library(purrr)

Part I. Multiple choice and multiple answer questions

Question 1.

Answer: hw4.Rmd; ./hw4.Rmd; ../hw4.

Question 2.

Answer:

The recycling rule of R is invoked here.
The odd entries of x are multiplied by 1, and the even entries are multiplied by -1.

x <- 1:8
x <- x * c(1, -1)
x

[1]  1 -2  3 -4  5 -6  7 -8

length(x)

[1] 8

Question 3.

Answer:

(pets <- c(cat = 10, dog = 15, lizard = 2))

   cat    dog lizard 
    10     15      2

pets[-1] ### Correct

   dog lizard 
    15      2

pets[2:3] ### Correct

   dog lizard 
    15      2

pets[c("dog", "lizard")] ### Correct

   dog lizard 
    15      2

pets[1:2]

cat dog 
 10  15

Question 4.

Answer:

x <- iris$Species
mean(x)

[1] NA

median(x)

Error in median.default(x): need numeric data

table(x)  ### Correct

x
    setosa versicolor  virginica 
        50         50         50

quantile(x)

Error in quantile.default(x): (unordered) factors are not allowed

Question 5.

Answer:

(x <- c(82, 93, 64, 80, 95))

[1] 82 93 64 80 95

sort(x)[1] ### Correct

[1] 64

sort(x, decreasing=TRUE)[1]

[1] 95

rank(x)[1]

[1] 3

order(x)[1]

[1] 3

Question 6.

Answer:

writeLines("\n") ### Correct

writeLines("A\nB")

A
B

writeLines("")

writeLines("AB")

AB

writeLines(" ")

writeLines("A B")

A B

writeLines("<CR>")

<CR>

writeLines("A<CR>B")

A<CR>B

Question 7.

Answer:

x <- c("Junior", "Junior", "Freshman", "Freshman", "Freshman", "Sophomore", "Sophomore", "Sophomore", "Sophomore", "Junior", "Freshman", "Junior", "Freshman", "Freshman")
(x <- as.integer(as.character(x)))

 [1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA

x <- c("Junior", "Junior", "Freshman", "Freshman", "Freshman", "Sophomore", "Sophomore", "Sophomore", "Sophomore", "Junior", "Freshman", "Junior", "Freshman", "Freshman")
(x <- factor(x))

 [1] Junior    Junior    Freshman  Freshman  Freshman  Sophomore Sophomore
 [8] Sophomore Sophomore Junior    Freshman  Junior    Freshman  Freshman 
Levels: Freshman Junior Sophomore

x <- c("Junior", "Junior", "Freshman", "Freshman", "Freshman", "Sophomore", "Sophomore", "Sophomore", "Sophomore", "Junior", "Freshman", "Junior", "Freshman", "Freshman")
(x <- as.factor(x))

 [1] Junior    Junior    Freshman  Freshman  Freshman  Sophomore Sophomore
 [8] Sophomore Sophomore Junior    Freshman  Junior    Freshman  Freshman 
Levels: Freshman Junior Sophomore

x <- c("Junior", "Junior", "Freshman", "Freshman", "Freshman", "Sophomore", "Sophomore", "Sophomore", "Sophomore", "Junior", "Freshman", "Junior", "Freshman", "Freshman")
(x <- factor(x, c("Freshman", "Sophomore", "Junior", "Senior"))) ### Correct

 [1] Junior    Junior    Freshman  Freshman  Freshman  Sophomore Sophomore
 [8] Sophomore Sophomore Junior    Freshman  Junior    Freshman  Freshman 
Levels: Freshman Sophomore Junior Senior

Question 8.

Answer:

(x <- c(1, 15, 3, 20, 5, 8, 9, 10, 1, 3))

 [1]  1 15  3 20  5  8  9 10  1  3

x >= 13 && x < 10

[1] FALSE

x >= 13 || x < 10

[1] TRUE

x >= 13 | x < 10 ### Correct

 [1]  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE FALSE  TRUE  TRUE

x >= 13 & x < 10

 [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE

Question 9.

Answer:

(x <- c(NA, 123))

[1]  NA 123

x == NA

[1] NA NA

(x <- c(NA, 123))

[1]  NA 123

complete(x)

Error in complete(x): could not find function "complete"

(x <- c(NA, 123))

[1]  NA 123

x = NA

(x <- c(NA, 123))

[1]  NA 123

is.na(x) ### Correct

[1]  TRUE FALSE

Question 10.

