12 Exploratory data analysis
For this chapter you’ll need the following files, which are available for download here: ucr2017.rda and offenses_known_yearly_1960_2020.rds.
When you first start working on new data it is important to spend some time getting familiar with the data. This includes understanding how many rows and columns it has, what each row means (is each row an offender? a victim? crime in a city over a day/month/year?, etc.), and what columns it has. Basically you want to know if the data is capable of answering the question you are asking.
While not a comprehensive list, the following is a good start for exploratory data analysis of new data sets.
- What are the units (what does each row represent?)?
- What variables are available?
- What time period does it cover?
- Are there outliers? How many?
- Are there missing values? How many?
For this lesson we will use a data set of FBI Uniform Crime Reporting (UCR) data for 2017. This data includes every agency that reported their data for all 12 months of the year. Throughout this lesson we will look at some summary statistics for the variables we are interested in and make some basic graphs to visualize the data.
First, we need to load the data. Make sure your working directory is set to the folder where the data is.
head() will print out the first 6 rows of every column in the data. Since we only have 9 columns, we will use this function. Be careful when you have many columns (100+) as printing all of them out makes it read to read.
head(ucr2017) #> ori year agency_name state population actual_murder #> 1 AK00101 2017 anchorage alaska 296188 27 #> 2 AK00102 2017 fairbanks alaska 32937 10 #> 3 AK00103 2017 juneau alaska 32344 1 #> 4 AK00104 2017 ketchikan alaska 8230 1 #> 5 AK00105 2017 kodiak alaska 6198 0 #> 6 AK00106 2017 nome alaska 3829 0 #> actual_rape_total actual_robbery_total #> 1 391 778 #> 2 24 40 #> 3 50 46 #> 4 19 0 #> 5 15 4 #> 6 7 0 #> actual_assault_aggravated #> 1 2368 #> 2 131 #> 3 206 #> 4 14 #> 5 41 #> 6 52
From these results it appears that each row is a single agency’s annual data for 2017 and the columns show the number of crimes for four crime categories included (the full UCR data contains many more crimes which we’ll see in a later lesson).
Finally, we can run
names() to print out every column name. We can already see every name from
head() but this is useful when we have many columns and don’t want to use
names(ucr2017) #>  "ori" "year" #>  "agency_name" "state" #>  "population" "actual_murder" #>  "actual_rape_total" "actual_robbery_total" #>  "actual_assault_aggravated"
12.1 Summary and Table
An important function in understanding the data you have is
summary() which, as discussed in Section 3.5, provides summary statistics on the numeric columns you have. Let’s take a look at the results before seeing how to do something similar for categorical columns.
summary(ucr2017) #> ori year agency_name #> Length:15764 Min. :2017 Length:15764 #> Class :character 1st Qu.:2017 Class :character #> Mode :character Median :2017 Mode :character #> Mean :2017 #> 3rd Qu.:2017 #> Max. :2017 #> state population actual_murder #> Length:15764 Min. : 0 Min. : 0.000 #> Class :character 1st Qu.: 914 1st Qu.: 0.000 #> Mode :character Median : 4460 Median : 0.000 #> Mean : 19872 Mean : 1.069 #> 3rd Qu.: 15390 3rd Qu.: 0.000 #> Max. :8616333 Max. :653.000 #> actual_rape_total actual_robbery_total #> Min. : -2.000 Min. : -1.00 #> 1st Qu.: 0.000 1st Qu.: 0.00 #> Median : 1.000 Median : 0.00 #> Mean : 8.262 Mean : 19.85 #> 3rd Qu.: 5.000 3rd Qu.: 4.00 #> Max. :2455.000 Max. :13995.00 #> actual_assault_aggravated #> Min. : -1.00 #> 1st Qu.: 1.00 #> Median : 5.00 #> Mean : 49.98 #> 3rd Qu.: 21.00 #> Max. :29771.00
table() function returns every unique value in a category and how often that value appears. Unlike
summary() we can’t just put the entire data set into the (), we need to specify a single column. To specify a column you use the dollar sign notation which is
data$column. For most functions we use to examine the data as a whole, you can do the same for a specific column.
