount of crime around a location, but you rather have a weighted crime count, where crimes closer to the location have a greater weight than those further away.
• You want to measure attributes nearby a location (so things that predict crime), but give a higher weight to those closer to a location.

The second is actually more common in academic literature — see John Hipp’s Egohoods, or Liz Groff’s work on measuring nearby to bars, or Joel Caplan and using kernel density to estimate the effect of crime generators. Jerry Ratcliffe and colleagues work on the buffer intensity calculator is actually the motivation for the original request. So here are some quick code snippets in R to accomplish either. Here is the complete code and original data to replicate.

Here I use over 250,000 reported Part 1 crimes in DC from 08 through 2015, 173 school locations, and 21,506 street units (street segment midpoints and intersections) I constructed for various analyses in DC (all from open data sources) as examples.

## Example 1: Crime Buffer Intensities Around Schools

First, lets define where our data is located and read in the CSV files (don’t judge me setting the directory, I do not use RStudio!)

``````MyDir <- 'C:\\Users\\axw161530\\Dropbox\\Documents\\BLOG\\buffer_stuff_R\\Code' #Change to location on your machine!
setwd(MyDir)

Now there are several ways to do this, but here is the way I think will be most useful in general for folks in the crime analysis realm. Basically the workflow is this:

• For a given school, calculate the distance between all of the crime points and that school
• Apply whatever function to that distance to get your weight

For the function to the distance there are a bunch of choices (see Jerry’s buffer intensity I linked to previously for some example discussion). I’ve written previously about using the bi-square kernel. So I will illustrate with that.

Here is an example for the first school record in the dataset.

``````#Example for crimes around school, weighted by Bisquare kernel
BiSq_Fun <- function(dist,b){
ifelse(dist < b, ( 1 - (dist/b)^2 )^2, 0)
}

S1 <- t(SchoolLoc[1,2:3])
Dis <- sqrt( (CrimeData\$BLOCKXCOORD - S1[1])^2 + (CrimeData\$BLOCKYCOORD - S1[2])^2 )
Wgh <- sum( BiSq_Fun(Dis,b=2000) )``````

Then repeat that for all of the locations that you want the buffer intensities, and stuff it in the original `SchoolLoc` data frame. (Takes less than 30 seconds on my machine.)

``````SchoolLoc\$BufWeight <- -1 #Initialize field

#Takes about 30 seconds on my machine
for (i in 1:nrow(SchoolLoc)){
S <- t(SchoolLoc[i,2:3])
Dis <- sqrt( (CrimeData\$BLOCKXCOORD - S[1])^2 + (CrimeData\$BLOCKYCOORD - S[2])^2 )
SchoolLoc[i,'BufWeight'] <- sum( BiSq_Fun(Dis,b=2000) )
}``````

In this example there are 173 schools and 276,621 crimes. It is too big to create all of the pairwise comparisons at once (which will generate nearly 50 million records), but the looping isn’t too cumbersome and slow to worry about building a KDTree.

One thing to note about this technique is that if the buffers are large (or you have locations nearby one another), one crime can contribute to weighted crimes for multiple places.

## Example 2: Weighted School Counts for Street Units

To extend this idea to estimating attributes at places just essentially swaps out the crime locations with whatever you want to calculate, ala Liz Groff and her inverse distance weighted bars paper. I will show something alittle different though, in using the weights to create a weighted sum, which is related to John Hipp and Adam Boessen’s idea about Egohoods.

So here for every street unit I’ve created in DC, I want an estimate of the number of students nearby. I not only want to count the number of kids in attendance in schools nearby, but I also want to weight schools that are closer to the street unit by a higher amount.

So here I read in the street unit data. Also I do not have school attendance counts in this dataset, so I just simulate some numbers to illustrate.

``````StreetUnits <- read.csv('DC_StreetUnits.csv')
StreetUnits\$SchoolWeight <- -1 #Initialize school weight field

