An Analytical Detective
Crime is an international concern, but it is documented and handled in very different ways in different countries. In the United States, violent crimes and property crimes are recorded by the Federal Bureau of Investigation (FBI). Additionally, each city documents crime, and some cities release data regarding crime rates. The city of Chicago, Illinois releases crime data from 2001 onward online.
Chicago is the third most populous city in the United States, with a population of over 2.7 million people. The city of Chicago is shown in the map below, with the state of Illinois highlighted in red.
There are two main types of crimes: violent crimes, and property crimes. In this problem, we’ll focus on one specific type of property crime, called “motor vehicle theft” (sometimes referred to as grand theft auto). This is the act of stealing, or attempting to steal, a car. In this problem, we’ll use some basic data analysis in R to understand the motor vehicle thefts in Chicago.
Please download the file mvtWeek1 (CSV) for this problem (do not open this file in any spreadsheet software before completing this problem because it might change the format of the Date field). Here is a list of descriptions of the variables:
- ID: a unique identifier for each observation.
- Date: the date the crime occurred.
- LocationDescription: the location where the crime occurred.
- Arrest: whether or not an arrest was made for the crime (TRUE if an arrest was made, and FALSE if an arrest was not made).
- Domestic: whether or not the crime was a domestic crime, meaning that it was committed against a family member (TRUE if it was domestic, and FALSE if it was not domestic).
- Beat: the area, or “beat” in which the crime occurred. This is the smallest regional division defined by the Chicago police department.
- District: the police district in which the crime occured. Each district is composed of many beats, and are defined by the Chicago Police Department.
- CommunityArea: the community area in which the crime occurred. Since the 1920s, Chicago has been divided into what are called “community areas”, of which there are now 77. The community areas were devised in an attempt to create socially homogeneous regions.
- Year: the year in which the crime occurred.
- Latitude: the latitude of the location at which the crime occurred.
- Longitude: the longitude of the location at which the crime occurred.
Problem 1.1 - Loading the Data
Read the dataset mvtWeek1 (CSV) into R, using the read.csv function, and call the data frame “mvt”. Remember to navigate to the directory on your computer containing the file mvtWeek1.csv first. It may take a few minutes to read in the data, since it is pretty large. Then, use the str and summary functions to answer the following questions.
How many rows of data (observations) are in this dataset?
Exercise 1
Numerical Response
Explanation
If you type str(mvt) in the R console, the first row of output says that this is a data frame with 191,641 observations.
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Problem 1.2 - Loading the Data
How many variables are in this dataset?
Exercise 2
Numerical Response
Explanation
If you type str(mvt) in the R console, the first row of output says that this is a data frame with 11 variables.
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Problem 1.3 - Loading the Data
Using the “max” function, what is the maximum value of the variable “ID”?
Exercise 3
Numerical Response
Explanation
You can compute the maximum value of the ID variable with max(mvt$ID).
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Problem 1.4 - Loading the Data
What is the minimum value of the variable “Beat”?
Exercise 4
Numerical Response
Explanation
If you type summary(mvt) in your R console, you can see the summary statistics for each variable. This shows that the minimum value of Beat is 111. Alternatively, you could use the min function by typing min(mvt$Beat).
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Problem 1.5 - Loading the Data
How many observations have value TRUE in the Arrest variable (this is the number of crimes for which an arrest was made)?
Exercise 5
Numerical Response
Explanation
If you type summary(mvt) in your R console, you can see the summary statistics for each variable. This shows that 15,536 observations fall under the category TRUE for the variable Arrest.
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Problem 1.6 - Loading the Data
How many observations have a LocationDescription value of ALLEY?
Exercise 6
Numerical Response
Explanation
If you type summary(mvt) in your R console, you can see the summary statistics for each variable. This shows that 2,308 observations fall under the category ALLEY for the variable LocationDescription. You can also read this from table(mvt$LocationDescription).
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Problem 2.1 - Understanding Dates in R
In many datasets, like this one, you have a date field. Unfortunately, R does not automatically recognize entries that look like dates. We need to use a function in R to extract the date and time. Take a look at the first entry of Date (remember to use square brackets when looking at a certain entry of a variable).
In what format are the entries in the variable Date?
Exercise 7
Month/Day/Year Hour:Minute
Day/Month/Year Hour:Minute
Hour:Minute Month/Day/Year
Hour:Minute Day/Month/Year
Explanation
If you type mvt$Date[1] in your R console, you can see that the first entry is 12/31/12 23:15. This must be in the format Month/Day/Year Hour:Minute.
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Problem 2.2 - Understanding Dates in R
Now, let’s convert these characters into a Date object in R. In your R console, type
DateConvert = as.Date(strptime(mvt$Date, “%m/%d/%y %H:%M”))
This converts the variable “Date” into a Date object in R. Take a look at the variable DateConvert using the summary function.