head(ucr2017$agency_name) #>  "anchorage" "fairbanks" "juneau" "ketchikan" #>  "kodiak" "nome"
There are only two columns in our data with categorical values that we can use - year and state so let’s use
table() on both of them. The columns ori and agency_name are also categorical but as each row of data has a unique ORI and name, running
table() on those columns would not be helpful.
table(ucr2017$year) #> #> 2017 #> 15764
We can see that every year in our data is 2017, as expected based on the data name. year is a numerical column so why can we use
table() on it? R doesn’t differentiate between numbers and characters when seeing how often each value appears. If we ran
table() on the column “actual_murder” it would tell us how many times each unique value in the column appeared in the data. That wouldn’t be very useful as we don’t really care how many times an agency has 7 murders, for example (though looking for how often a numeric column has the value 0 can be helpful in finding likely erroneous data). As numeric variables often have many more unique values than character variables, it also leads to many values being printed, making it harder to understand. For columns where the number of categories is important to us, such as years, states, neighborhoods, we should use
table(ucr2017$state) #> #> alabama alaska #> 305 32 #> arizona arkansas #> 107 273 #> california colorado #> 732 213 #> connecticut delaware #> 107 63 #> district of columbia florida #> 3 603 #> georgia guam #> 522 1 #> hawaii idaho #> 4 95 #> illinois indiana #> 696 247 #> iowa kansas #> 216 309 #> kentucky louisiana #> 352 192 #> maine maryland #> 135 152 #> massachusetts michigan #> 346 625 #> minnesota mississippi #> 397 71 #> missouri montana #> 580 108 #> nebraska nevada #> 225 59 #> new hampshire new jersey #> 176 576 #> new mexico new york #> 116 532 #> north carolina north dakota #> 310 108 #> ohio oklahoma #> 532 409 #> oregon pennsylvania #> 172 1473 #> rhode island south carolina #> 49 427 #> south dakota tennessee #> 92 466 #> texas utah #> 999 125 #> vermont virginia #> 85 407 #> washington west virginia #> 250 200 #> wisconsin wyoming #> 433 57
This shows us how many times each state is present in the data. States with a larger population tend to appear more often, this makes sense as those states have more agencies to report. Right now the results are in alphabetical order, but when knowing how frequently something appears, we usually want it ordered by frequency. We can use the
sort() function to order the results from
table(). Just put the entire
table() function inside of the () in
sort(table(ucr2017$state)) #> #> guam district of columbia #> 1 3 #> hawaii alaska #> 4 32 #> rhode island wyoming #> 49 57 #> nevada delaware #> 59 63 #> mississippi vermont #> 71 85 #> south dakota idaho #> 92 95 #> arizona connecticut #> 107 107 #> montana north dakota #> 108 108 #> new mexico utah #> 116 125 #> maine maryland #> 135 152 #> oregon new hampshire #> 172 176 #> louisiana west virginia #> 192 200 #> colorado iowa #> 213 216 #> nebraska indiana #> 225 247 #> washington arkansas #> 250 273 #> alabama kansas #> 305 309 #> north carolina massachusetts #> 310 346 #> kentucky minnesota #> 352 397 #> virginia oklahoma #> 407 409 #> south carolina wisconsin #> 427 433 #> tennessee georgia #> 466 522 #> new york ohio #> 532 532 #> new jersey missouri #> 576 580 #> florida michigan #> 603 625 #> illinois california #> 696 732 #> texas pennsylvania #> 999 1473
And if we want to sort it in decreasing order of frequency, we can use the parameter
sort() and set it to TRUE. A parameter is just an option used in an R function to change the way the function is used or what output it gives. Almost all functions have these parameters and they are useful if you don’t want to use the default setting in the function. This parameter,
decreasing changes the
sort() output to print from largest to smallest. By default this parameter is set to FALSE and here we say it is equal to TRUE.