SchoolLoc\$StudentNum <- round(runif(nrow(SchoolLoc),100,2000)) ``````

Now it is very similar to the previous example, you just do a weighted sum of the attribute, instead of just counting up the weights. Here for illustration purposes I use a different weighting function, inverse distance weighting with a distance cut-off. (I figured this would need a better data management strategy to be timely, but this loop works quite fast as well, again under a minute on my machine.)

``````#Will use inverse distance weighting with cut-off instead of bi-square
Inv_CutOff <- function(dist,cut){
ifelse(dist < cut, 1/dist, 0)
}

for (i in 1:nrow(StreetUnits)){
SU <- t(StreetUnits[i,2:3])
Dis <- sqrt( (SchoolLoc\$XMeters - SU[1])^2 + (SchoolLoc\$YMeters - SU[2])^2 )
Weights <- Inv_CutOff(Dis,cut=8000)
StreetUnits[i,'SchoolWeight'] <- sum( Weights*SchoolLoc\$StudentNum )
}   ``````

The same idea could be used for other attributes, like sales volume for restaurants to get a measure of the business of the location (I think more recent work of John Hipp’s uses the number of employees).

Some attributes you may want to do the weighted mean instead of a weighted sum. For example, if you were using estimates of the proportion of residents in poverty, it makes more sense for this measure to be a spatially smoothed mean estimate than a sum. In this case it works exactly the same but you would replace `sum( Weights*SchoolLoc\$StudentNum )` with `sum( Weights*SchoolLoc\$StudentNum )/sum(Weights)`. (You could use the centroid of census block groups in place of the polygon data.)

## Some Wrap-Up

Using these buffer weights really just swaps out one arbitrary decision for data analysis (the buffer distance) with another (the distance weighting function). Although the weighting function is more complicated, I think it is probably closer to reality for quite a few applications.

Many of these different types of spatial estimates are all related to another (kernel density estimation, geographically weighted regression, kriging). So there are many different ways that you could go about making similar estimates. Not letting the perfect be the enemy of the good, I think what I show here will work quite well for many crime analysis applications.

# Monitoring Use of Force in New Jersey

Recently ProPublica published a map of uses-of-force across different jurisdictions in New Jersey. Such information can be used to monitor whether agencies are overall doing a good or bad job.

I’ve previously discussed the idea of using funnel charts to spot outliers, mostly around homicide rates but the idea is the same when examining any type of rate. For example in another post I illustrated its use for examining rates of officer involved shootings.

Here is another example applying it to lesser uses of force in New Jersey. Below is the rate of use of force reports per the total number of arrests. (Code to replicate at the end of the post.)

The average use of force per arrests in the state is around 3%. So the error bars show relative to the state average. Here is an interactive chart in which you can use tool tips to see the individual jurisdictions.

Now the original press release noted by Seth Stoughton on twitter noted that several towns have ratio’s of black to white use of force that are very high. Scott Wolfe suspected that was partly a function of smaller towns will have more variable rates. Basically as one is comparing the ratio between two rates with error, the error bars around the rate ratio will also be quite large.

Here is the chart showing the same type of funnel around the rate ratio of black to white use-of-force relative to the average over the whole sample (the black percent use of force is 3.2 percent of arrests, and the white percent use of force is 2.4, and the rate ratio between the two is 1.35). I show in the code how I constructed this, which I should write a blog post about itself, but in short there are decisions I could make to make the intervals wider. So the points that are just slightly above a ratio of 2 at around 10,000 arrests are arguably not outliers, those more to the top-right of the plot though are much better evidence. (I’d note that if one group is very small, you could always make these error bars really large, so to construct them you need to make reasonable assumptions about the size of the two groups you are comparing.)

And here is another interactive chart in which you can view the outliers again. The original press release, Millville, Lakewood, and South Orange are noted as outliers. Using arrests as the denominator instead of population, they each have a rate ratio of around 2. In this chart Millville and Lakewood are outside the bounds, but just barely. South Orange is within the bounds. So those aren’t the places I would have called out according to this chart.

That same twitter thread other folks noted the potential reliability/validity of such data (Pete Moskos and Kyle McLean). These charts cannot say why individual agencies are outliers — either high or low. It could be their officers are really using force at different rates, it could also be though they are using different definitions to reporting force. There are also potential other individual explanations that explain the use of force distribution as well as the ratio differences in black vs white — no doubt policing in Princeton vs Camden are substantively different. Also even if all individual agencies are doing well, it does not mean there are no potential problem officers (as noted by David Pyrooz, often a few officers contribute to most UoF).

Despite these limitations, I still think there is utility in this type of monitoring though. It is basically a flag to dig deeper when anomalous patterns are spotted. Those unaccounted for factors contribute to more points being pushed outside of my constructed limits (overdispersion), but more clearly indicate when a pattern is so far outside the norm of what is expected the public deserves some explanation of the pattern. Also it highlights when agencies are potentially doing good, and so can be promoted according to their current practices.

This is a terrific start to effectively monitoring police agencies by ProPublica — state criminal justice agencies should be doing this themselves though.

Here is the code to replicate the analysis.

# Projecting spatial data in Python and R

I use my blog as sort of a scholarly notebook. I often repeatedly do a task, and then can’t find where I did it previously. One example is projecting crime data, so here are my notes on how to do that in python and R.

Commonly I want to take public crime data that is in spherical lat/lon coordinates and project it to some local projection. Most of the time so I can do simply euclidean geometry (like buffers within X feet, or distance to the nearest crime generator in meters). Sometimes you need to do the opposite — if I have the projected data and I want to plot the points on a webmap it is easier to work with the lat/lon coordinates. As a note, if you import your map data and then your points are not on the map (or in a way off location), there is some sort of problem with the projection.

I used to do this in ArcMap (toolbox -> Data Management -> Projections), but doing it these programs are faster. Here are examples of going back and forth for some Dallas coordinates. Here is the data and code to replicate the post.

# Python

In python there is a library `pyproj` that does all the work you need. It isn’t part of the default python packages, so you will need to install it using pip or whatever. Basically you just need to define the to/from projections you want. Also it always returns the projected coordinates in meters, so if you want feet you need to do a conversions from meters to feet (or whatever unit you want). For below `p1` is the definition you want for lat/lon in webmaps (which is not a projection at all). To figure out your local projection though takes a little more work.

To figure out your local projection I typically use this online tool, prj2epsg. You can upload a `prj` file, which is the locally defined projection file for shapefiles. (It is plain text as well, so you can just open in a text editor and paste into that site as well.) It will then tell you want EPSG code corresponds to your projection.

Below illustrates putting it all together and going back and forth for an example area in Dallas. I tend to write the functions to take one record at a time for use in various workflows, but I am sure someone can write a vectorized version though that will take whole lists that is a better approach.

``````import pyproj

#These functions convert to/from Dallas projection
#In feet to lat/lon
p1 = pyproj.Proj(proj='latlong',datum='WGS84')
p2 = pyproj.Proj(init='epsg:2276') #show how to figure this out, http://spatialreference.org/ref/epsg/ and http://prj2epsg.org/search
met_to_feet = 3.280839895 #http://www.meters-to-feet.com/

#This converts Lat/Lon to projected coordinates
def DallConvProj(Lat,Lon):
#always returns in meters
if abs(Lat) > 180 or abs(Lon) > 180:
return (None,None)
else:
x,y = pyproj.transform(p1, p2, Lon, Lat)
return (x*met_to_feet, y*met_to_feet)

#This does the opposite, coverts projected to lat/lon
def DallConvSph(X,Y):
if abs(X) < 2000000 or abs(Y) < 6000000:
return (None,None)
else:
Lon,Lat = pyproj.transform(p2, p1, X/met_to_feet, Y/met_to_feet)
return (Lon, Lat)

#check coordinates
x1 = -96.828295; y1 = 32.832521
print DallConvProj(Lat=y1,Lon=x1)

x2 = 2481939.934525765; y2 = 6989916.200679892
print DallConvSph(X=x2, Y=y2)``````

# R

In R I use the library `proj4` to do the projections for point data. R can read in the projection data from a file as well using the `rgdal` library.

``````library(proj4)
library(rgdal)

MyDir <- "C:\\Users\\axw161530\\Dropbox\\Documents\\BLOG\\Projections_R_Python"
setwd(MyDir)
DalProj <- proj4string(DalBound)

ProjData <- data.frame(x=c(2481939.934525765),
y=c(6989916.200679892),
lat=c(32.832521),
lon=c(-96.828295))

LatLon <- proj4::project(as.matrix(ProjData[,c('x','y')]), proj=DalProj, inverse=TRUE)
#check to see if true
cbind(ProjData[,c('lon','lat')],as.data.frame(LatLon))

XYFeet <- proj4::project(as.matrix(ProjData[,c('lon','lat')]), proj=DalProj)
cbind(ProjData[,c('x','y')],XYFeet)

plot(DalBound)
points(ProjData\$x,ProjData\$y,col='red',pch=19,cex=2)``````

The last plot function shows that the XY point is within the Dallas basemap for the projected boundary. But if you want to project the boundary file as well, you can use the `spTransform` function. Here I have a simple example of tacking the projected boundary file and transforming to lat/lon, so can be superimposed on a leaflet map.

Additionally I show a trick I sometimes use for maps by transforming the boundary polygon to a polyline, as it provides easier styling options sometimes.

``````#transform boundary to lat/lon
DalLatLon <- spTransform(DalBound,CRS("+init=epsg:4326") )
plot(DalLatLon)
points(ProjData\$lon,ProjData\$lat,col='red',pch=19,cex=2)

#Leaflet useful for boundaries to be lines instead of areas
DallLine <- as(DalLatLon, 'SpatialLines')
library(leaflet)

BaseMapDallas <- leaflet() %>%
addProviderTiles(providers\$OpenStreetMap, group = "Open Street Map") %>%
addProviderTiles(providers\$CartoDB.Positron, group = "CartoDB Lite") %>%
addPolylines(data=DallLine, color='black', weight=4, group="Dallas Boundary Lines") %>%
addPolygons(data=DalLatLon,color = "#1717A1", weight = 1, smoothFactor = 0.5,
opacity = 1.0, fillOpacity = 0.5, group="Dallas Boundary Area") %>%
addLayersControl(baseGroups = c("Open Street Map","CartoDB Lite"),
overlayGroups = c("Dallas Boundary Area","Dallas Boundary Lines"),
options = layersControlOptions(collapsed = FALSE)) %>%
hideGroup("Dallas Boundary Lines")

BaseMapDallas``````

I have too much stuff in the blog queue at the moment, but hopefully I get some time to write up my notes on using leaflet maps in R soon.

