Transforming columns every nth row to multiple rows - r

Question

I have a data frame that I successfully separated every nth row of a column and arranged the snippets into rows:
list = data.frame(x = c("A", "1", "2", "3", "B", "1", "2", "3"))
x
1 A
2 1
3 2
4 3
5 B
6 1
7 2
8 3
to
w x y z
1 A 1 2 3
2 B 1 2 3
I managed to achieve this with:
table <- data.frame(matrix(list$x, ncol = 4, byrow = TRUE))
In the next step I would like to do this with multiple columns and place the resulting tables beneath each other.
From something like this:
x y z
1 A D G
2 3 5 6
3 3 7 4
4 6 7 5
5 B E H
6 4 5 2
7 3 3 4
8 7 7 6
9 C F I
10 7 4 3
11 3 6 9
12 6 8 0
to
w x y z
1 A 3 3 6
2 B 4 3 7
3 C 7 3 6
4 D 5 7 6
5 E 5 3 7
6 F 4 6 8
7 G 6 4 7
8 H 2 4 6
9 I 3 9 0
I am really stuck with this one.
If someone has an idea I would highly appreciate some help.

Answer 1

Making use of lapply and dplyr::bind_rows() this could be achieved like so:
library(dplyr)
df_list <- lapply(list, function(x) data.frame(matrix(x, ncol = 4, byrow = TRUE)))
df_list %>%
dplyr::bind_rows() %>%
setNames(c("w", "x", "y", "z"))
#> w x y z
#> 1 A 3 3 6
#> 2 B 4 3 7
#> 3 C 7 3 6
#> 4 D 5 7 7
#> 5 E 5 3 7
#> 6 F 4 6 8
#> 7 G 6 4 5
#> 8 H 2 4 6
#> 9 I 3 9 0
Or using do.call and rbind:
df_list <- do.call(rbind, c(df_list, list(make.row.names = FALSE)))
setNames(df_list, c("w", "x", "y", "z"))
#> w x y z
#> 1 A 3 3 6
#> 2 B 4 3 7
#> 3 C 7 3 6
#> 4 D 5 7 7
#> 5 E 5 3 7
#> 6 F 4 6 8
#> 7 G 6 4 5
#> 8 H 2 4 6
#> 9 I 3 9 0
DATA
list <- read.table(text = " x y z
1 A D G
2 3 5 6
3 3 7 4
4 6 7 5
5 B E H
6 4 5 2
7 3 3 4
8 7 7 6
9 C F I
10 7 4 3
11 3 6 9
12 6 8 0", header = TRUE)

Answer 2

Here is another solution. You could use good old base::by function in order to split a data set into groups and apply a function on each chunk. (Here is to dear #Henrik who taught me this valuable trick):
do.call(rbind, by(df, rep(seq_len(nrow(df)/4), each = 4), FUN = \(x) {
{setNames(as.data.frame(t(x[-4])), c("w", "x", "y", "z")) |>
`rownames<-`(NULL)}
})) |> `rownames<-`(NULL)
w x y z
1 A 3 3 6
2 D 5 7 7
3 G 6 4 5
4 B 4 3 7
5 E 5 3 7
6 H 2 4 6
7 C 7 3 6
8 F 4 6 8
9 I 3 9 0

Answer 3

Because the number of rows in each chunk seems to be known and constant, you may unlist the data frame, and use modulo (%%) to distinguish between the characters belonging to the "sub headers" and the numeric values.
v = unlist(d)
i = (1:nrow(d) - 1) %% 4 == 0
data.frame(w = v[i],
matrix(v[!i], ncol = 3, byrow = TRUE, dimnames = list(NULL, names(d))))
w x y z
x1 A 3 3 6
x5 B 4 3 7
x9 C 7 3 6
y1 D 5 7 7
y5 E 5 3 7
y9 F 4 6 8
z1 G 6 4 5
z5 H 2 4 6
z9 I 3 9 0
Wrap v[i] in as.integer or as.numeric if that's the desired class.
Add row.names = NULL to the data.frame call if you happen to care about row names.

Answer 4

Base R Split-Apply-Combine, a bit-ugly but is generic enough to handle data having more or less than 4 rows per matrix:
res <- data.frame(
do.call(
rbind,
lapply(
with(
lst,
split(
lst,
cumsum(
apply(
lst,
1,
function(row){
all(
grepl(
"[a-zA-Z]",
row
)
)
}
)
)
)
),
function(x){
type.convert(
setNames(
data.frame(
t(x),
stringsAsFactors = FALSE
),
letters[23:26]
)
)
}
)
),
row.names = NULL
)