sort(table(ucr2017$state), decreasing = TRUE) #> #> pennsylvania texas #> 1473 999 #> california illinois #> 732 696 #> michigan florida #> 625 603 #> missouri new jersey #> 580 576 #> new york ohio #> 532 532 #> georgia tennessee #> 522 466 #> wisconsin south carolina #> 433 427 #> oklahoma virginia #> 409 407 #> minnesota kentucky #> 397 352 #> massachusetts north carolina #> 346 310 #> kansas alabama #> 309 305 #> arkansas washington #> 273 250 #> indiana nebraska #> 247 225 #> iowa colorado #> 216 213 #> west virginia louisiana #> 200 192 #> new hampshire oregon #> 176 172 #> maryland maine #> 152 135 #> utah new mexico #> 125 116 #> montana north dakota #> 108 108 #> arizona connecticut #> 107 107 #> idaho south dakota #> 95 92 #> vermont mississippi #> 85 71 #> delaware nevada #> 63 59 #> wyoming rhode island #> 57 49 #> alaska hawaii #> 32 4 #> district of columbia guam #> 3 1
We often want to make quick plots of our data to get a visual understanding of the data. We will learn a different - and in my opinion a superior - way to make graphs in Chapters 15 but for now let’s use the function
plot() function is built into R so we don’t need to use any packages for it.
Let’s make a few scatterplots showing the relationship between two variables. With
plot() the syntax (how you write the code) is
plot(x_axis_variable, y_axis_variable). So all we need to do is give it the variable for the x- and y-axis. Each dot will represent a single agency (a single row in our data).
Above we are telling R to plot the number of murders on the x-axis and the number of robberies on the y-axis. This shows the relationship between a city’s number of murders and number of robberies. We can see that there is a relationship where more murders is correlated with more robberies. However, there are a huge number of agencies in the bottom-left corner which have very few murders or robberies. This makes sense as - as we see in the
summary() above - most agencies are small, with the median population under 5,000 people.
To try to avoid that clump of small agencies at the bottom, let’s make a new data set of only agencies with a population over 1 million. We will use the
filter() function from the
dplyr package that was introduced in Chapter @ref(#subsetting-intro). For
filter() we need to first include our dataset name, which is ucr2017, and then say our conditional statement. Our conditional statement is that rows in the “population” column have a value of over 1 million. For the
dplyr functions we don’t put our column name in quotes.
And we’ll save our results into a new object called ucr2017_big_cities Since we’re using the
dplyr package we need to tell R that we want to use it by using
library(dplyr) #> #> Attaching package: 'dplyr' #> The following objects are masked from 'package:stats': #> #> filter, lag #> The following objects are masked from 'package:base': #> #> intersect, setdiff, setequal, union <- filter(ucr2017, population > 1000000) ucr2017_big_cities
Now we have 18 agencies with a population of over 1 million people.
Now we can do the same graph as above but using this new data set.
The problem is somewhat solved. There is still a small clumping of agencies with few robberies or aggravated assaults but the issue is much better. And interestingly the trend is similar with this small subset of data as with all agencies included.
To make our graph look better, we can add labels for the axes and a title (there are many options for changing the appears of this graph, we will just use these three).
- xlab - X-axis label
- ylab - Y-axis label
- main - Graph title
Like all parameters, we add them in the () of
plot() and separate each parameter by a comma. Since we are adding text to write in the plot, all of these parameter inputs must be in quotes.
plot(ucr2017_big_cities$actual_murder, ucr2017_big_cities$actual_robbery_total, xlab = "Murder", ylab = "Robberies", main = "Relationship between murder and robbery")
12.3 Aggregating (summaries of groups)
Right now we have the number of crimes in each agency. For many policy analyses we’d be looking at the effect on the state as a whole, rather than at the agency-level. If we wanted to do this in our data, we would need to aggregate up to the state level. Aggregating data means that we group values at some higher level than they currently are (e.g. from agency to state, from day to month, from city street to city neighborhood) and then do some mathematical operation of our choosing (in our case usually sum) to that group.