# Some more testing coefficient contrasts: Multinomial models and indirect effects

Testing the equality of two coefficients is one of my more popular posts. This is a good thing — often more interesting hypotheses are to test two parameters against each other, as opposed to a strict null hypothesis of a coefficient against zero. Every now an then I get questions about applying this idea to new situations in which it is not always straightforward how to figure out. So here are a few examples using demonstration R code.

# Multinomial Models

One question I received about applying the advice was to test coefficients across different contrasts in multinomial models. It may not seem obvious, but the general approach of extracting out the coefficients and the covariance between those estimates works the same way as most regression equations.

So in a quick example in R:

``````library(nnet)
data(mtcars)
library(car)

mtcars\$cyl <- as.factor(mtcars\$cyl)
mtcars\$am <- as.factor(mtcars\$am)
mod <- multinom(cyl ~ am + hp, data=mtcars, Hess=TRUE)
summary(mod)``````

And the estimates for `mod` are:

``````> summary(mod)
Call:
multinom(formula = cyl ~ am + hp, data = mtcars)

Coefficients:
(Intercept)       am1        hp
6   -42.03847  -3.77398 0.4147498
8   -92.30944 -26.27554 0.7836576

Std. Errors:
(Intercept)       am1        hp
6    27.77917  3.256003 0.2747842
8    31.93525 46.854100 0.2559052

Residual Deviance: 7.702737
AIC: 19.70274 ``````

So say we want to test whether the `hp` effect is the same for `6` cylinders vs `8` cylinders. To test that, we just grab the covariance and construct our test:

``````#Example constructing test by hand
v <- vcov(mod)
c <- coef(mod)
dif <- c[1,3] - c[2,3]
se <- sqrt( v[3,3] + v[6,6] - 2*v[3,6])
z <- dif/se
#test stat, standard error, and two-tailed p-value
dif;se;2*(1 - pnorm(abs(z)))``````

Which we end up with a p-value of `0.0002505233`, so we would reject the null that these two effects are equal to one another. Note to get the variance-covariance estimates for the parameters you need to set `Hess=TRUE` in the `multinom` call.

Another easier way though is to use the `car` libraries function `linearHypothesis` to conduct the same test:

``````> linearHypothesis(mod,c("6:hp = 8:hp"),test="Chisq")
Linear hypothesis test

Hypothesis:
6:hp - 8:hp = 0

Model 1: restricted model
Model 2: cyl ~ am + hp

Df  Chisq Pr(>Chisq)
1
2  1 13.408  0.0002505 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1``````

You can see although this is in terms of a Chi-square test, it results in the same p-value. The Wald test however can be extended to testing multiple coefficient equalities, and a popular one for multinomial models is to test if any coefficients change across different levels of the dependent categories. The idea behind that test is to see if you can collapse that category with another that is equivalent.

To do that test, I created a function that does all of the contrasts at once:

``````#Creating function to return tests for all coefficient equalities at once
all_tests <- function(model){
v <- colnames(coef(model))
d <- rownames(coef(model))
allpairs <- combn(d,2,simplify=FALSE)
totn <- length(allpairs) + length(d)
results <- data.frame(ord=1:totn)
results\$contrast <- ""
results\$test <- ""
results\$Df <- NULL
results\$Chisq <- NULL
results\$pvalue <- NULL
iter <- 0
for (i in allpairs){
iter <- iter + 1
l <- paste0(i[1],":",v)
r <- paste0(i[2],":",v)
test <- paste0(l," = ",r)
temp_res <- linearHypothesis(model,test,test="Chisq")
results\$contrast[iter] <- paste0(i[1]," vs ",i[2])
results\$test[iter] <- paste(test,collapse=" and ")
results\$Df[iter] <- temp_res\$Df[2]
results\$Chisq[iter] <- temp_res\$Chisq[2]
results\$pvalue[iter] <- temp_res\$Pr[2]
}
ref <- model\$lab[!(model\$lab %in% d)]
for (i in d){
iter <- iter + 1
test <- paste0(i,":",v," = 0")
temp_res <- linearHypothesis(model,test,test="Chisq")
results\$contrast[iter] <- paste0(i," vs ",ref)
results\$test[iter] <- paste(test,collapse=" and ")
results\$Df[iter] <- temp_res\$Df[2]
results\$Chisq[iter] <- temp_res\$Chisq[2]
results\$pvalue[iter] <- temp_res\$Pr[2]
}
return(results)
}``````

Not only does this construct the test of the observed categories, but also tests whether each set of coefficients is simultaneously zero, which is the appropriate contrast for the referent category.

``````> all_tests(mod)
ord contrast                                                            test Df        Chisq       pvalue
1   1   6 vs 8 6:(Intercept) = 8:(Intercept) and 6:am1 = 8:am1 and 6:hp = 8:hp  3    17.533511 0.0005488491
2   2   6 vs 4                    6:(Intercept) = 0 and 6:am1 = 0 and 6:hp = 0  3     5.941417 0.1144954481
3   3   8 vs 4                    8:(Intercept) = 0 and 8:am1 = 0 and 8:hp = 0  3 44080.662112 0.0000000000``````

User beware of multiple testing with this, as I am not sure as to the appropriate post-hoc correction here when examining so many hypotheses. This example with just three is obviously not a big deal, but with more categories you get n choose 2, or `(n*(n-1))/2` total contrasts.

# Testing the equality of multiple indirect effects

Another example I was asked about recently was testing whether you could use the same procedure to calculate indirect effects (popular in moderation and mediation analysis). Those end up being a bit more tricky, as to define the variance and covariance between those indirect effects we are not just dealing with adding and subtracting values of the original parameters, but are considering multiplications.

Thus to estimate the standard error and covariance parameters of indirect effects folks often use the delta method. In R using the `lavaan` library, here is an example (just taken from a code snippet Yves Rosseel posted himself), to estimate the variance-covariance matrix model defined indirect parameters.

``````#function taken from post in
library(lavaan)
vcov.def <- function(model){
m <- model
orig <- vcov(m)
free <- m@Fit@x
jac <- lavaan:::lavJacobianD(func = m@Model@def.function, x = free)
vcov_def <- jac %*% orig %*% t(jac)
estNames <- subset(parameterEstimates(m),op==":=")
row.names(vcov_def) <- estNames\$lhs
colnames(vcov_def) <- estNames\$lhs
#I want to print the covariance table estimates to make sure the
#labels are in the correct order
estNames\$se2 <- sqrt(diag(vcov_def))
estNames\$difSE <- estNames\$se - estNames\$se2
print(estNames[,c('lhs','se','se2','difSE')])
print('If difSE is not zero, labels are not in right order')
return(vcov_def)
}``````

Now here is an example of testing individual parameter estimates for indirect effects.

