Lab 1: Looking Under the Hood of R

A1. Predict

typeof(c(1, TRUE))

## [1] "double"

typeof(c(1L, TRUE))

## [1] "integer"

typeof(c(1, "2", TRUE))

## [1] "character"

Why are the three answers not the same? What did TRUE become in each case?

A2. Predict

TRUE + TRUE + TRUE

## [1] 3

c(TRUE, FALSE, TRUE) * 2

## [1] 2 0 2

If TRUE is a logical value, why is arithmetic legal?

A3. Predict

1 == 1L

## [1] TRUE

typeof(1)

## [1] "double"

typeof(1L)

## [1] "integer"

is.numeric(1L)

## [1] TRUE

1 and 1L compare as equal. Are they the same type? What does is.numeric() say about an integer?

A4. Predict

f <- factor(c("10", "2", "2"))
as.numeric(f)

## [1] 1 2 2

as.numeric(as.character(f))

## [1] 10  2  2

Note 1.4 treats factors as categorical data with levels. What is as.numeric() looking at when the object is a factor?

B1. Predict

x <- c(NA, NaN, Inf, NULL)
x

## [1]  NA NaN Inf

length(x)

## [1] 3

typeof(x)

## [1] "double"

You passed four things to c(). Why does x not have length 4?

B2. Predict

NA == NA

## [1] NA

NaN == NaN

## [1] NA

NULL == NULL

## logical(0)

Then:

is.na(NA)

## [1] TRUE

is.nan(NaN)

## [1] TRUE

is.null(NULL)

## [1] TRUE

Why is == a poor way to test for these values? Which is.*() function matches which kind of nothing?

B3. Predict

is.na(NaN)

## [1] TRUE

is.nan(NA)

## [1] FALSE

typeof(NA)

## [1] "logical"

typeof(c(NA, 1))

## [1] "double"

typeof(c(NA, "a"))

## [1] "character"

NA looks like one object. Use the coercion hierarchy from note 1.4. What happened to NA when it entered those two vectors?

C1. Predict

x <- 10:13
x[c(1, 1, 1)]

## [1] 10 10 10

x[-c(1, 1)]

## [1] 11 12 13

Positive indices select. Negative indices drop. Why does repeating 1 mean two different things in these two lines?

C2. Predict

x <- c(10, 20, 30, 40)
x[c(TRUE, FALSE)]

## [1] 10 30

x[c(TRUE, FALSE, TRUE)]

## [1] 10 30 40

Note 1.4 says R recycles the shorter vector in arithmetic. What rule is R using here, and where does the extra TRUE go?

C3. Predict

record <- list(
  info = list(name = "Ada", year = 1L),
  scores = c(90, 85)
)
record[1]

## $info
## $info$name
## [1] "Ada"
## 
## $info$year
## [1] 1

record[[1]]

## $name
## [1] "Ada"
## 
## $year
## [1] 1

record$info$name

## [1] "Ada"

record[["info"]][1]

## $name
## [1] "Ada"

record[["info"]][[1]]

## [1] "Ada"

Note 1.4 distinguishes [, [[, and $. info is a list inside a list. What does each line return, and which results are still lists?

C4. Predict

students <- data.frame(name = c("Ada", "Bob"), grade = c(91, 74))
typeof(students)

## [1] "list"

is.list(students)

## [1] TRUE

is.data.frame(students)

## [1] TRUE

class(students[, "grade"])

## [1] "numeric"

class(students[, "grade", drop = FALSE])

## [1] "data.frame"

class(students["grade"])

## [1] "data.frame"

class(students[["grade"]])

## [1] "numeric"

typeof() and is.list() are from note 1.4. Why can a data frame be a list and a data frame at the same time? Which extractions are still data frames, and what is drop for?

C5. Predict

m <- matrix(1:6, nrow = 2)
m

##      [,1] [,2] [,3]
## [1,]    1    3    5
## [2,]    2    4    6

m[2, 1]

## [1] 2

Fill in the matrix on paper before you run this, using matrix(1:6, nrow = 2) from note 1.4. In what order does R fill the matrix, and why is m[2, 1] equal to 2 rather than 4?