In Section 11.3.3 we started to see if marijuana legalization affected murder in Colorado. We subsetted the data to only include agencies in Colorado from 2011-2017. Now we can continue to answer the question by aggregating to the state-level to see the total number of murders per year.
Let’s think about how our data are and how we would (theoretically, before we write any code) find that out.
Our data is a single row for each agency and we have a column indicating the year the agency reported. So how would be find out how many murders happened in Colorado for each year? Well, first we take all the agencies in 2011 (the first year available) and add up the murders for all agencies that reported that year. Then take all the rows in 2012 and add up their murders. And so on for all the years.
To do this in R we’ll be using two new functions from the
These functions do the aggregation process in two steps. First we use
group_by() to tell R which columns we want to group our data by - these are the higher level of aggregation columns so in our case will be the year of data (as we will already subset data to only Colorado and only the years 2011 through 2017). Then we need to sum up the number of murders each year. We do this using
summarize() and we’ll specify in the function that we want to sum up the data, rather than use some other math on it like finding the average number of murders each year.
First, let’s load back in the data and then repeat the subsetting code we did in Chapter 11.3.3 to keep only data for Colorado from 2011 through 2017. We’ll also include the “actual_robbery_total” column that we excluded in Chapter 11.3.3 so we can see how easy it is to aggregate multiple columns at once using this method.
<- readRDS("data/offenses_known_yearly_1960_2020.rds") offenses_known_yearly_1960_2020 <- filter(offenses_known_yearly_1960_2020, state == "colorado", year %in% 2011:2017) colorado <- select(colorado, actual_murder, actual_robbery_total, state, year, population, ori, agency_name)colorado
First we must group the data by using the
group() by function. Here we’re just grouping the data by year, but we could group it by multiple columns if we want by adding a comma and then the next column we want.
<- group_by(colorado, year)colorado
Now we can summarize the data using the
summarize() function. As with other
dplyr functions the first input is the dataset name. Then we choose our math function (sum, mean, median, etc.) and just apply that function on the column we want. So in our case we want the sum of murders so we use
sum() and include the column we want to aggregate inside of
summarize(colorado, sum(actual_murder)) #> # A tibble: 7 x 2 #> year `sum(actual_murder)` #> <dbl> <dbl> #> 1 2011 154 #> 2 2012 163 #> 3 2013 172 #> 4 2014 148 #> 5 2015 173 #> 6 2016 203 #> 7 2017 218
If we want to aggregate another column we just add a comma after our initial column and add another math operation function and the column we want. Here we’re also using
sum() but we could use different math operations if we want - they don’t need to be the same.
summarize(colorado, sum(actual_murder), sum(actual_robbery_total)) #> # A tibble: 7 x 3 #> year `sum(actual_murder)` `sum(actual_robbery_total)` #> <dbl> <dbl> <dbl> #> 1 2011 154 3287 #> 2 2012 163 3369 #> 3 2013 172 3122 #> 4 2014 148 3021 #> 5 2015 173 3305 #> 6 2016 203 3513 #> 7 2017 218 3811
We could even do different math operations on the same column and we’d get multiple columns from it. Let’s add another column showing the average number of robberies as an example.
summarize(colorado, sum(actual_murder), sum(actual_robbery_total), mean(actual_robbery_total)) #> # A tibble: 7 x 4 #> year `sum(actual_murd~ `sum(actual_robb~ `mean(actual_ro~ #> <dbl> <dbl> <dbl> <dbl> #> 1 2011 154 3287 11.2 #> 2 2012 163 3369 11.2 #> 3 2013 172 3122 10.3 #> 4 2014 148 3021 9.94 #> 5 2015 173 3305 10.9 #> 6 2016 203 3513 11.6 #> 7 2017 218 3811 12.5
summarize() calls the columns it makes using what we include in the parentheses. Since we said “sum(actual_murder)” to get the sum of the murder column, it names that new column “sum(actual_murder).” Usually we’ll want to name the columns ourselves. We can do this by assigning the summarized column to a name using “name =” before it. For example, we could write “murders = sum(actual_murder)” and it will name that column “murders” instead of “sum(actual_murder).” Like other things in
dplyr functions, we don’t need to put quotes around our new column name. We’ll save this final summarized data into an object called “colorado_agg” so we can use it to make graphs. And to be able to create crime rates per population, we’ll also find the sum of the population for each year.