``````set.seed(10)
n <- 100
X1 <- rnorm(n)
X2 <- rnorm(n)
X3 <- rnorm(n)
M <- 0.5*X1 + 0.4*X2 + 0.3*X3 + rnorm(n)
Y <- 0.1*X1 + 0.2*X2 + 0.3*X3 + 0.7*M + rnorm(n)
Data <- data.frame(X1 = X1, X2 = X2, X3 = X3, Y = Y, M = M)
model <- ' # direct effect
Y ~ X1 + X2 + X3 + d*M
# mediator
M ~ a*X1 + b*X2 + c*X3
# indirect effects
bd := b*d
cd := c*d
'
model_SP.fit <- sem(model, data = Data)
summary(model_SP.fit)

#now apply to your own sem model
defCov <- vcov.def(model_SP.fit)``````

Unfortunately as far as I know, the `linearHypothesis` function does not work for `lavaan` objects, so if we want to test whether the indirect effect of whether `ad = bd` we need to construct it by hand. But with the `vcov.def` function we have those covariance terms we needed.

``````#testing hypothesis that "ad = bd"
#so doing "ad - bd = 0"
model_SP.param <- parameterEstimates(model_SP.fit)
model_SP.defined <- subset(model_SP.param, op==":=")
dif <- model_SP.defined\$est[1] - model_SP.defined\$est[2]
var_dif <- defCov[1,1] + defCov[2,2] - 2*defCov[1,2]
#so the test standard error of the difference is
se_dif <- sqrt(var_dif)
#and the test statistic is
tstat <- dif/se_dif
#two tailed p-value
dif;se_dif;2*(1 - pnorm(abs(tstat)))``````

To test whether all three indirect parameters are equal to each other at once, one way is to estimate a restricted model, and then use a likelihood ratio test of the restricted vs the full model. It is pretty easy in `lavaan` to create coefficient restrictions, just set what was varying to only be one parameter:

``````restrict_model <- ' # direct effect
Y ~ X1 + X2 + X3 + d*M
# mediator
M ~ a*X1 + a*X2 + a*X3
# indirect effects
'

model_SP.restrict <- sem(restrict_model, data = Data)
lavTestLRT(model_SP.fit, model_SP.restrict)``````

If folks know of an easier way to do the Wald tests via lavaan models let me know, I would be interested!

# Making interactive plots with R and Plotly

I wrote a small op-ed based on the homicide studies work I recently published about interpreting crime trends. Unfortunately that op-ed was not picked up by anyone (I missed the timing abit, maybe next year when the UCR stats come out I can just update the numbers and make the same point). I’ve posted that op-ed here, and I wanted to make a quick blog post detailing how I made the interactive graphs in that post using R and the Plotly library. All the data and code to replicate this can be downloaded from here.

Unfortunately with my free wordpress blog I cannot embed the actual interactive graphics, but I will provide links to online versions at my UT Dallas page that work and show a screenshot of each. So first, lets load all of the libraries that you will need, as well as set the working directory. (Of course change it to where you have your data saved on your local machine.)

``````#########################################################
#Making a shiny app for homicide rate chart
library(shiny)
library(ggplot2)
library(plotly)
library(htmlwidgets)
library(scales)

mydir <- "C:\\Users\\axw161530\\Box Sync\\Projects\\HomicideGraphs\\Analysis\\Analysis"
setwd(mydir)
#########################################################``````

Now I just read in the data. I have two datasets, the funnel rates just has additional columns to draw the funnel graphs already created. (See here or here, or the data in the original Homicide Studies paper linked at the top, on how to construct these.)

``````############################################################
#Get the data

summary(FunnRates)
FunnRates\$Population <- FunnRates\$Pop1 #These are just to make nicer labels
FunnRates\$HomicideRate <- FunnRates\$HomRate

summary(IntRates)
############################################################``````

## Funnel Chart for One Year

First, plotly makes it dead easy to take graphs you created via ggplot and turn them into an interactive graph. So here is a link to the interactive chart, and below is a screenshot.

To walk through the code, first you make your (almost) plane Jane ggplot object. Here I name it `p`. You will get an error for an “unknown aesthetics: text”, but this will be used by plotly to create tooltips. Then you use the `ggplotly` function to turn the original ggplot graph `p` into an interactive graph. By default the plotly object has more stuff in the tooltip than I want, which you can basically just go into the innards of the plotly object and strip out. Then the final part is just setting the margins to be alittle larger than default, as the axis labels were otherwise slightly cut-off.

``````############################################################
#Make the funnel chart
year_sel <- 2015
p <- ggplot(data = FunnRates[FunnRates\$Year == year_sel,]) + geom_point(aes(x=Population, y=HomicideRate, text=NiceLab), pch=21) +
geom_line(aes(x=Population,y=LowLoc99)) + geom_line(aes(x=Population,y=HighLoc99)) +
labs(title = paste0("Homicide Rates per 100,000 in ",year_sel)) +
scale_x_log10("Population", limits=c(10000,10000000), breaks=c(10^4,10^5,10^6,10^7), labels=comma) +
scale_y_continuous("Homicide Rate", breaks=seq(0,110,10)) +
theme_bw() #+ theme(text = element_text(size=20), axis.title.y=element_text(margin=margin(0,10,0,0)))

pl <- ggplotly(p, tooltip = c("HomicideRate","text"))
#pl <- plotly_build(p, width=1000, height=900)
#See https://stackoverflow.com/questions/45801389/disable-hover-information-for-a-specific-layer-geom-of-plotly
pl\$x\$data[[2]]\$hoverinfo <- "none"
pl\$x\$data[[3]]\$hoverinfo <- "none"
pl <- pl %>% layout(margin = list(l = 75, b = 65))
############################################################``````

After this point you can just type `pl` into the console and it will open up an interactive window. Or you can use the `saveWidget` function from the htmlwidgets package, something like `saveWidget(as_widget(pl), "FunnelChart_2015.html", selfcontained=TRUE)` to save the graph to an html file.

Now there are a couple of things. You can edit various parts of the graph, such as its size and label text size, but depending on your application these might not be a good idea. If you need to take into account smaller screens, I think it is best to use some of the defaults, as they adjust per the screen that is in use. For the size of the graph if you are embedding it in a webpage using iframe’s you can set the size at that point. If you look at my linked op-ed you can see I make the funnel chart taller than wider — that is through the iframe specs.

## Funnel Chart over Time

Ok, now onto the fun stuff. So we have a funnel chart for one year, but I have homicide years from 1965 through 2015. Can we examine those over time. Plotly has an easy to use additional argument to ggplot graphs, named `Frame`, that allows you to add a slider to the interactive chart for animation. The additional argument `ids` links one object over time, ala Hans Rosling bubble chart over time. Here is a link to the interactive version, and below is a screen shot:

``````############################################################
#Making the funnel chart where you can select the year
py <- ggplot(data = FunnRates) + geom_point(aes(x=Population, y=HomicideRate, text=NiceLab, frame=Year,ids=ORI), pch=21) +
geom_line(aes(x=Population,y=LowLoc99,frame=Year)) + geom_line(aes(x=Population,y=HighLoc99,frame=Year)) +
labs(title = paste0("Homicide Rates per 100,000")) +
scale_x_log10("Population", limits=c(10000,10000000), breaks=c(10^4,10^5,10^6,10^7), labels=comma) +
scale_y_continuous("Homicide Rate", breaks=seq(0,110,10), limits=c(0,110)) +
theme_bw() #+ theme(text = element_text(size=20), axis.title.y=element_text(margin=margin(0,10,0,0)))

ply <- ggplotly(py, tooltip = c("text")) %>% animation_opts(0, redraw=FALSE)
ply\$x\$data[[2]]\$hoverinfo <- "none"
ply\$x\$data[[3]]\$hoverinfo <- "none"
saveWidget(as_widget(ply), "FunnelChart_YearSelection.html", selfcontained=FALSE)
############################################################``````

The way I created the data it does not make sense to do a smooth animation for the funnel line, so this just flashes to each new year (via the `animation_opts` spec). (I could make the data so it would look nicer in an animation, but will wait for someone to pick up the op-ed before I bother too much more with this.) But it accomplishes via the slider the ability for you to pick which year you want.

