# Exploring Open Data - Predicting the Amoung of Violations

### Introduction

In my last post, I went over some of the highlights of the open data set of all Philadelphia Parking Violations. In this post, I’ll go through the steps to build a model to predict the amount of violations the city issues on a daily basis. I’ll walk you through cleaning and building the data set, selecting and creating the important features, and building predictive models using Random Forests and Linear Regression.

### Step 1: Load Packages and Data

Just an initial step to get the right libraries and data loaded in R.

``````library(plyr)
library(randomForest)

## DATA FILE FROM OPENDATAPHILLY

## READ IN THE WEATHER DATA (FROM NCDC)

## LIST OF ALL FEDERAL HOLIDAYS DURING THE
## RANGE OF THE DATA SET
holidays <- as.Date(c("2012-01-02", "2012-01-16",
"2012-02-20", "2012-05-28",
"2012-07-04", "2012-09-03",
"2012-10-08", "2012-11-12",
"2012-11-22", "2012-12-25",
"2013-01-01", "2013-01-21",
"2013-02-18", "2013-05-27",
"2013-07-04", "2013-09-02",
"2013-10-14", "2013-11-11",
"2013-11-28", "2013-12-25",
"2014-01-01", "2014-01-20",
"2014-02-17", "2014-05-26",
"2014-07-04", "2014-09-01",
"2014-10-13", "2014-11-11",
"2014-11-27", "2014-12-25",
"2015-01-01", "2015-01-09",
"2015-02-16", "2015-05-25",
"2015-07-03", "2015-09-07"))``````

### Step 2: Formatting the Data

First things first, we have to total the amount of tickets per day from the raw data. For this, I use the plyr command ddply. Before I can use the ddply command, I need to format the Issue.Date.and.Time column to be a Date variable in the R context.

``````days <- as.data.frame(as.Date(
ptix\$Issue.Date.and.Time,
format = "%m/%d/%Y"))
names(days) <- "DATE"
count_by_day <- ddply(days, .(DATE), summarize,
count = length(DATE)) ``````

Next, I do the same exact date formatting with the weather data.

``````weather_data\$DATE <- as.Date(as.POSIXct(strptime(as.character(weather_data\$DATE),
format = "%Y%m%d")),
format = "%m/%d/%Y")``````

Now that both the ticket and weather data have the same date format (and name), we can use the join function from the plyr package.

``count_by_day <- join(count_by_day, weather_data, by = "DATE")``

With the data joined by date, it is time to clean. There are a number of columns with unneeded data (weather station name, for example) and others with little or no data in them, which I just flatly remove. The data has also been coded with negative values representing that data had not been collected for any number of reasons (I’m not surprised that snow was not measured in the summer); for that data, I’ve made any values coded -9999 into 0. There are some days where the maximum or minimum temperature was not gathered (I’m not sure why). As this is the main variable I plan to use to predict daily violations, I drop the entire row if the temperature data is missing.

``````## I DON'T CARE ABOUT THE STATION OR ITS NAME -
## GETTING RID OF IT
count_by_day\$STATION <- NULL
count_by_day\$STATION_NAME <- NULL

## A BUNCH OF VARIABLE ARE CODED WITH NEGATIVE VALUES
## IF THEY WEREN'T COLLECTED - CHANGING THEM TO 0s
count_by_day\$MDPR[count_by_day\$MDPR < 0] <- 0
count_by_day\$DAPR[count_by_day\$DAPR < 0] <- 0
count_by_day\$PRCP[count_by_day\$PRCP < 0] <- 0
count_by_day\$SNWD[count_by_day\$SNWD < 0] <- 0
count_by_day\$SNOW[count_by_day\$SNOW < 0] <- 0
count_by_day\$WT01[count_by_day\$WT01 < 0] <- 0
count_by_day\$WT03[count_by_day\$WT03 < 0] <- 0
count_by_day\$WT04[count_by_day\$WT04 < 0] <- 0

## REMOVING ANY ROWS WITH MISSING TEMP DATA
count_by_day <- count_by_day[
count_by_day\$TMAX > 0, ]
count_by_day <- count_by_day[
count_by_day\$TMIN > 0, ]

## GETTING RID OF SOME NA VALUES THAT POPPED UP
count_by_day <- count_by_day[!is.na(
count_by_day\$TMAX), ]

## REMOVING COLUMNS THAT HAVE LITTLE OR NO DATA
## IN THEM (ALL 0s)
count_by_day\$TOBS <- NULL
count_by_day\$WT01 <- NULL
count_by_day\$WT04 <- NULL
count_by_day\$WT03 <- NULL

## CHANGING THE DATA, UNNECESSARILY, FROM 10ths OF
## DEGREES CELCIUS TO JUST DEGREES CELCIUS
count_by_day\$TMAX <- count_by_day\$TMAX / 10
count_by_day\$TMIN <- count_by_day\$TMIN / 10``````

### Step 3: Visualizing the Data

At this point, we have joined our data sets and gotten rid of the unhelpful “stuff.” What does the data look like?