<- summarize(colorado, murders = sum(actual_murder), robberies = sum(actual_robbery_total), population = sum(population)) colorado_agg colorado_agg#> # A tibble: 7 x 4 #> year murders robberies population #> <dbl> <dbl> <dbl> <dbl> #> 1 2011 154 3287 5155993 #> 2 2012 163 3369 5227884 #> 3 2013 172 3122 5308236 #> 4 2014 148 3021 5402555 #> 5 2015 173 3305 5505856 #> 6 2016 203 3513 5590124 #> 7 2017 218 3811 5661529
Now we can see that the total number of murders increased over time. So can we conclude that marijuana legalization increases murder? No, all this analysis shows is that the years following marijuana legalization, murders increased in Colorado. But that can be due to many reasons other than marijuana. For a proper analysis you’d need a comparison area that is similar to Colorado prior to legalization (and didn’t legalize marijuana) and see if their murders changes following Colorado’s legalization.
To control for population, we’ll standardize our murder data by creating a murder rate per 100,000 people. We can do this by dividing the murder column by the population column and then multiplying by 100,000. Let’s do that and save the result into a new column called “murder_rate.”
$murder_rate <- colorado_agg$murders / colorado_agg$population * 100000colorado_agg
If we also wanted a robbery rate we’d do the same with the robberies column.
$robbery_rate <- colorado_agg$robberies / colorado_agg$population * 100000colorado_agg
dplyr package has a helpful function that can do this too, and allows us to do it while writing less code. The
mutate() function lets us create or alter columns in our data. Like other
dplyr functions we start by including our dataset in the parentheses, and then we can follow standard assignment (covered in Section @ref(#assignment)) though we must use
= here and not
<-. A benefit of using
mutate() is that we don’t have to write out our dataset name each time. So we’d write
murder_rate = murders / population * 100000. And if we wanted to make two (or more) columns at the same time we just add a comma after our first assignment and then do the next assignment.
mutate(colorado_agg, murder_rate = murders / population * 100000, robbery_rate = robberies / population * 100000) #> # A tibble: 7 x 6 #> year murders robberies population murder_rate #> <dbl> <dbl> <dbl> <dbl> <dbl> #> 1 2011 154 3287 5155993 2.99 #> 2 2012 163 3369 5227884 3.12 #> 3 2013 172 3122 5308236 3.24 #> 4 2014 148 3021 5402555 2.74 #> 5 2015 173 3305 5505856 3.14 #> 6 2016 203 3513 5590124 3.63 #> 7 2017 218 3811 5661529 3.85 #> # ... with 1 more variable: robbery_rate <dbl>
Mpw let’s make a plot of this data showing the murder rate over time. With time-series graphs we want the time variable to be on the x-axis and the numeric variable we are measuring to the on the y-axis.
plot(x = colorado_agg$year, y = colorado_agg$murder_rate)
plot() makes a scatterplot. If we set the parameter
type to “l” it will be a line plot.
plot(x = colorado_agg$year, y = colorado_agg$murder_rate, type = "l")
We can add some labels and a title to make this graph easier to read.
plot(x = colorado_agg$year, y = colorado_agg$murder_rate, type = "l", xlab = "Year", ylab = "Murders per 100k Population", main = "Murder Rate in Colorado, 2011-2017")
12.4 Pipes in
To end this chapter we’ll talk about something called a pipe that is a very useful and powerful part of
Think about the math equation 1 + 2 + 3 + 4. Here we know that we add 1 and 2 together, and then add the result to 3 and then add the result of that to 4. This is much simpler to write than splitting everything up and summing each value together in a different line. In terms of R, we have so far been doing things as if we could only add two numbers together and then need a separate line to add the third (and another line to add the fourth) number. For example, below are the two lines of code we used to subset the data to just the right state and years we wanted, and the columns we wanted. We did this in two separate lines. In our math example, we did 1 + 2. And then found the answer, and separately did 3 + 3. And then again found the answer and did 6 + 4.