## Fan Chart Just One City

Next we are onto the fan charts for each individual city with the prediction intervals. Again you can just create this simple chart in ggplot, and then use plotly to make a version with tooltips. Here is a link to an interactive version, and below is a screenshot.

``````###################################################
#Making the fan graph for New Orleans
titleLab <- unique(IntRates[,c("ORI","NiceLab","AgencyName","State")])
p2 <- ggplot(data=IntRates[IntRates\$ORI == "LANPD00",], aes(x=Year, y=HomRate)) +
geom_ribbon(aes(ymin=LowB, ymax=HighB), alpha=0.2) +
geom_ribbon(aes(ymin=LagLow25, ymax=LagHigh25), alpha=0.5) +
geom_point(shape=21, color="white", fill="red", size=2) +
labs(x = "Year", y="Homicide Rate per 100,000") +
#scale_x_continuous(breaks=seq(1960,2015,by=5)) +
ggtitle(paste0("Prediction Intervals for ",titleLab[titleLab\$ORI == "LANPD00",c("NiceLab")])) +
theme_bw() #+ theme(text = element_text(size=20), axis.title.y=element_text(margin=margin(0,10,0,0)))
#p2
pl2 <- ggplotly(p2, tooltip = c("Year","HomRate"), dynamicTicks=TRUE)
pl2\$x\$data[[1]]\$hoverinfo <- "none"
pl2\$x\$data[[2]]\$hoverinfo <- "none"
pl2 <- pl2 %>% layout(margin = list(l = 100, b = 65))
#pl2
saveWidget(as_widget(pl2), "FanChart_NewOrleans.html", selfcontained=FALSE)
###################################################``````

Note when you save the widget to `selfcontained=FALSE`, it hosts several parts of the data into separate folders. I always presumed this was more efficient than making one huge html file, but I don’t know for sure.

## Fan Chart with Dropdown Selection

Unfortunately the `frame` type animation does not make as much sense here. It would be hard for someone to find a particular city of interest in that slider (as a note though the slider can have nominal data, if I only had a few cities it would work out ok, with a few hundred it will not though). So feature request if anyone from plotly is listening — please have a dropdown type option for ggplot graphs! In the meantime though there is an alternative using a tradition `plot_ly` type chart. Here is that interactive fan chart with a police agency dropdown, and below is a screenshot.

``````###################################################
#Making the fan graph where you can select the city of interest
#Need to have a dropdown for the city

titleLab <- unique(IntRates[,c("ORI","NiceLab","State")])
nORI <- length(titleLab[,1])
choiceP <- vector("list",nORI)
for (i in 1:nORI){
choiceP[[i]] <- list(method="restyle", args=list("transforms[0].value", unique(IntRates\$NiceLab)[i]), label=titleLab[i,c("NiceLab")])
}

trans <- list(list(type='filter',target=~NiceLab, operation="=", value=unique(IntRates\$NiceLab)[1]))
textLab <- ~paste("Homicide Rate:",HomRate,'\$
Year:',Year,'\$
Homicides:',Homicide,'\$
Population:',Pop1,'\$
Agency Name:',NiceLab)

#Lets try with the default plotly
#See https://community.plot.ly/t/need-help-on-using-dropdown-to-filter/6596
ply4 <- IntRates %>%
plot_ly(x= ~Year,y= ~HighB, type='scatter', mode='lines', line=list(color='transparent'), showlegend=FALSE, name="90%", hoverinfo="none", transforms=trans) %>%
add_trace(y=~LowB,  type='scatter', mode='lines', line=list(color='transparent'), showlegend=FALSE, name='10%', hoverinfo="none", transforms=trans,
fill = 'tonexty', fillcolor='rgba(105,105,105,0.3)') %>%
add_trace(x=~Year,y=~HomRate, text=~NiceLab, type='scatter', mode='markers', marker = list(size=10, color = 'rgba(255, 182, 193, .9)', line = list(color = 'rgba(152, 0, 0, .8)', width = 1)),
hoverinfo='text', text=textLab, transforms=trans) %>%
layout(title = "Homicide Rates and 80% Prediction Intervals by Police Department",
xaxis = list(title="Year"),
yaxis = list(title="Homicide Rate per 100,000"),

saveWidget(as_widget(ply4), "FanChart_Dropdown.html", selfcontained=FALSE)
###################################################``````

So in short plotly makes it super-easy to make interactive graphs with tooltips. Long term goal I would like to make a visual supplement to the traditional UCR report (I find the complaint of what tables to include to miss the point — there are much better ways to show the information that worrying about the specific tables). So if you would like to work on that with me always feel free to get in touch!

# Geocoding with census data and the Census API

For my online GIS class I have a tutorial on creating an address locator using street centerline data in ArcGIS. Eventually I would like to put all of my class online, but for now I am just sharing that one, as I’ve forwarded it alot recently.

That tutorial used local street centerline data in Dallas that you can download from Dallas’s open data site. It also gives directions on how to use an online ESRI geocoding service — which Dallas has. But what if those are not an option? A student recently wanted to geocode data from San Antonio, and the only street data file they publicly provide lacks the beginning and ending street number.

That data is insufficient to create an address locator. It is also the case that the road data you can download from the census’s web interface lacks this data. But you can download street centerline data with beginning and end addresses from the census from the FTP site. For example here is the url that contains the streets with the address features. To use that you just have to figure out what state and county you are interested in downloaded. The census even has ESRI address locators already made for you using 2012 data at the state level. Again you just need to figure out your states number and download it.

Once you download the data with the begin and ending street numbers you can follow along with that tutorial the same as the public data.

Previously I’ve written about using the Google geocoding API. If you just have crime data from one jurisdiction, it is simple to make a geocoder for just that locality. But if you have data for many cities (say if you were geocoding home addresses) this can be more difficult. An alternative online API to google that does not have daily limits is the Census Geocoding API.