There are clearly two populations here. With the benefit of hindsight, the small population on the left of the histogram is mainly Sundays. The larger population with the majority of the data is all other days of the week.

Let’s make some new features to explore this idea.

### Step 4: New Feature Creation

As we see in the histogram above, there are obviously a few populations in the data - I know that day of the week, holidays, and month of the year likely have some strong influence on how many violations are issued. If you think about it, most parking signs include the clause: “Except Sundays and Holidays.” Plus, spending more than a few summers in Philadelphia at this point, I know that from Memorial Day until Labor Day the city relocates to the South Jersey Shore (emphasis on the South part of the Jersey Shore). That said - I add in those features as predictors.

``````## FEATURE CREATION - ADDING IN THE DAY OF WEEK
count_by_day\$DOW <- as.factor(weekdays(count_by_day\$DATE))

## FEATURE CREATION - ADDING IN IF THE DAY WAS A HOLIDAY
count_by_day\$HOL <- 0
count_by_day\$HOL[as.character(count_by_day\$DATE) %in%
as.character(holidays)] <- 1
count_by_day\$HOL <- as.factor(count_by_day\$HOL)

## FEATURE CREATION - ADDING IN THE MONTH
count_by_day\$MON <- as.factor(months(count_by_day\$DATE))``````

Now - let’s see if the Sunday thing is real. Here is a scatterplot of the data. The circles represent Sundays; triangles are all other days of the week.

You can clearly see that Sunday’s tend to do their own thing in a very consistent manner that is similar to the rest of the week. In other words, the slope for Sundays is very close to that of the slope for all other days of the week. There are some points that don’t follow those trends, which are likely due to snow, holidays, and/or other man-made or weather events.

Let’s split the data into a training and test set (that way we can see how well we do with the model). I’m arbitrarily making the test set the last year of data; everything before that is the training set.

``````train <- count_by_day[count_by_day\$DATE < "2014-08-01", ]
test <- count_by_day[count_by_day\$DATE >= "2014-08-01", ]``````

### Step 5: Feature Identification

We now have a data set that is ready for some model building! The problem to solve next is figuring out which features best explain the count of violations issued each day. My preference is to use Random Forests to tell me which features are the most important. We’ll also take a look to see which, if any, variables are highly correlated. High correlation amongst input variables will lead to high variability due to multicollinearity issues.

``````featForest <- randomForest(count ~ MDPR + DAPR + PRCP +
SNWD + SNOW + TMAX +
TMIN + DOW + HOL + MON,
data = train, importance = TRUE,
ntree = 10000)

## PLOT THE VARIABLE TO SEE THE IMPORTANCE
varImpPlot(featForest)``````

In the Variable Importance Plot below, you can see very clearly that the day of the week (DOW) is by far the most important variable in describing the amount of violations written per day. This is followed by whether or not the day was a holiday (HOL), the minimum temperature (TMIN), and the month (MON). The maximum temperature is in there, too, but I think that it is likely highly correlated with the minimum temperature (we’ll see that next). The rest of the variables have very little impact.

``cor(count_by_day[,c(3:9)])``

I’ll skip the entire output of the correlation table, but TMIN and TMAX have a correlation coefficient of 0.940379171. Because TMIN has a higher variable importance and there is a high correlation between the TMIN and TMAX, I’ll leave TMAX out of the model.

### Step 6: Building the Models

The goal here was to build a multiple linear regression model - since I’ve already started down the path of Random Forests, I’ll do one of those, too, and compare the two. To build the models, we do the following:

``````## BUILD ANOTHER FOREST USING THE IMPORTANT VARIABLES
predForest <- randomForest(count ~ DOW + HOL + TMIN + MON,
data = train, importance = TRUE,
ntree = 10000)

## BUILD A LINEAR MODEL USING THE IMPORTANT VARIABLES
linmod_with_mon <- lm(count ~ TMIN + DOW + HOL + MON,
data = train)``````