What is the month and year of the median date in our dataset? Enter your answer as “Month Year”, without the quotes. (Ex: if the answer was 2008-03-28, you would give the answer “March 2008”, without the quotes.)
Exercise 8
Text Response Answer:May 2006
Explanation
If you type summary(DateConvert), you can see that the median date is 2006-05-21.
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Problem 2.3 - Understanding Dates in R
Now, let’s extract the month and the day of the week, and add these variables to our data frame mvt. We can do this with two simple functions. Type the following commands in R:
mvt$Month = months(DateConvert)
mvt$Weekday = weekdays(DateConvert)
This creates two new variables in our data frame, Month and Weekday, and sets them equal to the month and weekday values that we can extract from the Date object. Lastly, replace the old Date variable with DateConvert by typing:
mvt$Date = DateConvert
Using the table command, answer the following questions.
In which month did the fewest motor vehicle thefts occur?
Exercise 9
JanuaryFebruaryMarchAprilMayJuneJulyAugustSeptemberOctoberNovemberDecember February
Explanation
If you type table(mvt$Month), you can see that the month with the smallest number of observations is February.
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Problem 2.4 - Understanding Dates in R
On which weekday did the most motor vehicle thefts occur?
Exercise 10
MondayTuesdayWednesdayThursdayFridaySaturdaySunday Friday
Explanation
If you type table(mvt$Weekday), you can see that the weekday with the largest number of observations is Friday.
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Problem 2.5 - Understanding Dates in R
Each observation in the dataset represents a motor vehicle theft, and the Arrest variable indicates whether an arrest was later made for this theft. Which month has the largest number of motor vehicle thefts for which an arrest was made?
Exercise 11
JanuaryFebruaryMarchAprilMayJuneJulyAugustSeptemberOctoberNovemberDecember January
Explanation
If you type table(mvt$Arrest,mvt$Month), you can see that the largest number of observations with Arrest=TRUE occurs in the month of January.
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Problem 3.1 - Visualizing Crime Trends
Now, let’s make some plots to help us better understand how crime has changed over time in Chicago. Throughout this problem, and in general, you can save your plot to a file. For more information, this website very clearly explains the process.
First, let’s make a histogram of the variable Date. We’ll add an extra argument, to specify the number of bars we want in our histogram. In your R console, type
hist(mvt$Date, breaks=100)
Looking at the histogram, answer the following questions.
In general, does it look like crime increases or decreases from 2002 - 2012?
Exercise 12
Increases
Decreases
Explanation
While there is not a clear trend, it looks like crime generally decreases.
In general, does it look like crime increases or decreases from 2005 - 2008?
Exercise 13
Increases
Decreases
Explanation
In this time period, there is a clear downward trend in crime.
In general, does it look like crime increases or decreases from 2009 - 2011?
Exercise 14
Increases
Decreases
Explanation
In this time period, there is a clear upward trend in crime.
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Problem 3.2 - Visualizing Crime Trends
Now, let’s see how arrests have changed over time. Create a boxplot of the variable “Date”, sorted by the variable “Arrest” (if you are not familiar with boxplots and would like to learn more, check out this tutorial). In a boxplot, the bold horizontal line is the median value of the data, the box shows the range of values between the first quartile and third quartile, and the whiskers (the dotted lines extending outside the box) show the minimum and maximum values, excluding any outliers (which are plotted as circles). Outliers are defined by first computing the difference between the first and third quartile values, or the height of the box. This number is called the Inter-Quartile Range (IQR). Any point that is greater than the third quartile plus the IQR or less than the first quartile minus the IQR is considered an outlier.
Does it look like there were more crimes for which arrests were made in the first half of the time period or the second half of the time period? (Note that the time period is from 2001 to 2012, so the middle of the time period is the beginning of 2007.)
Exercise 15
First half
Second half
Explanation
You can create the boxplot with the command boxplot(mvt$Date ~ mvt$Arrest). If you look at the boxplot, the one for Arrest=TRUE is definitely skewed towards the bottom of the plot, meaning that there were more crimes for which arrests were made in the first half of the time period.
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Problem 3.3 - Visualizing Crime Trends
Let’s investigate this further. Use the table function for the next few questions.
For what proportion of motor vehicle thefts in 2001 was an arrest made?
Note: in this question and many others in the course, we are asking for an answer as a proportion. Therefore, your answer should take a value between 0 and 1.
Exercise 16
Numerical Response
Explanation
If you create a table using the command table(mvt$Arrest, mvt$Year), the column for 2001 has 2152 observations with Arrest=TRUE and 18517 observations with Arrest=FALSE. The fraction of motor vehicle thefts in 2001 for which an arrest was made is thus 2152/(2152+18517) = 0.1041173.
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Problem 3.4 - Visualizing Crime Trends
For what proportion of motor vehicle thefts in 2007 was an arrest made?