Here is a simple example in R of calling the census API and geocoding a list of addresses.

``````library(httr)
library(jsonlite)

soup <- GET(url=base,query=list(street=street,city=city,state=state,zip=zip,format='json',benchmark=benchmark))
dat <- fromJSON(content(soup,as='text'), simplifyVector=TRUE)
if (length(D_dat) > 1){
return(c(D_dat['matchedAddress'],D_dat['coordinates'][[1]])) #error will just return null, x[1] is lon, x[2] is lat
}
else {return(c('',NA,NA))}
}

#now create function to loop over data frame and return set of addresses
geo_CensusTIGER <- function(street,city,state,zip,sleep=1,benchmark=4){
#make empy matrix
l <- length(street)
MyDat <- data.frame(matrix(nrow=l,ncol=3))
for (i in 1:l){
if (length(x) > 0){
MyDat[i,1] <- x[1]
MyDat[i,2] <- x[2]
MyDat[i,3] <- x[3]
}
Sys.sleep(sleep)
}
MyDat\$street <- street
MyDat\$city <- city
MyDat\$zip <- zip
MyDat\$state <- state
return(MyDat)
}

## Arbitrary dataframe for an exercise
IdNum = c(1,2,3,4,5),
Address = c("450 W Harwood Rd", "2878 Fake St", "2775 N Collin St", "2775 N Collins St", "Lakewood Blvd and W Shore Dr"),
City = c("Hurst", "Richardson", "Arlington", "Arlington", "Dallas"),
State = c("TX", "TX", "TX", "TX", "TX")
)

If you check out the results, you will see that this API does not appear to do fuzzy matching. 2775 N Collin St failed, whereas 2775 N Collins St was able to return a match. You can also see though it will return an intersection, but in my tests "/" did not work (so in R you can simply use `gsub` to replace different intersection types with `and`). I haven’t experimented with it too much, so let me know if you have any other insight into this API.

I will follow up in another post a python function to use the Census geocoding API, as well as using the Nominatim online geocoding API, which you can use for addresses outside of the United States.

# Testing the equality of coefficients – Same Independent, Different Dependent variables

As promised earlier, here is one example of testing coefficient equalities in SPSS, Stata, and R.

Here we have different dependent variables, but the same independent variables. This is taken from Dallas survey data (original data link, survey instrument link), and they asked about fear of crime, and split up the questions between fear of property victimization and violent victimization. Here I want to see if the effect of income is the same between the two. People in more poverty tend to be at higher risk of victimization, but you may also expect people with fewer items to steal to be less worried. Here I also control for the race and the age of the respondent.

The dataset has missing data, so I illustrate how to select out for complete case analysis, then I estimate the model. The fear of crime variables are coded as Likert items with a scale of 1-5, (higher values are more safe) but I predict them using linear regression (see the Stata code at the end though for combining ordinal logistic equations using `suest`). Race is of course nominal, and income and age are binned as well, but I treat the income bins as a linear effect. I pasted the codebook for all of the items at the end of the post.

These models with multiple dependent variables have different names, economists call them seemingly unrelated regression, psychologists will often just call them multivariate models, those familiar with structural equation modeling can get the same results by allowing residual covariances between the two outcomes — they will all result in the same coefficient estimates in the end.

# SPSS

In SPSS we can use the `GLM` procedure to estimate the model. Simultaneously we can specify particular contrasts to test whether the income coefficient is different for the two outcomes.

``````*Grab the online data.
SPSSINC GETURI DATA URI="https://dl.dropbox.com/s/r98nnidl5rnq5ni/MissingData_DallasSurv16.sav?dl=0" FILETYPE=SAV DATASET=MissData.

*Conducting complete case analysis.
COUNT MisComplete = Safety_Violent Safety_Prop Gender Race Income Age (MISSING).
COMPUTE CompleteCase = (MisComplete = 0).
FILTER BY CompleteCase.

*This treats the different income categories as a continuous variable.
*Can use GLM to estimate seemingly unrelated regression in SPSS and test.
*equality of the two coefficients.
GLM Safety_Violent Safety_Prop BY Race Age WITH Income
/DESIGN=Income Race Age
/PRINT PARAMETER
/LMATRIX Income 1
/MMATRIX ALL 1 -1.

FILTER OFF.  ``````

In the output you can see the coefficient estimates for the two equations. The income effect for violent crime is 0.168 (0.023) and for property crime is 0.114 (0.022).

And then you get a separate table for the contrast estimates.

You can see that the contrast estimate, 0.054, equals 0.168 – 0.114. The standard error in this output (0.016) takes into account the covariance between the two estimates. Here you would reject the null that the effects are equal across the two equations, and the effect of income is larger for violent crime. Because higher values on these Likert scales mean a person feels more safe, this is evidence that those with higher incomes are more likely to be fearful of property victimization, controlling for age and race.

Unfortunately the different matrix contrasts are not available in all the different types of regression models in SPSS. You may ask whether you can fit two separate regressions and do this same test. The answer is you can, but that makes assumptions about how the two models are independent — it is typically more efficient to estimate them at once, and here it allows you to have the software handle the Wald test instead of constructing it yourself.

# R

As I stated previously, seemingly unrelated regression is another name for these multivariate models. So we can use the R libraries `systemfit` to estimate our seemingly unrelated regression model, and then use the library `multcomp` to test the coefficient contrast. This does not result in the exact same coefficients as SPSS, but devilishly close. You can download the csv file of the data here.

``````library(systemfit) #for seemingly unrelated regression
library(multcomp)  #for hypothesis tests of models coefficients

names(SurvData)[1] <- "Safety_Violent" #name gets messed up because of BOM

#Need to recode the missing values in R, use NA
NineMis <- c("Safety_Violent","Safety_Prop","Race","Income","Age")
#summary(SurvData[,NineMis])
for (i in NineMis){
SurvData[SurvData[,i]==9,i] <- NA
}

#Making a complete case dataset
SurvComplete <- SurvData[complete.cases(SurvData),NineMis]
#Now changing race and age to factor variables, keep income as linear
SurvComplete\$Race <- factor(SurvComplete\$Race, levels=c(1,2,3,4), labels=c("Black","White","Hispanic","Other"))
SurvComplete\$Age <- factor(SurvComplete\$Age, levels=1:5, labels=c("18-24","25-34","35-44","45-54","55+"))
summary(SurvComplete)

#now fitting seemingly unrelated regression
viol <- Safety_Violent ~ Income + Race + Age
prop <- Safety_Prop ~ Income + Race + Age
fitsur <- systemfit(list(violreg = viol, propreg= prop), data=SurvComplete, method="SUR")
summary(fitsur)

#testing whether income effect is equivalent for both models
viol_more_prop <- glht(fitsur,linfct = c("violreg_Income - propreg_Income = 0"))
summary(viol_more_prop) ``````

Here is a screenshot of the results then:

This is also the same as estimating a structural equation model in which the residuals for the two regressions are allowed to covary. We can do that in R with the `lavaan` library.