In looking at the summary, I have questions on whether or not the month variable (MON) is significant to the model or not. Many of the variables have rather high p-values.

``````summary(linmod_with_mon)

Call:
lm(formula = count ~ TMIN + DOW + HOL + MON, data = train)

Residuals:
Min     1Q Median      3Q    Max
-4471.5 -132.1   49.6   258.2 2539.8

Coefficients:
Estimate Std. Error  t value Pr(>|t|)
(Intercept)  5271.4002    89.5216   58.884  < 2e-16 ***
TMIN          -15.2174     5.6532   -2.692 0.007265 **
DOWMonday    -619.5908    75.2208   -8.237 7.87e-16 ***
DOWSaturday  -788.8261    74.3178  -10.614  < 2e-16 ***
DOWSunday   -3583.6718    74.0854  -48.372  < 2e-16 ***
DOWThursday   179.0975    74.5286    2.403 0.016501 *
DOWTuesday   -494.3059    73.7919   -6.699 4.14e-11 ***
DOWWednesday -587.7153    74.0264   -7.939 7.45e-15 ***
HOL1        -3275.6523   146.8750  -22.302  < 2e-16 ***
MONAugust     -99.8049   114.4150   -0.872 0.383321
MONDecember  -390.2925   109.4594   -3.566 0.000386 ***
MONFebruary  -127.8091   112.0767   -1.140 0.254496
MONJanuary    -73.0693   109.0627   -0.670 0.503081
MONJuly      -346.7266   113.6137   -3.052 0.002355 **
MONJune       -30.8752   101.6812   -0.304 0.761481
MONMarch       -1.4980    94.8631   -0.016 0.987405
MONMay          0.1194    88.3915    0.001 0.998923
MONNovember   170.8023    97.6989    1.748 0.080831 .
MONOctober    125.1124    92.3071    1.355 0.175702
MONSeptember  199.6884   101.9056    1.960 0.050420 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 544.2 on 748 degrees of freedom
Multiple R-squared: 0.8445, Adjusted R-squared: 0.8405
F-statistic: 213.8 on 19 and 748 DF, p-value: < 2.2e-16``````

To verify this, I build the model without the MON term and then do an F-Test to compare using the results of the ANOVA tables below.

``````## FIRST ANOVA TABLE (WITH THE MON TERM)
anova(linmod_with_mon)
Analysis of Variance Table

Response: count
Df     Sum Sq   Mean Sq  F value    Pr(>F)
TMIN        1   16109057  16109057  54.3844 4.383e-13 ***
DOW         6 1019164305 169860717 573.4523 < 2.2e-16 ***
HOL         1  147553631 147553631 498.1432 < 2.2e-16 ***
MON        11   20322464   1847497   6.2372 6.883e-10 ***
Residuals 748  221563026    296207

## SECOND ANOVA TABLE (WITHOUT THE MON TERM)
anova(linmod_wo_mon)
Analysis of Variance Table

Response: count
Df     Sum Sq   Mean Sq F value    Pr(>F)
TMIN        1   16109057  16109057  50.548 2.688e-12 ***
DOW         6 1019164305 169860717 532.997 < 2.2e-16 ***
HOL         1  147553631 147553631 463.001 < 2.2e-16 ***
Residuals 759  241885490    318690

## Ho: B9 = B10 = B11 = B12 = B13 = B14 = B15 = B16 =
##     B17 = B18 = B19 = 0
## Ha: At least one is not equal to 0

## F-Stat = MSdrop / MSE =
##          ((SSR1 - SSR2) / (DF(R)1 - DF(R)2)) / MSE

f_stat <- ((241885490 - 221563026) / (759 - 748)) / 296207

## P_VALUE OF THE F_STAT CALCULATED ABOVE
p_value <- 1 - pf(f_stat, 11, 748)``````

Since the P-Value 6.8829e-10 is MUCH MUCH less than 0.05, I can reject the null hypothesis and conclude that at least one of the parameters associated with the MON term is not zero. Because of this, I’ll keep the term in the model.

### Step 7: Apply the Models to the Test Data

Below I call the predict function to see how the Random Forest and Linear Model predict the test data. I am rounding the prediction to the nearest integer. To determine which model performs better, I am calculating the difference in absolute value of the predicted value from the actual count.

``````## PREDICT THE VALUES BASED ON THE MODELS
test\$RF <- round(predict(predForest, test), 0)
test\$LM <- round(predict.lm(linmod_with_mon, test), 0)

## SEE THE ABSOLUTE DIFFERENCE FROM THE ACTUAL
difOfRF <- sum(abs(test\$RF - test\$count))
difOfLM <- sum(abs(test\$LM - test\$count))``````

### Conclusion

As it turns out, the Linear Model performs better than the Random Forest model. I am relatively pleased with the Linear Model - an R-Squared value of 0.8445 ain’t nothin’ to shake a stick at. You can see that Random Forests are very useful in identifying the important features. To me, it tends to be a bit more of a “black box” in comparison the linear regression - I hesitate to use it at work for more than a feature identification tool.

Overall - a nice little experiment and a great dive into some open data. I now know that PPA rarely takes a day off, regardless of the weather. I’d love to know how much of the fines they write are actually collected. I may also dive into predicting what type of ticket you received based on your location, time of ticket, etc. All in another day’s work!