<- filter(offenses_known_yearly_1960_2020, state == "colorado", year %in% 2011:2017) colorado <- select(colorado, actual_murder, actual_robbery_total, state, year, population, ori, agency_name) colorado head(colorado) #> actual_murder actual_robbery_total state year #> 1 7 80 colorado 2017 #> 2 11 93 colorado 2016 #> 3 6 68 colorado 2015 #> 4 6 58 colorado 2014 #> 5 7 44 colorado 2013 #> 6 7 55 colorado 2012 #> population ori agency_name #> 1 99940 CO00100 adams #> 2 100526 CO00100 adams #> 3 100266 CO00100 adams #> 4 98569 CO00100 adams #> 5 97146 CO00100 adams #> 6 93542 CO00100 adams
In R we actually do have a way to chain together functions; to do the programming equivalent of 1 + 2 + 3 + 4 all at once. We do this through what is called a pipe, which allows us to take the result of one function and immediately put it into another function without having to save the initial result or start a new line of code. To use a pipe we put the following code after the end of a function:
%>%. These three characters,
%>% are the pipe and they must be written exactly like this. The pipe is itself actually a function, but is a special type of function we won’t go into detail about. Personally I don’t think this really looks like a pipe at all but it is called a pipe so that’s the terminology I’ll be using. How a pipe technically works is that it takes the output of the initial function (which is usually a data.frame) and puts it automatically is the first input in the next function. This won’t work for all functions but nearly all functions from the tidyverse collection of packages have a dataset as the first input so it will work here. The benefit is that we don’t need to keep saving out output from functions or specify which dataset to include in each function.
As an example, we’ll rewrite the above code using a pipe. We start with our data.frame which is normally the first input the the function, and then immediately have a pipe
%>% into a
dplyr function, which here is
filter(). Now we don’t need to say what the dataset is because we it takes the last thing that was piped into the function, which in our case is the entire data.frame offenses_known_yearly_1960_2020. After our
filter() is done we have another pipe and go into
select() will use as its first input (which is the data it is working with) as whatever is outputting from the
filter(). So the input to
select() will be the subsetted data output from
filter(). We can have as many pipes as we wish, and chain many different
dplyr functions together, but we just use two functions here so we’ll end after our
<- offenses_known_yearly_1960_2020 %>% filter(state == "colorado", year %in% 2011:2017) %>% select(actual_murder, actual_robbery_total, state, year, population, ori, agency_name)colorado
If we check results using
head(), we can see that this code is exactly the same as not using pipes.
head(colorado) #> actual_murder actual_robbery_total state year #> 1 7 80 colorado 2017 #> 2 11 93 colorado 2016 #> 3 6 68 colorado 2015 #> 4 6 58 colorado 2014 #> 5 7 44 colorado 2013 #> 6 7 55 colorado 2012 #> population ori agency_name #> 1 99940 CO00100 adams #> 2 100526 CO00100 adams #> 3 100266 CO00100 adams #> 4 98569 CO00100 adams #> 5 97146 CO00100 adams #> 6 93542 CO00100 adams
The normal way to write code using pipes is to have a new line after the pipe and after each comma in
select(). This doesn’t change how the code works at all, but is easier to read now because it has less code bunched together in a single line.
<- offenses_known_yearly_1960_2020 %>% colorado filter(state == "colorado", %in% 2011:2017) %>% year select(actual_murder, actual_robbery_total, state, year, population, ori, agency_name)