``````library(lavaan)

#for this need to convert factors into dummy variables for lavaan
DumVars <- data.frame(model.matrix(~Race+Age-1,data=SurvComplete))
names(DumVars) <- c("Black","White","Hispanic","Other","Age2","Age3","Age4","Age5")

SurvComplete <- cbind(SurvComplete,DumVars)

model <- '
#regressions
Safety_Prop    ~ Income + Black + Hispanic + Other + Age2 + Age3 + Age4 + Age5
Safety_Violent ~ Income + Black + Hispanic + Other + Age2 + Age3 + Age4 + Age5
#residual covariances
Safety_Violent ~~ Safety_Prop
Safety_Violent ~~ Safety_Violent
Safety_Prop ~~ Safety_Prop
'

fit <- sem(model, data=SurvComplete)
summary(fit)``````

I’m not sure offhand though if there is an easy way to test the coefficient differences with a lavaan object, but we can do it manually by grabbing the variance and the covariances. You can then see that the differences and the standard errors are equal to the prior output provided by the `glht` function in `multcomp`.

``````#Grab the coefficients I want, and test the difference
PCov <- inspect(fit,what="vcov")
PEst <- inspect(fit,what="list")
Diff <- PEst[9,'est'] - PEst[1,'est']
SE <- sqrt( PEst[1,'se']^2 + PEst[9,'se']^2 - 2*PCov[9,1] )
Diff;SE``````

# Stata

In Stata we can replicate the same prior analyses. Here is some code to simply replicate the prior results, using Stata’s postestimation commands (additional examples using postestimation commands here). Again you can download the csv file used here. The results here are exactly the same as the R results.

``````*Load in the csv file
import delimited MissingData_DallasSurvey.csv, clear

*BOM problem again
rename ïsafety_violent safety_violent

*we need to specify the missing data fields.
*for Stata, set missing data to ".", not the named missing value types.
foreach i of varlist safety_violent-ownhome {
tab `i'
}

*dont specify district
mvdecode safety_violent-race income-age ownhome, mv(9=.)
mvdecode yearsdallas, mv(999=.)

*making a variable to identify the number of missing observations
egen miscomplete = rmiss(safety_violent safety_prop race income age)
tab miscomplete
*even though any individual question is small, in total it is around 20% of the cases

*lets conduct a complete case analysis
preserve
keep if miscomplete==0

*Now can estimate multivariate regression, same as GLM in SPSS
mvreg safety_violent safety_prop = income i.race i.age

*test income coefficient is equal across the two equations
lincom _b[safety_violent:income] - _b[safety_prop:income]

*same results as seemingly unrelated regression
sureg (safety_violent income i.race i.age)(safety_prop income i.race i.age)

*To use lincom it is the exact same code as with mvreg
lincom _b[safety_violent:income] - _b[safety_prop:income]

*with sem model
tabulate race, generate(r)
tabulate age, generate(a)
sem (safety_violent <- income r2 r3 r4 a2 a3 a4 a5)(safety_prop <- income r2 r3 r4 a2 a3 a4 a5), cov(e.safety_violent*e.safety_prop)

*can use the same as mvreg and sureg
lincom _b[safety_violent:income] - _b[safety_prop:income]``````

You will notice here it is the exact some post-estimation `lincom` command to test the coefficient equality across all three models. (Stata makes this the easiest of the three programs IMO.)

Stata also allows us to estimate seemingly unrelated regressions combining different generalized outcomes. Here I treat the outcome as ordinal, and then combine the models using seemingly unrelated regression.

``````*Combining generalized linear models with suest
ologit safety_violent income i.race i.age
est store viol

ologit safety_prop income i.race i.age
est store prop

suest viol prop

*lincom again!
lincom _b[viol_safety_violent:income] - _b[prop_safety_prop:income]``````

An application in spatial criminology is when you are estimating the effect of something on different crime types. If you are predicting the number of crimes in a spatial area, you might separate Poisson regression models for assaults and robberies — this is one way to estimate the models jointly. Cory Haberman and Jerry Ratcliffe have an application of this as well estimate the effect of different crime types at different times of day – e.g. the effect of bars in the afternoon versus the effect of bars at nighttime.

# Codebook

Here is the codebook for each of the variables in the database.

``````Safety_Violent
1   Very Unsafe
2   Unsafe
3   Neither Safe or Unsafe
4   Safe
5   Very Safe
9   Do not know or Missing
Safety_Prop
1   Very Unsafe
2   Unsafe
3   Neither Safe or Unsafe
4   Safe
5   Very Safe
9   Do not know or Missing
Gender
1   Male
2   Female
9   Missing
Race
1   Black
2   White
3   Hispanic
4   Other
9   Missing
Income
1   Less than 25k
2   25k to 50k
3   50k to 75k
4   75k to 100k
5   over 100k
9   Missing
Edu
1   Less than High School
2   High School
3   Some above High School
9   Missing
Age
1   18-24
2   25-34
3   35-44
4   45-54
5   55+
9   Missing
OwnHome
1   Own
2   Rent
9   Missing
YearsDallas
999 Missing``````

# Identifying near repeat crime strings in R or Python

People in criminology should be familiar with repeats or near-repeats for crimes such as robbery, burglaries, or shootings. An additional neat application of this idea though is to pull out strings of incidents that are within particular distance and time thresholds. See this example analysis by Haberman and Ratcliffe, The Predictive Policing Challenges of Near Repeat Armed Street Robberies. This is particularly useful to an analyst interested in crime linkage — to see if those particular strings of incidents are likely to be committed by the same offender.

Here I will show how to pluck out those near-repeat strings in R or Python. The general idea is to transform the incidents into a network, where two incidents are connected only if they meet the distance and time requirements. Then you can identify the connected components of the graph, and those are your strings of near-repeat events.

To follow along, here is the data and the code used in the analysis. I will be showing this on an example set of thefts from motor vehicles (aka burglaries from motor vehicles) in Dallas in 2015. In the end I take two different approaches to this problem — in R the solution will only work for smaller datasets (say n~5000 or less), but the python code should scale to much larger datasets.

# Near-repeat strings in R

The approach I take in R does the steps as follows:

1. compute the distance matrix for the spatial coordinates
2. convert this matrix to a set of 0’s and 1’s, 1’s correspond to if the distance is below the user specified distance threshold (call it S)
3. compute the distance matrix for the times
4. convert this matrix to a set of 0’1 and 1’s, 1’s correspond to if the distance is below the user specified time threshold (call it T)
5. use element-wise multiplication on the S and T matrices, call the result A, then set the diagonal of A to zero
6. A is now an adjacency matrix, which can be converted into a network
7. extract the connected components of that network

So here is an example of reading in the thefts from motor vehicle data, and defining my function, `NearStrings`, to grab the strings of incidents. Note you need to have the `igraph` R library installed for this code to work.

``````library(igraph)

MyDir <- "C:\\Users\\axw161530\\Dropbox\\Documents\\BLOG\\SourceNearRepeats"
setwd(MyDir)

summary(BMV)

#make a function
NearStrings <- function(data,id,x,y,time,DistThresh,TimeThresh){
library(igraph) #need igraph to identify connected components
MyData <- data
SpatDist <- as.matrix(dist(MyData[,c(x,y)])) < DistThresh  #1's for if under distance
TimeDist <-  as.matrix(dist(MyData[,time])) < TimeThresh #1's for if under time
AdjMat <- SpatDist * TimeDist #checking for both under distance and under time
diag(AdjMat) <- 0 #set the diagonal to zero
row.names(AdjMat) <- MyData[,id] #these are used as labels in igraph
colnames(AdjMat) <- MyData[,id] #ditto with row.names
CompInfo <- components(G) #assigning the connected components
return(data.frame(CompId=CompInfo\$membership,CompNum=CompInfo\$csize[CompInfo\$membership]))
}``````

So here is a quick example run on the first ten records. Note I have a field that is named `DateInt` in the csv, which is just the integer number of days since the first of the year. In R though if the dates are actual date objects you can submit them to the `dist` function though as well.