D1. Explain, then try only the quoted form for installation

Why does this fail (do not keep running it):

install.packages(ggplot2)

while this is the intended form:

install.packages("ggplot2")

and yet both of these can succeed:

library(ggplot2)
library("ggplot2")

What is library() doing with an unquoted name that install.packages() will not do?

D2. Predict, in a session where you have not yet loaded dplyr

students <- data.frame(name = c("Ada", "Bob"), grade = c(91, 74))
filter

## function (.data, ..., .by = NULL, .preserve = FALSE) 
## {
##     check_by_typo(...)
##     check_not_both_by_and_preserve({
##         {
##             .by
##         }
##     }, .preserve)
##     UseMethod("filter")
## }
## <bytecode: 0x000001950d577fc8>
## <environment: namespace:dplyr>

filter(students, grade > 80)

##   name grade
## 1  Ada    91

Then load dplyr and run the same two lines again.

library(dplyr)
filter

## function (.data, ..., .by = NULL, .preserve = FALSE) 
## {
##     check_by_typo(...)
##     check_not_both_by_and_preserve({
##         {
##             .by
##         }
##     }, .preserve)
##     UseMethod("filter")
## }
## <bytecode: 0x000001950d577fc8>
## <environment: namespace:dplyr>

filter(students, grade > 80)

##   name grade
## 1  Ada    91

Note 1.2 covers library(), search(), and dplyr::filter(). There is already a filter() in Base R / stats. What changed after library(dplyr)? How can two functions share a name? How do you call the older one anyway?

D3. Predict

x <- c(10, 20, 30)
x |> mean()

## [1] 20

x %>% mean()

## [1] 20

Run this before loading tidyverse. Why does one pipe exist in Base R and the other does not? After library(tidyverse), run both lines again.

D4. Predict

search()

##  [1] ".GlobalEnv"        "package:lubridate" "package:forcats"  
##  [4] "package:stringr"   "package:dplyr"     "package:purrr"    
##  [7] "package:readr"     "package:tidyr"     "package:tibble"   
## [10] "package:ggplot2"   "package:tidyverse" "package:stats"    
## [13] "package:graphics"  "package:grDevices" "package:utils"    
## [16] "package:datasets"  "package:methods"   "Autoloads"        
## [19] "package:base"

library(ggplot2)
search()

##  [1] ".GlobalEnv"        "package:lubridate" "package:forcats"  
##  [4] "package:stringr"   "package:dplyr"     "package:purrr"    
##  [7] "package:readr"     "package:tidyr"     "package:tibble"   
## [10] "package:ggplot2"   "package:tidyverse" "package:stats"    
## [13] "package:graphics"  "package:grDevices" "package:utils"    
## [16] "package:datasets"  "package:methods"   "Autoloads"        
## [19] "package:base"

Where did ggplot2 go on the search path, and why does that position matter when two packages contain the same function name?

E1. Predict

seq(from = 1, to = 3) + c(10, 20, 30, 40, 50, 60)

## [1] 11 22 33 41 52 63

seq(1, 6, 2) * c(TRUE, FALSE)

## [1] 1 0 5

Note 1.4 covers seq(), named arguments, and recycling. Write the recycled vectors out by hand. When does R warn, and when does it stay silent? What did TRUE and FALSE become?

E2. Predict

1 / 0

## [1] Inf

0 / 0

## [1] NaN

TRUE / FALSE

## [1] Inf

Inf - Inf

## [1] NaN

NA + 1

## [1] NA

NULL + 1

## numeric(0)

Which of these are Inf, NaN, NA, and which is an error? What coercion makes TRUE / FALSE legal?

F1. Run and compare

file.exists("students.csv")
file.exists(file.path(".", "students.csv"))

base_students <- read.csv("students.csv")
tidy_students <- readr::read_csv("students.csv")

class(base_students)
class(tidy_students)
str(base_students)
str(tidy_students)

Note 1.3: . is the current directory, and file.path() builds a path. Why are the two existence checks the same? Both read the same file. Why are the objects not the same class? What extra classes does read_csv() attach, and what is a tibble claiming to be besides a data frame?

F2. Predict

base_students$grade * 0.01

## Error:
## ! object 'base_students' not found

base_students[, "grade"]

## Error:
## ! object 'base_students' not found

base_students["grade"]

## Error:
## ! object 'base_students' not found

One of these uses the vectorized arithmetic from note 1.4. Which extraction still has two dimensions?