Introduction to data - KEY

Departure delays in flights to Raleigh-Durham (RDU)

We can examine the distribution of departure delays of all flights with a histogram.

ggplot(data = nycflights, aes(x = dep_delay)) +geom_histogram()

## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

This function says to plot the dep_delay variable from the nycflights data frame on the x-axis. It also defines a geom (short for geometric object), which describes the type of plot you will produce.

Histograms are generally a very good way to see the shape of a single distribution, but that shape can change depending on how the data is split between the different bins. You can easily define the binwidth you want to use:

ggplot(data = nycflights, aes(x = dep_delay)) +geom_histogram(binwidth = 15)

ggplot(data = nycflights, aes(x = dep_delay)) +geom_histogram(binwidth = 150)

Exercise: How do these three histograms with the various binwidths compare?

If we want to focus on departure delays of flights headed to RDU only, we need to first filter the data for flights headed to RDU (dest == "RDU") and then make a histogram of only departure delays of only those flights.

rdu_flights <-nycflights %>%filter(dest == "RDU")ggplot(data = rdu_flights, aes(x = dep_delay)) +geom_histogram()

## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

Let’s decipher these three lines of code:

• Line 1: Take the nycflights data frame, filter for flights headed to RDU, and save the result as a new data frame called rdu_flights.
  • == means “if it’s equal to”.
  • RDU is in quotation marks since it is a character string.
• Line 2: Basically the same ggplot call from earlier for making a histogram, except that it uses the data frame for flights headed to RDU instead of all flights.

Logical operators: Filtering for certain observations (e.g. flights from a particular airport) is often of interest in data frames where we might want to examine observations with certain characteristics separately from the rest of the data. To do so we use the filter function and a series of logical operators. The most commonly used logical operators for data analysis are as follows:

• == means “equal to”
• != means “not equal to”
• > or < means “greater than” or “less than”
• >= or <= means “greater than or equal to” or “less than or equal to”

We can also obtain numerical summaries for these flights:

rdu_flights %>%summarise(mean_dd = mean(dep_delay), sd_dd = sd(dep_delay), n = n())

## Source: local data frame [1 x 3]
##
## mean_dd sd_dd n
## (dbl) (dbl) (int)
## 1 11.69913 35.55567 801

Note that in the summarise function we created a list of two elements. The names of these elements are user defined, like mean_dd, sd_dd, n, and you could customize these names as you like (just don’t use spaces in your names). Calculating these summary statistics also require that you know the function calls. Note that n() reports the sample size.

Summary statistics: Some useful function calls for summary statistics for a single numerical variable are as follows:

• mean
• median
• sd
• var
• IQR
• range
• min
• max

We can also filter based on multiple criteria. Suppose we are interested in flights headed to San Francisco (SFO) in February:

sfo_feb_flights <-nycflights %>%filter(dest == "SFO", month ==2)

Note that we can separate the conditions using commas if we want flights that are both headed to SFO and in February. If we are interested in either flights headed to SFO or in February we can use the | instead of the comma.

1. Create a new data frame that includes flights headed to SFO in February, and save this data frame as sfo_feb_flights. How many flights meet these criteria?
   1. 68 - CORRECT ANSWER
   2. 1345
   3. 2286
   4. 3563
   5. 32735

dim(sfo_feb_flights)

## [1] 68 16

2. Make a histogram and calculate appropriate summary statistics for arrival delays of sfo_feb_flights. Which of the following is false?
   1. The distribution is unimodal.
   2. The distribution is right skewed.
   3. No flight is delayed more than 2 hours. - CORRECT ANSWER
   4. The distribution has several extreme values on the right side.
   5. More than 50% of flights arrive on time or earlier than scheduled.

ggplot(data = sfo_feb_flights, aes(x = arr_delay)) +geom_histogram(binwidth = 10)

sfo_feb_flights %>%summarize(mean(arr_delay), median(arr_delay), max(arr_delay))

## Source: local data frame [1 x 3]
##
## mean(arr_delay) median(arr_delay) max(arr_delay)
## (dbl) (dbl) (dbl)
## 1 -4.5 -11 196

Another useful functionality is being able to quickly calculate summary statistics for various groups in your data frame. For example, we can modify the above command using the group_by function to get the same summary stats for each origin airport:

rdu_flights %>%group_by(origin) %>%summarise(mean_dd = mean(dep_delay), sd_dd = sd(dep_delay), n = n())

## Source: local data frame [3 x 4]
##
## origin mean_dd sd_dd n
## (chr) (dbl) (dbl) (int)
## 1 EWR 13.365517 32.08492 145
## 2 JFK 15.396667 40.30535 300
## 3 LGA 7.904494 32.18620 356

Here, we first grouped the data by origin, and then calculated the summary statistics.