``````#Quick example with the first ten records
BMVSub <- BMV[1:10,]
ExpStrings <- NearStrings(data=BMVSub,id='incidentnu',x='xcoordinat',y='ycoordinat',time='DateInt',DistThresh=30000,TimeThresh=3)
ExpStrings``````

So here we can see this prints out:

``````> ExpStrings
CompId CompNum
000036-2015      1       3
000113-2015      2       4
000192-2015      2       4
000251-2015      1       3
000360-2015      2       4
000367-2015      3       1
000373-2015      4       2
000378-2015      4       2
000463-2015      2       4
000488-2015      1       3``````

The `CompId` field is a unique Id for every string of events. The `CompNum` field states how many events are within the string. So we have one string of events that contains 4 records in this subset.

Now this R function comes with a big caveat, it will not work on large datasets. I’d say your pushing it with 10,000 incidents. The issue is holding the distance matrices in memory. But if you can hold the matrices in memory this will still run quite fast. For 5,000 incidents it takes around ~15 seconds on my machine.

``````#Second example alittle larger, with the first 5000 records
BMVSub2 <- BMV[1:5000,]
BigStrings <- NearStrings(data=BMVSub2,id='incidentnu',x='xcoordinat',y='ycoordinat',time='DateInt',DistThresh=1000,TimeThresh=3)``````

The elements in the returned matrix will line up with the original dataset, so you can simply add those fields in, and do subsequent analysis (such as exporting back into a mapping program and digging into the strings).

``````#Add them into the original dataset
BMVSub2\$CompId <- BigStrings\$CompId
BMVSub2\$CompNum <- BigStrings\$CompNum   ``````

You can check out the number of chains of different sizes by using aggregate and table.

``````#Number of chains
table(aggregate(CompNum ~ CompId, data=BigStrings, FUN=max)\$CompNum)``````

This prints out:

``````   1    2    3    4    5    6    7    9
3814  405   77   27    3    1    1    1``````

So out of our first 1,000 incidents, using the distance threshold of 1,000 feet and the time threshold of 3 days, we have 3,814 isolates. Thefts from vehicles with no other incidents nearby. We have 405 chains of 2 incidents, 77 chains of 3 incidents, etc. You can pull out the 9 incident like this since there is only one chain that long:

``````#Look up the 9 incident
BMVSub2[BMVSub2\$CompNum == 9,]  ``````

Which prints out here:

``````> BMVSub2[BMVSub2\$CompNum == 9,]
incidentnu xcoordinat ycoordinat StartDate DateInt CompId CompNum
2094 043983-2015    2460500    7001459 2/25/2015      56   1842       9
2131 044632-2015    2460648    7000542 2/26/2015      57   1842       9
2156 045220-2015    2461162    7000079 2/27/2015      58   1842       9
2158 045382-2015    2460154    7000995 2/27/2015      58   1842       9
2210 046560-2015    2460985    7000089  3/1/2015      60   1842       9
2211 046566-2015    2460452    7001457  3/1/2015      60   1842       9
2260 047544-2015    2460154    7000995  3/2/2015      61   1842       9
2296 047904-2015    2460452    7001457  3/3/2015      62   1842       9
2337 048691-2015    2460794    7000298  3/4/2015      63   1842       9``````

Or you can look up a particular chain by its uniqueid. Here is an example of a 4-chain set.

``````> #Looking up a particular incident chains
> BMVSub2[BMVSub2\$CompId == 4321,]
incidentnu xcoordinat ycoordinat StartDate DateInt CompId CompNum
4987 108182-2015    2510037    6969603 5/14/2015     134   4321       4
4988 108183-2015    2510037    6969603 5/14/2015     134   4321       4
4989 108184-2015    2510037    6969603 5/14/2015     134   4321       4
4993 108249-2015    2510037    6969603 5/14/2015     134   4321       4``````

Again, only use this function on smaller crime datasets.

# Near-repeat strings in Python

Here I show how to go about a similar process in Python, but the algorithm does not calculate the whole distance matrix at once, so can handle much larger datasets. An additional note is that I exploit the fact that this list is sorted by dates. This makes it so I do not have to calculate all pair-wise distances – I will basically only compare distances within a moving window under the time threshold – this makes it easily scale to much larger datasets.

So first I use the `csv` python library to read in the data and assign it to a list with a set of nested tuples. Also you will need the `networkx` library to extract the connected components later on.

``````import networkx as nx
import csv
import math

dir = r'C:\Users\axw161530\Dropbox\Documents\BLOG\SourceNearRepeats'

BMV_tup = []
with open(dir + r'\TheftFromMV.csv') as f:
for row in z:
BMV_tup.append(tuple(row))``````

The `BMV_tup` list has the column headers, so I extract that row and then figure out where all the elements I need, such as the XY coordinates, the unique Id’s, and the time column are located in the nested tuples.

``````colnames = BMV_tup.pop(0)
print colnames
print BMV_tup[0:10]

xInd = colnames.index('xcoordinat')
yInd = colnames.index('ycoordinat')
dInd = colnames.index('DateInt')
IdInd = colnames.index('incidentnu')``````

Now the magic — here is my function to extract those near-repeat strings. Again, the list needs to be sorted by dates for this to work.

``````def NearStrings(CrimeData,idCol,xCol,yCol,tCol,DistThresh,TimeThresh):
G = nx.Graph()
n = len(CrimeData)
for i in range(n):
for j in range(i+1,n):
if (float(CrimeData[j][tCol]) - float(CrimeData[i][tCol])) > TimeThresh:
break
else:
xD = math.pow(float(CrimeData[j][xCol]) - float(CrimeData[i][xCol]),2)
yD = math.pow(float(CrimeData[j][yCol]) - float(CrimeData[i][yCol]),2)
d = math.sqrt(xD + yD)
if d < DistThresh:
comp = nx.connected_components(G)
finList = []
compId = 0
for i in comp:
compId += 1
for j in i:
finList.append((j,compId))
return finList``````

We can then do the same test on the first ten records that we did in R.

``print NearStrings(CrimeData=BMV_tup[0:10],idCol=IdInd,xCol=xInd,yCol=yInd,tCol=dInd,DistThresh=30000,TimeThresh=3)``

And this subsequently prints out:

``````[('000378-2015', 1), ('000373-2015', 1), ('000113-2015', 2), ('000463-2015', 2), ('000192-2015', 2), ('000360-2015', 2),
('000251-2015', 3), ('000488-2015', 3), ('000036-2015', 3)]``````

The component Id’s wont be in the same order as in R, but you can see we have the same results. E.g. the string with three incidents contains the Id’s 000251, 000488, and 000036. Note that this approach does not return isolates — incidents which have no nearby space-time examples.

Running this on the full dataset of over 14,000 incidents takes around 20 seconds on my machine.

``BigResults = NearStrings(CrimeData=BMV_tup,idCol=IdInd,xCol=xInd,yCol=yInd,tCol=dInd,DistThresh=1000,TimeThresh=3)``

And that should scale pretty well for really big cities and really big datasets. I will let someone who knows R better than me figure out workarounds to scale to bigger datasets in that language.