Create a 3 x 3 x 2 array using the dim attribute with the vector below.
x <- c(5, 1, 5, 5, 1, 1, 5, 3, 2, 3, 2, 6, 4, 4, 1, 2, 1, 3)Try to create the same array using function array(). What do you notice about how the array object is populated?
x <- c(5, 1, 5, 5, 1, 1, 5, 3, 2, 3, 2, 6, 4, 4, 1, 2, 1, 3)
# set dim attribute
attr(x = x, which = "dim") <- c(3, 3, 2)
x#> , , 1
#>
#> [,1] [,2] [,3]
#> [1,] 5 5 5
#> [2,] 1 1 3
#> [3,] 5 1 2
#>
#> , , 2
#>
#> [,1] [,2] [,3]
#> [1,] 3 4 2
#> [2,] 2 4 1
#> [3,] 6 1 3attributes(x)#> $dim
#> [1] 3 3 2array(x, dim = c(3, 3, 2))#> , , 1
#>
#> [,1] [,2] [,3]
#> [1,] 5 5 5
#> [2,] 1 1 3
#> [3,] 5 1 2
#>
#> , , 2
#>
#> [,1] [,2] [,3]
#> [1,] 3 4 2
#> [2,] 2 4 1
#> [3,] 6 1 3The array is populated using column-major ordering.
| vector index | row | column | matrix |
|---|---|---|---|
| 1 | 1 | 1 | 1 |
| 2 | 2 | 1 | 1 |
| 3 | 3 | 1 | 1 |
| 4 | 1 | 2 | 1 |
| 5 | 2 | 2 | 1 |
| 6 | 3 | 2 | 1 |
| 7 | 1 | 3 | 1 |
| 8 | 2 | 3 | 1 |
| 9 | 3 | 3 | 1 |
| 10 | 1 | 1 | 2 |
| 11 | 2 | 1 | 2 |
| 12 | 3 | 1 | 2 |
| 13 | 1 | 2 | 2 |
| 14 | 2 | 2 | 2 |
| 15 | 3 | 2 | 2 |
| 16 | 1 | 3 | 2 |
| 17 | 2 | 3 | 2 |
| 18 | 3 | 3 | 2 |
Create a factor vector based on the vector of airport codes below. Try to do it without using function factor().
airports <- c("RDU", "ABE", "DTW", "GRR", "RDU", "GRR", "GNV",
"JFK", "JFK", "SFO", "DTW")Assume all the possible levels are
c("RDU", "ABE", "DTW", "GRR", "GNV", "JFK", "SFO")Hint: Think about what type of object factors are built on.
What if the possible levels are
c("RDU", "ABE", "DTW", "GRR", "GNV", "JFK", "SFO", "GSO", "ORD", "PHL")Factors are built on atomic integer vectors. First we’ll create an integer vector that serves as a mapping to the unique airport codes.
z <- as.integer(c(1,2,3,4,1,4,5,6,6,7,3))
# set levels
attr(x = z, which = "levels") <- c("RDU", "ABE", "DTW",
"GRR", "GNV", "JFK", "SFO")
# set class to be factor
attr(x = z, which = "class") <- "factor"
z#> [1] RDU ABE DTW GRR RDU GRR GNV JFK JFK SFO DTW
#> Levels: RDU ABE DTW GRR GNV JFK SFOattributes(z)#> $levels
#> [1] "RDU" "ABE" "DTW" "GRR" "GNV" "JFK" "SFO"
#>
#> $class
#> [1] "factor"If the levels change, then we need to adjust our integer vector to have 9 integer values.
Consider the vectors x and y below.
x <- letters[1:5]
y <- list(i = 1:5, j = -3:3, k = rep(0, 4))What is difference between subsetting with [ and [[ using integers? Try various indices.
Let’s look at atomic vectors first.
x[[4]]#> [1] "d"x[4]#> [1] "d"x[[1:3]]#> Error in x[[1:3]]: attempt to select more than one element in vectorIndexx[1:3]#> [1] "a" "b" "c"Subsetting with a vector of length one appears to be the same using [ versus [[. We cannot use [[ and subset with a vector of length greater than one.
Doing the same for y yields
y[[2]]#> [1] -3 -2 -1 0 1 2 3y[2]#> $j
#> [1] -3 -2 -1 0 1 2 3y[[1:2]]#> [1] 2y[1:2]#> $i
#> [1] 1 2 3 4 5
#>
#> $j
#> [1] -3 -2 -1 0 1 2 3y[[1:3]]#> Error in y[[1:3]]: recursive indexing failed at level 2y[1:3]#> $i
#> [1] 1 2 3 4 5
#>
#> $j
#> [1] -3 -2 -1 0 1 2 3
#>
#> $k
#> [1] 0 0 0 0We see similar results for the generic vector y. We still should not use [[ when subsetting with a vector of length greater than one. The result of y[[1:2]] seems to work, but it is not doing what you think. That code is equivalent to y[[1]][2].
Consider the atomic vector x.
set.seed(73961)
x <- sample(1:100, size = 100, replace = TRUE)
x#> [1] 13 99 21 78 5 2 11 99 46 97 57 53 97 12 41 50 59
#> [18] 58 1 63 43 89 31 23 51 35 22 34 82 29 92 44 67 62
#> [35] 59 48 49 20 9 36 77 94 86 54 11 79 49 37 29 52 30
#> [52] 4 23 65 20 15 28 4 43 16 49 69 23 36 25 93 2 21
#> [69] 68 6 37 94 99 40 11 26 93 56 78 9 61 60 91 20 74
#> [86] 55 97 98 45 98 74 46 94 38 26 78 49 25 21 100Use subsetting to
select every third value from x beginning at position 6;
remove all values with an odd index;
remove all numbers divisible by 3 or 5 and replace them with 0.
Recall that x is
#> [1] 13 99 21 78 5 2 11 99 46 97 57 53 97 12 41 50 59
#> [18] 58 1 63 43 89 31 23 51 35 22 34 82 29 92 44 67 62
#> [35] 59 48 49 20 9 36 77 94 86 54 11 79 49 37 29 52 30
#> [52] 4 23 65 20 15 28 4 43 16 49 69 23 36 25 93 2 21
#> [69] 68 6 37 94 99 40 11 26 93 56 78 9 61 60 91 20 74
#> [86] 55 97 98 45 98 74 46 94 38 26 78 49 25 21 100x[seq(6, length(x), by = 3)]#> [1] 2 46 53 41 58 43 23 22 29 67 48 9 94 11 37 30 65 28 16 23 93 68 94
#> [24] 11 56 61 20 97 98 94 78 21x[-seq(1, length(x), 2)]#> [1] 99 78 2 99 97 53 12 50 58 63 89 23 35 34 29 44 62
#> [18] 48 20 36 94 54 79 37 52 4 65 15 4 16 69 36 93 21
#> [35] 6 94 40 26 56 9 60 20 55 98 98 46 38 78 25 100x[!(x %% 3) | !(x %% 5)] <- 0
x#> [1] 13 0 0 0 0 2 11 0 46 97 0 53 97 0 41 0 59 58 1 0 43 89 31
#> [24] 23 0 0 22 34 82 29 92 44 67 62 59 0 49 0 0 0 77 94 86 0 11 79
#> [47] 49 37 29 52 0 4 23 0 0 0 28 4 43 16 49 0 23 0 0 0 2 0 68
#> [70] 0 37 94 0 0 11 26 0 56 0 0 61 0 91 0 74 0 97 98 0 98 74 46
#> [93] 94 38 26 0 49 0 0 0