Chapter 3.1.1 Exercises

In [22]:

i39 <- sapply(3:9, seq) # list of vectors
sapply(i39, fivenum)

1.0	1.0	1	1.0	1.0	1.0	1
1.5	1.5	2	2.0	2.5	2.5	3
2.0	2.5	3	3.5	4.0	4.5	5
2.5	3.5	4	5.0	5.5	6.5	7
3.0	4.0	5	6.0	7.0	8.0	9

In [7]:

vapply(i39, fivenum,
c(Min. = 0, "1st Qu." = 0, Median = 0, "3rd Qu." = 0, Max. = 0))

Min.	1.0	1.0	1	1.0	1.0	1.0	1
1st Qu.	1.5	1.5	2	2.0	2.5	2.5	3
Median	2.0	2.5	3	3.5	4.0	4.5	5
3rd Qu.	2.5	3.5	4	5.0	5.5	6.5	7
Max.	3.0	4.0	5	6.0	7.0	8.0	9

In [63]:

# sapply(*, "array") -- artificial example

(v <- structure(10*(5:8), names = LETTERS[1:4]))

f2 <- function(x, y) outer(rep(x, length.out = 3), y)
(a2 <- sapply(v, f2, y = 2*(1:5), simplify = "array"))

# wtf?

A

50

B

60

C

70

D

80

A	B	C	D
100	120	140	160
100	120	140	160
100	120	140	160
200	240	280	320
200	240	280	320
200	240	280	320
300	360	420	480
300	360	420	480
300	360	420	480
400	480	560	640
400	480	560	640
400	480	560	640
500	600	700	800
500	600	700	800
500	600	700	800

In [14]:

a.2 <- vapply(v, f2, outer(1:3, 1:5), y = 2*(1:5))

stopifnot(dim(a2) == c(3,5,4), all.equal(a2, a.2),
identical(dimnames(a2), list(NULL,NULL,LETTERS[1:4])))

hist(replicate(100, mean(rexp(10))))

In [19]:

# However, we can do what we want to do by using mapply().

mapply(sumsq, 1:10, 1:10, MoreArgs = list(x = x))

1. 223.829798225267
2. 134.436004138874
3. 116.850914987302
4. 110.348278325997
5. 107.179466861847
6. 105.372246941011
7. 104.230920575646
8. 103.456708227138
9. 102.903006726754
10. 102.490577631973

In [5]:

mapply(function(x, y) seq_len(x) + y,
c(a = 1, b = 2, c = 3), # names from first
c(A = 10, B = 0, C = -10))

$a

11

$b

1. 1
2. 2

$c

1. -9
2. -8
3. -7

In [6]:

word <- function(C, k) paste(rep.int(C, k), collapse = "")
utils::str(mapply(word, LETTERS[1:6], 6:1, SIMPLIFY = FALSE))

List of 6
 $ A: chr "AAAAAA"
 $ B: chr "BBBBB"
 $ C: chr "CCCC"
 $ D: chr "DDD"
 $ E: chr "EE"
 $ F: chr "F"

In [14]:

# Unfortunately, there are NAs in the data so we cannot simply take the means of those variables. However, we can tell the colMeans function to remove the NAs before computing the mean.

sapply(s, function(x) {
         colMeans(x[, c("Ozone", "Solar.R", "Wind")], 
                  na.rm = TRUE)
 })

	5	6	7	8	9
Ozone	23.61538	29.44444	59.115385	59.961538	31.44828
Solar.R	181.29630	190.16667	216.483871	171.857143	167.43333
Wind	11.62258	10.26667	8.941935	8.793548	10.18000

In [16]:

# With multiple factors and many levels, creating an interaction can result in many levels that are empty.

str(split(x, list(f1, f2)))

List of 10
 $ 1.1: num [1:2] -1.342 -0.283
 $ 2.1: num(0) 
 $ 1.2: num [1:2] 1.757 0.654
 $ 2.2: num(0) 
 $ 1.3: num -0.34
 $ 2.3: num -1.51
 $ 1.4: num(0) 
 $ 2.4: num [1:2] 0.316 0.733
 $ 1.5: num(0) 
 $ 2.5: num [1:2] 2.089 -0.111

In [17]:

# We can also take the group means without simplifying the result, which will give us a list. For functions that return a single value, usually, this is not what we want, but it can be done.

tapply(x, f, mean, simplify = FALSE)

$`1`

0.199351661827255

$`2`

0.486338333622552

$`3`

1.04280692788798