Answer:

### Example y and yHat
n <- 10
y <- matrix(rnorm(n))
yHat <- matrix(rnorm(n))
str(y)

 num [1:10, 1] 1.263 -0.326 1.33 1.272 0.415 ...

str(yHat)

 num [1:10, 1] 0.764 -0.799 -1.148 -0.289 -0.299 ...

### Desired result
sum((y - yHat)^2)

[1] 26.04449

(y - yHat) %*% (y - yHat)

Error in (y - yHat) %*% (y - yHat): non-conformable arguments

(y - yHat) * (y - yHat)

            [,1]
 [1,]  0.2493612
 [2,]  0.2235170
 [3,]  6.1377896
 [4,]  2.4395032
 [5,]  0.5095912
 [6,]  1.2733750
 [7,]  1.3942662
 [8,]  0.3566487
 [9,]  0.1948785
[10,] 13.2655612

y %*% yHat

Error in y %*% yHat: non-conformable arguments

t(y - yHat) %*% (y - yHat) ### Correct

         [,1]
[1,] 26.04449

Question 11.

Answer:

l <- list(A = c("Red", "Green", "Black"),
          B = matrix(1:6, 2),
          C = FALSE)
str(l)

List of 3
 $ A: chr [1:3] "Red" "Green" "Black"
 $ B: int [1:2, 1:3] 1 2 3 4 5 6
 $ C: logi FALSE

l[2]

$B
     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6

l$B ### Correct

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

l[names(l) == "B"]

$B
     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6

l["B"]

$B
     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6

Question 12.

Answer:

### Example dat
dat <- data.frame(A = 1:10, B = rep("A", 10), C = rep(TRUE, 10))
str(dat)

'data.frame':   10 obs. of  3 variables:
 $ A: int  1 2 3 4 5 6 7 8 9 10
 $ B: chr  "A" "A" "A" "A" ...
 $ C: logi  TRUE TRUE TRUE TRUE TRUE TRUE ...

map(dat, class) ### Correct

$A
[1] "integer"

$B
[1] "character"

$C
[1] "logical"

dim(dat)

[1] 10  3

class(dat)

[1] "data.frame"

class(dat[1:ncol(dat)])

[1] "data.frame"

Question 13.

Answer:

### correct
test1 <- function(gpa) {
  if (gpa >= 3 && gpa <= 3.5) {
    admit <- "yes"
  } else {
    admit <- "no"
  }
  admit
}

test2 <- function(gpa) {
  if (gpa >= 3 || gpa <= 3.5) {
    admit <- "yes"
  } else {
    admit <- "no"
  }
  admit
}
test3 <- function(gpa) {
  if (gpa >= 3 && gpa <= 3.5) {
    admit <- "yes"
  }
  admit <- "no"
  admit
}

test4 <- function(gpa) {
  if (gpa >= 3) {
    admit <- "yes"
  } else if (gpa <= 3.5) {
    admit <- "no"
  }
  admit
}


### Sample gpa
(gpa <- seq(1, 4, by = 0.2))

 [1] 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0

### Desired output
(admit0 <- ifelse((gpa >= 3) & (gpa <= 3.5), "yes", "no"))

 [1] "no"  "no"  "no"  "no"  "no"  "no"  "no"  "no"  "no"  "no"  "yes" "yes"
[13] "yes" "no"  "no"  "no"

identical(map_chr(gpa, test1), admit0)

[1] TRUE

identical(map_chr(gpa, test2), admit0)

[1] FALSE

identical(map_chr(gpa, test3), admit0)

[1] FALSE

identical(map_chr(gpa, test4), admit0)

[1] FALSE

Question 14.

Answer:

A while loop is used when the number of iterations is undetermined before starting the loop, e.g., to calculate the smallest n required for \(\sum_{i = 1}^{n} i^2\) to exceed 100000
while (TRUE) is equivalent to repeat
A for loop is used when the number of iterations is determined before starting the loop, e.g., to calculate a summary statistic for each file in a given list
When using a repeat loop, you must use a break statement within the loop because otherwise there will be an infinite loop

Question 15.

Answer: - When performing a repetitive task, it is a good practice to write a function for performing the task once, and then call the function repeatedly - The length of a function should be manageable. If a function is too long, it should be broken up into smaller functions

Question 16.