Exercise 17
Numerical Response
Explanation
If you create a table using the command table(mvt$Arrest, mvt$Year), the column for 2007 has 1212 observations with Arrest=TRUE and 13068 observations with Arrest=FALSE. The fraction of motor vehicle thefts in 2007 for which an arrest was made is thus 1212/(1212+13068) = 0.08487395.
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Problem 3.5 - Visualizing Crime Trends
For what proportion of motor vehicle thefts in 2012 was an arrest made?
Exercise 18
Numerical Response
Explanation
If you create a table using the command table(mvt$Arrest, mvt$Year), the column for 2012 has 550 observations with Arrest=TRUE and 13542 observations with Arrest=FALSE. The fraction of motor vehicle thefts in 2012 for which an arrest was made is thus 550/(550+13542) = 0.03902924.
Since there may still be open investigations for recent crimes, this could explain the trend we are seeing in the data. There could also be other factors at play, and this trend should be investigated further. However, since we don’t know when the arrests were actually made, our detective work in this area has reached a dead end.
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Problem 4.1 - Popular Locations
Analyzing this data could be useful to the Chicago Police Department when deciding where to allocate resources. If they want to increase the number of arrests that are made for motor vehicle thefts, where should they focus their efforts?
We want to find the top five locations where motor vehicle thefts occur. If you create a table of the LocationDescription variable, it is unfortunately very hard to read since there are 78 different locations in the data set. By using the sort function, we can view this same table, but sorted by the number of observations in each category. In your R console, type:
sort(table(mvt$LocationDescription))
Which locations are the top five locations for motor vehicle thefts, excluding the “Other” category? You should select 5 of the following options.
Exercise 19
Bank
Gas Station
Hotel/Motel
Street
Car Wash
Restaurant
Parking Lot/Garage (Non-Residential)
Alley
Driveway (Residential)
Vacant Lot/Land
Explanation
If you type sort(table(mvt$LocationDescription)), the locations with the largest number of motor vehicle thefts are listed last. These are Street, Parking Lot/Garage (Non. Resid.), Alley, Gas Station, and Driveway - Residential.
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Problem 4.2 - Popular Locations
Create a subset of your data, only taking observations for which the theft happened in one of these five locations, and call this new data set “Top5”. To do this, you can use the | symbol. In lecture, we used the & symbol to use two criteria to make a subset of the data. To only take observations that have a certain value in one variable or the other, the | character can be used in place of the & symbol. This is also called a logical “or” operation.
Alternately, you could create five different subsets, and then merge them together into one data frame using rbind.
How many observations are in Top5?
Exercise 20
Numerical Response
Explanation
You can create this subset with the command:
Top5 = subset(mvt, LocationDescription==“STREET” | LocationDescription==“PARKING LOT/GARAGE(NON.RESID.)” | LocationDescription==“ALLEY” | LocationDescription==“GAS STATION” | LocationDescription==“DRIVEWAY - RESIDENTIAL”)
If you look at the structure of this data frame with str(Top5), you can see that there are 177510 observations.
Another way of doing this would be to use the %in% operator in R. This operator checks for inclusion in a set. You can create the same subset by typing the following two lines in your R console:
TopLocations = c(“STREET”, “PARKING LOT/GARAGE(NON.RESID.)”, “ALLEY”, “GAS STATION”, “DRIVEWAY - RESIDENTIAL”)
Top5 = subset(mvt, LocationDescription %in% TopLocations)
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Problem 4.3 - Popular Locations
R will remember the other categories of the LocationDescription variable from the original dataset, so running table(Top5$LocationDescription) will have a lot of unnecessary output. To make our tables a bit nicer to read, we can refresh this factor variable. In your R console, type:
Top5$LocationDescription = factor(Top5$LocationDescription)
If you run the str or table function on Top5 now, you should see that LocationDescription now only has 5 values, as we expect.
Use the Top5 data frame to answer the remaining questions.
One of the locations has a much higher arrest rate than the other locations. Which is it? Please enter the text in exactly the same way as how it looks in the answer options for Problem 4.1.
Exercise 21
Text Response Answer:Gas Station
Explanation
If you create a table of LocationDescription compared to Arrest, table(Top5$LocationDescription, Top5$Arrest), you can then compute the fraction of motor vehicle thefts that resulted in arrests at each location. Gas Station has by far the highest percentage of arrests, with over 20% of motor vehicle thefts resulting in an arrest.
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Problem 4.4 - Popular Locations
On which day of the week do the most motor vehicle thefts at gas stations happen?
Exercise 22
MondayTuesdayWednesdayThursdayFridaySaturdaySunday Saturday
Explanation
This can be read from table(Top5$LocationDescription, Top5$Weekday).
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Problem 4.5 - Popular Locations
On which day of the week do the fewest motor vehicle thefts in residential driveways happen?
Exercise 23
MondayTuesdayWednesdayThursdayFridaySaturdaySunday Saturday
Explanation
This can be read from table(Top5$LocationDescription, Top5$Weekday).
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