Extract every 3 column from a df to a list of df-s-CodePudding

I would like to make a loop that extracts every 3 column from my dataframe. The original dataframe length is not fix, so it could be 30 column or 12 long. The only thing what is fix that there will be 3 columns after each other I want to extract and put it into a list of dataframes.

For example:

KO5_1    KO5_2     KO5_3    KO9_1
1        1         3        3
2        0         0        3
2        2         3        0
0        0         1        2

and I want the KO5_1 KO5_2 KO5_3 to be extracted and put together in a separate df. The KO9 and the other 2 of its kind will go another df. The names are changing its just an example.

How can I make this loop?

Thank you!

CodePudding user response：

Let's use a sample data frame. The names aren't important.

set.seed(1)
df <- as.data.frame(replicate(10, sample(5)))

df
#>   V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
#> 1  1  5  3  2  1  1  2  1  2   4
#> 2  4  3  5  5  3  2  5  4  5   2
#> 3  3  4  1  4  5  5  1  3  3   1
#> 4  5  2  4  3  4  3  4  2  4   3
#> 5  2  1  2  1  2  4  3  5  1   5

To get this into a list of 3 data frames comprising every 3rd column, we can do:

n <- 3

lapply(seq(n) %% n, function(i) df[seq_along(df) %% n == i])
#> [[1]]
#>   V1 V4 V7 V10
#> 1  1  2  2   4
#> 2  4  5  5   2
#> 3  3  4  1   1
#> 4  5  3  4   3
#> 5  2  1  3   5
#> 
#> [[2]]
#>   V2 V5 V8
#> 1  5  1  1
#> 2  3  3  4
#> 3  4  5  3
#> 4  2  4  2
#> 5  1  2  5
#>
#> [[3]]
#>   V3 V6 V9
#> 1  3  1  2
#> 2  5  2  5
#> 3  1  5  3
#> 4  4  3  4
#> 5  2  4  1

You can change n to get every 4th column, 5th column, etc.

CodePudding user response：

Use split.default, with gl(n, 3) to split every three columns. Or gl(3, 1) if you wanna get three groups of one column every three columns.

df <- as.data.frame(replicate(10, sample(5)))
names(df) <- paste0("K0", 1:10)

n = ceiling(ncol(df) / 3)
split.default(df, gl(n, 3))

$`1`
  K01 K02 K03
1   2   3   3
2   4   1   5
3   3   2   2
4   1   5   1
5   5   4   4

$`2`
  K04 K05 K06
1   5   4   4
2   4   3   2
3   2   5   1
4   3   2   3
5   1   1   5

$`3`
  K07 K08 K09
1   3   4   1
2   5   5   3
3   1   1   2
4   4   3   5
5   2   2   4

$`4`
  K010
1    3
2    5
3    1
4    4
5    2