Answer:

l <- split(iris[, c("Sepal.Length", "Sepal.Width")], iris$Species)
str(l)

List of 3
 $ setosa    :'data.frame': 50 obs. of  2 variables:
  ..$ Sepal.Length: num [1:50] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
  ..$ Sepal.Width : num [1:50] 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
 $ versicolor:'data.frame': 50 obs. of  2 variables:
  ..$ Sepal.Length: num [1:50] 7 6.4 6.9 5.5 6.5 5.7 6.3 4.9 6.6 5.2 ...
  ..$ Sepal.Width : num [1:50] 3.2 3.2 3.1 2.3 2.8 2.8 3.3 2.4 2.9 2.7 ...
 $ virginica :'data.frame': 50 obs. of  2 variables:
  ..$ Sepal.Length: num [1:50] 6.3 5.8 7.1 6.3 6.5 7.6 4.9 7.3 6.7 7.2 ...
  ..$ Sepal.Width : num [1:50] 3.3 2.7 3 2.9 3 3 2.5 2.9 2.5 3.6 ...

pmap(list(l, 1:2), function(x, y) median(x[[y]]))

Error in `stop_bad_length()`:
! Element 2 of `.l` must have length 1 or 3, not 2

imap(l, function(x, y) median(x[[y]]))

$setosa
NULL

$versicolor
NULL

$virginica
NULL

map2(l, 1:2, function(x, y) median(x[[y]]))

Error: Mapped vectors must have consistent lengths:
* `.x` has length 3
* `.y` has length 2

map(l, function(x) map_dbl(x, median)) ### Correct

$setosa
Sepal.Length  Sepal.Width 
         5.0          3.4 

$versicolor
Sepal.Length  Sepal.Width 
         5.9          2.8 

$virginica
Sepal.Length  Sepal.Width 
         6.5          3.0

Question 17.

Answer:

l <- list(matrix(2:5, 2, 2),
          matrix(c(1, 0, 0, 2), 2, 2),
          matrix(c(-2, 0, 0, 1), 2, 2))
str(l)

List of 3
 $ : int [1:2, 1:2] 2 3 4 5
 $ : num [1:2, 1:2] 1 0 0 2
 $ : num [1:2, 1:2] -2 0 0 1

### Desired output
reduce(l, `+`)

     [,1] [,2]
[1,]    1    4
[2,]    3    8

sum(l[[1]], l[[2]], l[[3]])

[1] 16

l[[1]] + l[[2]] + l[[3]] ### Correct

     [,1] [,2]
[1,]    1    4
[2,]    3    8

list(l[[1]], l[[1]] + l[[2]], l[[1]] + l[[2]] + l[[3]])

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

[[2]]
     [,1] [,2]
[1,]    3    4
[2,]    3    7

[[3]]
     [,1] [,2]
[1,]    1    4
[2,]    3    8

list(l[[1]], l[[2]], l[[3]])

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

[[2]]
     [,1] [,2]
[1,]    1    0
[2,]    0    2

[[3]]
     [,1] [,2]
[1,]   -2    0
[2,]    0    1

Question 18.

Answer:

The output is 10, because f() searches for the variable x in the global environment, and when f() searches for x, the value is 10

x <- 1
f <- function() {
  x
}
x <- 10
f()

[1] 10

Question 19.

Only tripple(1) will be evaluated, and will be evaluated only once.

tripple <- function(x) x * 3
f <- function(x) {
  y <- 2
  c(x, x)
}
y <- 1
f(tripple(y))

[1] 3 3

Part II. Coding questions

Question 20.

dat <- iris[seq(2, nrow(iris), by = 2), ]
head(dat)

Question 21.

### Example x
x <- c("unknown", "female", "unknown", "male", "male", "female")

for (i in 1:length(x)){
  if (x[i] == "unknown"){
    x[i] <- NA
  }
}
x

[1] NA       "female" NA       "male"   "male"   "female"

Question 22.

### Example z
z <- c("cat", "cat", "dog", "cat", "dog", "dog")

### Method 1
w <- vector("numeric", length(z))
w <- ifelse(z == "cat", 1, 2)
w

[1] 1 1 2 1 2 2

### Method 2
map_dbl(z, function(x){
  if (x == "cat"){
    1
  } else {
    2
  }
})

[1] 1 1 2 1 2 2

### Method 3
w <- vector("numeric", length(z))
for (i in 1:length(z)){
  if (z[i] == "cat"){
    w[i] <- 1
  } else {
    w[i] <- 2
  }
}
w

[1] 1 1 2 1 2 2

Question 23.

SampleVar <- function(x){
  mean((x - mean(x))^2)
}
x <- 1:10
n <- length(x)
testthat::expect_equal(SampleVar(x), (n - 1) / n * var(x))

Midterm Exam

STAT 486/586 – Introduction to Statistical Computing

Yuhang (Tom) LIN

Due: March 24, 2022

Part I. Multiple choice and multiple answer questions

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.

Question 16.

Question 17.

Question 18.

Question 19.

Part II. Coding questions

Question 20.

Question 21.

Question 22.

Question 23.