3. Calculate the median and interquartile range for arr_delays of flights in the sfo_feb_flights data frame, grouped by carrier. Which carrier is the has the hights IQR of arrival delays?
   1. American Airlines
   2. JetBlue Airways
   3. Virgin America
   4. Delta and United Airlines - CORRECT ANSWER
   5. Frontier Airlines

sfo_feb_flights %>%group_by(carrier) %>%summarize(med_ad = median(arr_delay), iqr_ad = IQR(arr_delay)) %>%arrange(desc(iqr_ad))

## Source: local data frame [5 x 3]
##
## carrier med_ad iqr_ad
## (chr) (dbl) (dbl)
## 1 DL -15.0 22.00
## 2 UA -10.0 22.00
## 3 VX -22.5 21.25
## 4 AA 5.0 17.50
## 5 B6 -10.5 12.25

Departure delays over months

Which month would you expect to have the highest average delay departing from an NYC airport?

Let’s think about how we would answer this question:

• First, calculate monthly averages for departure delays. With the new language we are learning, we need to
  • group_by months, then
  • summarise mean departure delays.
• Then, we need to arrange these average delays in descending order

nycflights %>%group_by(month) %>%summarise(mean_dd = mean(dep_delay)) %>%arrange(desc(mean_dd))

## Source: local data frame [12 x 2]
##
## month mean_dd
## (int) (dbl)
## 1 7 20.754559
## 2 6 20.350293
## 3 12 17.368189
## 4 4 14.554477
## 5 3 13.517602
## 6 5 13.264800
## 7 8 12.619097
## 8 2 10.687227
## 9 1 10.233333
## 10 9 6.872436
## 11 11 6.103183
## 12 10 5.880374

4. Which month has the highest average departure delay from an NYC airport?
   1. January
   2. March
   3. July - CORRECT ANSWER
   4. October
   5. December

nycflights %>%group_by(month) %>%summarise(mean_dd = mean(dep_delay)) %>%arrange(desc(mean_dd))

## Source: local data frame [12 x 2]
##
## month mean_dd
## (int) (dbl)
## 1 7 20.754559
## 2 6 20.350293
## 3 12 17.368189
## 4 4 14.554477
## 5 3 13.517602
## 6 5 13.264800
## 7 8 12.619097
## 8 2 10.687227
## 9 1 10.233333
## 10 9 6.872436
## 11 11 6.103183
## 12 10 5.880374

5. Which month has the highest median departure delay from an NYC airport?
   1. January
   2. March
   3. July
   4. October
   5. December - COORECT ANSWER

nycflights %>%group_by(month) %>%summarise(med_dd = median(dep_delay)) %>%arrange(desc(med_dd))

## Source: local data frame [12 x 2]
##
## month med_dd
## (int) (dbl)
## 1 12 1
## 2 6 0
## 3 7 0
## 4 3 -1
## 5 5 -1
## 6 8 -1
## 7 1 -2
## 8 2 -2
## 9 4 -2
## 10 11 -2
## 11 9 -3
## 12 10 -3

6. Is the mean or the median a more reliable measure for deciding which month(s) to avoid flying if you really dislike delayed flights, and why?
   1. Mean would be more reliable as it gives us the true average. - CORRECT ANSWER
   2. Mean would be more reliable as the distribution of delays is symmetric.
   3. Median would be more reliable as the distribution of delays is skewed.
   4. Median would be more reliable as the distribution of delays is symmetric.
   5. Both give us useful information.

We can also visualize the distributions of departure delays across months using side-by-side box plots:

ggplot(nycflights, aes(x = factor(month), y = dep_delay)) +geom_boxplot()

There is some new syntax here: We want departure delays on the y-axis and the months on the x-axis to produce side-by-side box plots. Side-by-side box plots require a categorical variable on the x-axis, however in the data frame month is stored as a numerical variable (numbers 1 - 12). Therefore we can force R to treat this variable as categorical, what R calls a factor, variable with factor(month).

On time departure rate for NYC airports

Suppose you will be flying out of NYC and want to know which of the three major NYC airports has the best on time departure rate of departing flights. Suppose also that for you a flight that is delayed for less than 5 minutes is basically “on time”. You consider any flight delayed for 5 minutes of more to be “delayed”.

