I would prefer tidyverse solution!
The question is related to this post.
Example data
structure(list(A = c(79L, 42L, 74L, 49L, 82L, 22L, 88L, 13L,
54L, 68L), B = c(41L, 22L, 1L, 40L, 96L, 48L, 40L, 56L, 19L,
84L), C = c(20L, 10L, 1L, 27L, 34L, 27L, 35L, 3L, 78L, 36L),
D = c(40L, 92L, 76L, 81L, 73L, 30L, 10L, 57L, 19L, 18L),
G = c(50L, 74L, 37L, 60L, 23L, 42L, 22L, 94L, 28L, 68L),
H = c(54L, 62L, 92L, 61L, 91L, 76L, 51L, 60L, 89L, 36L),
J = c(64L, 59L, 1L, 99L, 36L, 26L, 15L, 16L, 83L, 39L), K = c(29L,
30L, 80L, 33L, 44L, 28L, 9L, 53L, 11L, 68L), L = c(42L, 29L,
10L, 75L, 24L, 68L, 56L, 77L, 23L, 92L), M = c(56L, 27L,
61L, 40L, 76L, 50L, 31L, 15L, 72L, 40L), N = c(45L, 33L,
37L, 32L, 5L, 20L, 45L, 38L, 25L, 32L), Z = c(52L, 88L, 74L,
91L, 86L, 43L, 4L, 6L, 61L, 69L), X = c(58L, 92L, 19L, 99L,
9L, 58L, 53L, 49L, 48L, 32L), Y = c(75L, 13L, 63L, 37L, 30L,
98L, 98L, 94L, 38L, 25L), S = c(99L, 64L, 27L, 30L, 100L,
40L, 76L, 2L, 10L, 57L), P = c(16L, 76L, 69L, 64L, 68L, 34L,
96L, 22L, 48L, 1L)), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -10L))
# A tibble: 10 x 16
A B C D G H J K L M N
<int> <int> <int> <int> <int> <int> <int> <int> <int> <int> <int>
1 79 41 20 40 50 54 64 29 42 56 45
2 42 22 10 92 74 62 59 30 29 27 33
3 74 1 1 76 37 92 1 80 10 61 37
4 49 40 27 81 60 61 99 33 75 40 32
5 82 96 34 73 23 91 36 44 24 76 5
6 22 48 27 30 42 76 26 28 68 50 20
7 88 40 35 10 22 51 15 9 56 31 45
8 13 56 3 57 94 60 16 53 77 15 38
9 54 19 78 19 28 89 83 11 23 72 25
10 68 84 36 18 68 36 39 68 92 40 32
# ... with 5 more variables: Z <int>, X <int>, Y <int>, S <int>,
# P <int>
Split up the dataframe in to multiple every 4 columns, into a list, such that one can loop/map export them individually.
Desired output:
[[1]]
# A tibble: 10 x 4
A B C D
<int> <int> <int> <int>
1 79 41 20 40
2 42 22 10 92
3 74 1 1 76
4 49 40 27 81
5 82 96 34 73
6 22 48 27 30
7 88 40 35 10
8 13 56 3 57
9 54 19 78 19
10 68 84 36 18
[[2]]
# A tibble: 10 x 4
G H J K
<int> <int> <int> <int>
1 50 54 64 29
2 74 62 59 30
3 37 92 1 80
4 60 61 99 33
5 23 91 36 44
6 42 76 26 28
7 22 51 15 9
8 94 60 16 53
9 28 89 83 11
10 68 36 39 68
And so on...
The intention is to export each of them into CSV individually.
>Solution :
This is a straightforward one-liner in base R:
lapply(seq(ncol(df)/4) - 1, function(x) df[4 * x + 1:4])
#> [[1]]
#> # A tibble: 10 x 4
#> A B C D
#> <int> <int> <int> <int>
#> 1 79 41 20 40
#> 2 42 22 10 92
#> 3 74 1 1 76
#> 4 49 40 27 81
#> 5 82 96 34 73
#> 6 22 48 27 30
#> 7 88 40 35 10
#> 8 13 56 3 57
#> 9 54 19 78 19
#> 10 68 84 36 18
#>
#> [[2]]
#> # A tibble: 10 x 4
#> G H J K
#> <int> <int> <int> <int>
#> 1 50 54 64 29
#> 2 74 62 59 30
#> 3 37 92 1 80
#> 4 60 61 99 33
#> 5 23 91 36 44
#> 6 42 76 26 28
#> 7 22 51 15 9
#> 8 94 60 16 53
#> 9 28 89 83 11
#> 10 68 36 39 68
#>
#> [[3]]
#> # A tibble: 10 x 4
#> L M N Z
#> <int> <int> <int> <int>
#> 1 42 56 45 52
#> 2 29 27 33 88
#> 3 10 61 37 74
#> 4 75 40 32 91
#> 5 24 76 5 86
#> 6 68 50 20 43
#> 7 56 31 45 4
#> 8 77 15 38 6
#> 9 23 72 25 61
#> 10 92 40 32 69
#>
#> [[4]]
#> # A tibble: 10 x 4
#> X Y S P
#> <int> <int> <int> <int>
#> 1 58 75 99 16
#> 2 92 13 64 76
#> 3 19 63 27 69
#> 4 99 37 30 64
#> 5 9 30 100 68
#> 6 58 98 40 34
#> 7 53 98 76 96
#> 8 49 94 2 22
#> 9 48 38 10 48
#> 10 32 25 57 1
Though if for some reason you need a tidyverse solution, the equivalent would be:
purrr::map(seq(ncol(df)/4) - 1, ~ df[4 * .x + 1:4])
Created on 2022-06-29 by the reprex package (v2.0.1)