Lab 5: Dplyr, Pipes, and More (2024)

Now we’ll load the tidyverse suite of packages. (You should already have tidyverse installed from the last lab; but if for some reason you still need to install again, then you can just look back at the last lab’s instructions.) This gives us access to the pipe operator %>% as well as the dplyr and tidyr packages needed to complete this lab.

library(tidyverse)

For each of the following code blocks, which are written with pipes, write equivalent code in base R (to do the same thing).

• 1a.

letters %>% toupper %>% paste(collapse="+") 

## [1] "A+B+C+D+E+F+G+H+I+J+K+L+M+N+O+P+Q+R+S+T+U+V+W+X+Y+Z"

# YOUR CODE GOES HERE

• 1b.

" Ceci n'est pas une pipe " %>% gsub("une", "un", .) %>% trimws

## [1] "Ceci n'est pas un pipe"

# YOUR CODE GOES HERE

• 1c.

rnorm(1000) %>% hist(breaks=30, main="N(0,1) draws", col="pink", prob=TRUE) 

# YOUR CODE GOES HERE

• 1d.

rnorm(1000) %>% hist(breaks=30, plot=FALSE) %>% `[[`("density") %>% max

## [1] 0.405

# YOUR CODE GOES HERE

For each of the following code blocks, which are written in base R, write equivalent code with pipes (to do the same thing).

• 2a. Hint: you’ll have to use the dot ., as seen above in Q1b, or in the lecture notes.

paste("Your grade is", sample(c("A","B","C","D","R"), size=1))

## [1] "Your grade is D"

# YOUR CODE GOES HERE

• 2b. Hint: you can use the dot . again, in order to index state.name directly in the last pipe command.

state.name[which.max(state.x77[,"Illiteracy"])] 

## [1] "Louisiana"

# YOUR CODE GOES HERE

• 2c. Note: str.url is defined for use in this and the next question; you can simply refer to it in your solution code (it is not part of the code you have to convert to pipes).

str.url = "http://www.stat.cmu.edu/~ryantibs/statcomp/data/king.txt"lines = readLines(str.url)text = paste(lines, collapse=" ")words = strsplit(text, split="[[:space:]]|[[:punct:]]")[[1]]wordtab = table(words)wordtab = sort(wordtab, decreasing=TRUE)head(wordtab, 10)

## words## of the to and a be will that is ## 203 98 98 58 40 37 32 25 24 23

# YOUR CODE GOES HERE

• 2d. Hint: the only difference between this and the last part is the line words = words[words != ""]. This is a bit tricky line to do with pipes: use the dot ., once more, and manipulate it as if were a variable name.

lines = readLines(str.url)text = paste(lines, collapse=" ")words = strsplit(text, split="[[:space:]]|[[:punct:]]")[[1]]words = words[words != ""]wordtab = table(words)wordtab = sort(wordtab, decreasing=TRUE)head(wordtab, 10)

## words## of the to and a be will that is we ## 98 98 58 40 37 32 25 24 23 21

# YOUR CODE GOES HERE

Below we read in the prostate cancer data set, as visited in previous labs.

pros.df = read.table("http://www.stat.cmu.edu/~ryantibs/statcomp/data/pros.dat")

In the following, use pipes and dplyr verbs to answer questions on pros.df.

• 3a. Among the men whose lcp value is equal to the minimum value (across the entire data set), report the range (min and max) of lpsa.

# YOUR CODE GOES HERE

• 3b. Order the rows by decreasing age, then display the rows from men who are older than 70 and without SVI.

# YOUR CODE GOES HERE

• 3c. Order the rows by decreasing age, then decreasing lpsa score, and display the rows from men who are older than 70 and without SVI, but only the age, lpsa, lcavol, and lweight columns. Hint: arrange() can take two arguments, and the order you pass in them specifies the priority.