In order to determine which airport has the best on time departure rate, we need to

• first classify each flight as “on time” or “delayed”,
• then group flights by origin airport,
• then calculate on time departure rates for each origin airport,
• and finally arrange the airports in descending order for on time departure percentage.

Let’s start with classifying each flight as “on time” or “delayed” by creating a new variable with the mutate function.

nycflights <-nycflights %>%mutate(dep_type = ifelse(dep_delay <5, "on time", "delayed"))

The first argument in the mutate function is the name of the new variable we want to create, in this case dep_type. Then if dep_delay < 5 we classify the flight as "on time" and "delayed" if not, i.e. if the flight is delayed for 5 or more minutes.

Note that we are also overwriting the nycflights data frame with the new version of this data frame that includes the new dep_type variable.

We can handle all the remaining steps in one code chunk:

nycflights %>%group_by(origin) %>%summarise(ot_dep_rate = sum(dep_type == "on time") /n()) %>%arrange(desc(ot_dep_rate))

## Source: local data frame [3 x 2]
##
## origin ot_dep_rate
## (chr) (dbl)
## 1 LGA 0.7279229
## 2 JFK 0.6935854
## 3 EWR 0.6369892

7. If you were selecting an airport simply based on on time departure percentage, which NYC airport would you choose to fly out of?
   1. EWR
   2. JFK
   3. LGA - CORRECT ANSWER

nycflights %>%group_by(origin) %>%summarise(ot_dep_rate = sum(dep_type == "on time") /n()) %>%arrange(desc(ot_dep_rate))

## Source: local data frame [3 x 2]
##
## origin ot_dep_rate
## (chr) (dbl)
## 1 LGA 0.7279229
## 2 JFK 0.6935854
## 3 EWR 0.6369892

We can also visualize the distribution of on on time departure rate across the three airports using a segmented bar plot.

ggplot(data = nycflights, aes(x = origin, fill = dep_type)) +geom_bar()

8. Mutate the data frame so that it includes a new variable that contains the average speed, avg_speed traveled by the plane for each flight (in mph). What is the tail number of the plane with the fastest avg_speed? Hint: Average speed can be calculated as distance divided by number of hours of travel, and note that air_time is given in minutes. If you just want to show the avg_speed and tailnum and none of the other variables, use the select function at the end of your pipe to select just these two variables with select(avg_speed, tailnum). You can Google this tail number to find out more about the aircraft.
   1. N666DN - CORRECT ANSWER
   2. N755US
   3. N779JB
   4. N947UW
   5. N959UW

nycflights <-nycflights %>%mutate(avg_speed = distance /(air_time/60))
nycflights %>%select(avg_speed, tailnum) %>%arrange(desc(avg_speed))

## Source: local data frame [32,735 x 2]
##
## avg_speed tailnum
## (dbl) (chr)
## 1 703.3846 N666DN
## 2 557.4419 N779JB
## 3 554.2197 N571JB
## 4 547.8857 N568JB
## 5 547.8857 N5EHAA
## 6 547.8857 N656JB
## 7 544.7727 N789JB
## 8 538.6517 N516JB
## 9 535.6425 N648JB
## 10 535.6425 N510JB
## .. ... ...

9. Make a scatterplot of avg_speed vs. distance. Which of the following is true about the relationship between average speed and distance.
   1. As distance increases the average speed of flights decreases.
   2. The relationship is linear.
   3. There is an overall postive association between distance and average speed. - CORRECT ANSWER
   4. There are no outliers.
   5. The distribution of distances are uniform over 0 to 5000 miles.

ggplot(nycflights, aes(x = distance, y = avg_speed)) +geom_point()

10. Suppose you define a flight to be “on time” if it gets to the destination on time or earlier than expected, regardless of any departure delays. Mutate the data frame to create a new variable called arr_type with levels "on time" and "delayed" based on this definition. Also mutate to create a new variable called del_type with levels "on time" and "delayed" depending on whether there was ny departure delay. Then, determine the on time arrival percentage based on whether the flight departed on time or not. What percent of flights that were "delayed" departing arrive "on time"? CORRECT NUMERIC INPUT: 0.27

nycflights %>%mutate(arr_type = ifelse(arr_delay <=0, "on time", "delayed")) %>%mutate(dep_type = ifelse(dep_delay <=0, "on time", "delayed")) %>%select(arr_type, dep_type) %>%table()

## dep_type
## arr_type delayed on time
## delayed 9291 4171
## on time 3508 15765

3508 /(3508 +9291)

## [1] 0.2740839