Data without Borders - Assignment 2

data-without-borders-hw2-part1.R

# Write code to return the percentage of people who were frisked for each 
# race.  In other words, count up the number of people who were frisked for a given race 
# divided by the number of people of that race stopped.  Which race leads to the highest 
# percentage of frisks?  Which one the lowest? 
 
#read in the data
snf <- read.csv("http://www.jakeporway.com/teaching/data/snf_2.csv", as.is=TRUE)

#race1 is black
race1.in.total = table(snf$race ==1)
race1.frisked = table(snf$race==1 & snf$frisked == 1)
race1.frisked / race1.in.total * 100
#    FALSE      TRUE 
#144.08737  59.30879 

#race2 is black Hispanic
table(snf$race==2 & snf$frisked == 1) / table(snf$race ==2) *100

#    FALSE      TRUE 
#103.21993  59.71156

#race 3 is white Hispanic
table(snf$race==3 & snf$frisked == 1) / table(snf$race ==3) *100

#    FALSE      TRUE 
#114.90219  55.76884 

#race 4 is white
table(snf$race==4 & snf$frisked == 1) / table(snf$race ==4) *100

#FALSE      TRUE 
#105.58145  41.90888 

#race 5 is Asian/Pacific Islander
table(snf$race==5 & snf$frisked == 1) / table(snf$race ==5) *100

#    FALSE      TRUE 
#101.76448  48.51948 

#race 6 is American Indian/Native Alaskan
table(snf$race==6 & snf$frisked == 1) / table(snf$race ==6) *100
FALSE      TRUE 
100.19523  45.93301 

#race -1 is undefined
table(snf$race==-1 & snf$frisked == 1) / table(snf$race ==-1) *100

#    FALSE      TRUE 
#101.56461  48.63133 


#Plot the number of times each crime occurs in descending order (we’ve learned a couple of ways to do this, #though using sort(), table() and that new type= parameter to plot() is your best bet).   What does this #distribution of crimes look like?  In other words, are there an equal number of every kind of crime or are there #a few that dominate?

plot(rev(sort(table(snf$crime.suspected))))


#Well I’m kind of answering that question for you here – let’s take the top 30 
#suspected crimes and look at those.  If we were to just look at stops where the 
#crime.suspected was one of the top 30 crimes, what percentage of the stops would that 
#cover?  Do you think that’s enough?

all.crimes.sorted = data.frame(rev(sort(table(snf$crime.suspected))))
top.30.crimes = all.crimes.sorted[1:30,]

#Well I’m kind of answering that question for you here – let’s take the top 30 
#suspected crimes and look at those.  If we were to just look at stops where the 
#crime.suspected was one of the top 30 crimes, what percentage of the stops would that 
#cover?

sum(top.30.crimes)
#[1] 53048
sum(top.30.crimes) / sum(all.crimes.sorted) * 100
#[1] 91.32194
#the top 30 crimes cover 91.3% of all crimes.


#Write code to create a variable called “crime.abbv” that consists of just the 
#first three letters of crime.suspected and show the code to add it to our main data frame.  
#Now what percentage of the stops do the top 30 crime.abbvs account for?
crime.abbr = substr(snf$crime.suspected, 1,3)
snf$crime.abbr = crime.abbr
all.crimes.sorted.abbr = data.frame(rev(sort(table(snf$crime.abbr))))
all.crimes.sorted.abbr[1:30,]
new.top.30.crimes = all.crimes.sorted.abbr[1:30,]
sum(new.top.30.crimes) / sum(all.crimes.sorted) * 100
#[1] 98.43172
#Now, using the abbreviated names for the crimes suspected, the top 30 make up 98.4% of all crimes.



#Write code to show the top 3 crimes each race is suspected of (rev(), 
#sort(), and table() are your friends here again, but you’ll have to subset the data by race 
#first).  Huh.  If you do this right, almost all the top 3’s should be the same, but a few are 
#different.  What are these differences? 

#top three crimes for race 1
race1 = snf[snf$race == 1,]
race1 = snf[snf$race == 1,]$crime.abbr
rev(sort(table(race1)))[1:3]
FEL   MIS   CPW 
10744  4487  4009 

#top 3 crimes suspected for race 2
rev(sort(table(snf[snf$race == 2,]$crime.abbr)))[1:3]

# FEL  MIS  CPW 
#1405  747  636 

#top 3 crimes suspected for race 3
rev(sort(table(snf[snf$race == 3,]$crime.abbr)))[1:3]

#FEL  MIS  CPW 
#5072 2423 1573 

#top 3 crimes suspected for race 4
rev(sort(table(snf[snf$race == 4,]$crime.abbr)))[1:3]
 FEL  MIS  BUR 
1647  769  485 

#top 3 crimes suspected for race 5
rev(sort(table(snf[snf$race == 5,]$crime.abbr)))[1:3]

FEL ROB MIS 
788 224 188

#top 3 crimes suspected for race 6
rev(sort(table(snf[snf$race == 6,]$crime.abbr)))[1:3]
FEL ROB GLA 
 91  28  21 

#top 3 crimes suspected for race -1
rev(sort(table(snf[snf$race == -1,]$crime.abbr)))[1:3]

FEL MIS CPW 
566 229 202 

data-without-borders-hw2-part2.R

#Let’s create an “hour” variable that tells us what hour of the day each stop
#happened during and add it to our dataset.  How do we do this?  Well we’ve got a 
#great column of “time” variables that always has the hour in the same place.  Use 
#the substr() function we learned about above to strip out the hour, then use 
#as.numeric() from lecture 2 to convert it to a number.


hour = substr(snf$time, 12, 13)
hour = as.numeric(hour)


#Create a line plot (i.e. a plot with type=”l”) of the stops by hour.  
stops.by.hour = table(hour)
plot(stops.by.hour, type="l")

Which hour of the day has the most stops?  
stops.by.hour[which(stops.by.hour == max(stops.by.hour))]
#20
#4607

Which hour has the fewest?
stops.by.hour[which(stops.by.hour == min(stops.by.hour))]
#  6 
#323 

#Create the same plot but with points instead of lines.  Use a different plotting 
#symbol than the default and color the max point and min points different colors.

#create a vector to hold the colors for each point
day.colors = rep(1, 24)
day.colors
# [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

#change the color of the max number to 6 (hot pink) and min number to blue
day.colors[which(stops.by.hour == max(stops.by.hour))] = 6
day.colors[which(stops.by.hour == min(stops.by.hour))] = 5
#plot
plot(stops.by.hour, type="p", col=day.colors)