# YOUR CODE GOES HERE

• 3d. We’re going to resolve Q2c from Lab 3 using the tidyverse. Using purrr and dplyr, perform t-tests for each variable in the data set, between SVI and non-SVI groups. To be precise, you will perform a t-test for each column excluding the SVI variable itself, by running the function t.test.by.ind() below (which is just as in Q2c in Lab 3). Print the returned t-test objects out to the console.

t.test.by.ind = function(x, ind) { stopifnot(all(ind %in% c(0, 1))) return(t.test(x[ind == 0], x[ind == 1]))} 

# YOUR CODE GOES HERE

• 3e. Extend your code from the last part (append just one more line of code, glued together by a pipe) to extract the p-values from each of the returned t-test objects, and print them out to the console.

# YOUR CODE GOES HERE

Below, we read in two data sets of the 1000 fastest times ever recorded for the 100m sprint, in men’s and women’s track., as seen in the last lab.

sprint.m.df = read.table( file="http://www.stat.cmu.edu/~ryantibs/statcomp/data/sprint.m.txt", sep="\t", quote="", header=TRUE)sprint.w.df = read.table( file="http://www.stat.cmu.edu/~ryantibs/statcomp/data/sprint.w.txt", sep="\t", quote="", header=TRUE)

In the following, use pipes and dplyr verbs to answer questions on sprint.w.df.

• 4a. Order the rows by increasing Wind value, and then display only the women who ran at most 10.7 seconds.

# YOUR CODE GOES HERE

• 4b. Order the rows by terms of increasing Time, then increasing Wind, and again display only the women who ran at most 10.7 seconds, but only the Time, Wind, Name, and Date columns.

# YOUR CODE GOES HERE

• 4c. Plot the Time versus Wind columns, but only using data where Wind values that are nonpositive. Hint: note that for a data frame, df with columns colX and colY, you can use plot(colY ~ colX, data=df), to plot df$colY (y-axis) versus df$colX (x-axis).

# YOUR CODE GOES HERE

• 4d. Extend your code from the last part (append just two more lines of code, glued together by a pipe) to plot the single fastest Time per Wind value. (That is, your plot should be as in the last part, but among points that share the same x value, only the point with the lowest y value should be drawn.)

# YOUR CODE GOES HERE

In the following, use pipes and dplyr and tidyr verbs to answer questions on sprint.m.df. In some parts, it might make more sense to use direct indexing, and that’s perfectly fine.

• 5a. Confirm that the Time column is stored as character data type. Why do you think this is? Convert the Time column to numeric. Hint: after converting to numeric, there will be NA values; look at the position of one such NA value and revisit the original Time column to see why it was stored as character type in the first place.

# YOUR CODE GOES HERE

• 5b. Define a reduced data frame dat.reduced as follows. For each athlete, and each city, keep the fastest of all times they recorded in this city. Then drop all rows with an NA value in the Time column Your new data frame dat.reduced should have 600 rows and 3 columns (Name, City, Time). Confirm that it has these dimensions, and display its first 10 rows. Hint: drop_na() in the tidyr package allows you to drop rows based on NA values.

# YOUR CODE GOES HERE

• 5c. The data frame dat.reduced is said to be in “long” format: it has observations on the rows, and variables (Name, City, Time) on the columns. Arrange the rows alphebtically by city; convert this data frame into “wide” format; and then order the rows so that they are alphabetical by sprinter name. Call the result dat.wide. To be clear, here the first column should be the athlete names, and the remaining columns should correspond to the cities. Confirm that your data frame has dimension 141 x 152 Do these dimensions make sense to you?

# YOUR CODE GOES HERE

• 5d. Not counting the names in the first column, how many non-NA values does dat.wide have? How could you have guessed this number ahead of time, directly from dat.reduced (before any pivoting at all)?

# YOUR CODE GOES HERE

• 5e. From dat.wide, look at the row for “Usain Bolt”, and determine the city names that do not have NA values. These should be the cities in which he raced. Determine these cities directly from dat.reduced, and confirm that they match.

# YOUR CODE GOES HERE

• 5f. Convert dat.wide back into “long” format, and call the result dat.long. Remove rows that have NA values (hint: you can do this by setting values_drop_na = TRUE in the call to the pivoting function), and order the rows alphabetically by athlete and city name. Once you’ve done this, dat.long should have matching entries to dat.reduced; confirm that this is the case.

# YOUR CODE GOES HERE