How can I stack columns per x columns in R

Question

I\'m looking to transform a data frame of 660 columns into 3 columns just by stacking them on each other per 3 columns without manually re-arranging (since I have 660 column

Answer 1

rbind.data.frame requires that all columns match up. So use setNames to replace the names of the C:D columns:

rbind( dat[1:2], setNames(dat[3:4], names(dat[1:2]))  )
  A  B
1 1  4
2 2  5
3 3  6
4 7 10
5 8 11
6 9 12

To generalize that to multiple columns use do.call and lapply:

 dat <- setNames( as.data.frame( matrix(1:36, ncol=12) ), LETTERS[1:12])
 dat
#----
  A B C  D  E  F  G  H  I  J  K  L
1 1 4 7 10 13 16 19 22 25 28 31 34
2 2 5 8 11 14 17 20 23 26 29 32 35
3 3 6 9 12 15 18 21 24 27 30 33 36


do.call( rbind, lapply( seq(1,12, by=3), function(x) setNames(dat[x:(x+2)], LETTERS[1:3])   ))
    A  B  C
1   1  4  7
2   2  5  8
3   3  6  9
4  10 13 16
5  11 14 17
6  12 15 18
7  19 22 25
8  20 23 26
9  21 24 27
10 28 31 34
11 29 32 35
12 30 33 36

The 12 would be replaced by 660 and everything else should work.

Answer 2

Here is another base R option

i1 <- c(TRUE, FALSE)
`row.names<-`(data.frame(A= unlist(df1[i1]), B = unlist(df1[!i1])), NULL)
#  A  B
#1 1  4
#2 2  5
#3 3  6
#4 7 10
#5 8 11
#6 9 12

---

Or another option is melt from data.table

library(data.table)
i1 <- seq(1, ncol(df1), by = 2)
i2 <- seq(2, ncol(df1), by = 2)
melt(setDT(df1), measure = list(i1, i2), value.name = c("A", "B"))

Answer 3

rbindlist from data.table package can also be used for the task and seems to be much more efficient.

# EXAMPLE DATA 
df1 <- read.table(text = '
Col1 Col2 Col3 Col4
1    2    3    4
5    6    7    8
1    2    3    4
5    6    7    8', header = TRUE)

library(data.table)
library(microbenchmark)
library(purrr)
microbenchmark(
    Map = as.data.frame(Map(c, df1[,1:2], df1[, 3:4])),
    Reshape = reshape(df1, direction="long", varying=split(names(df1), rep(seq_len(ncol(df1)/2), 2))),
    Purr = df1 %>% unclass() %>%    # convert to non-data.frame list
        split(names(.)[seq(length(.) / 2)]) %>%     # split columns by indexed names
        map_df(simplify), 
    DataTable = rbindlist(list(df1[,1:2], df1[, 3:4])),
    Mapply = data.frame(mapply(c, df1[,1:2], df1[, 3:4], SIMPLIFY=FALSE)),
    Rbind = rbind(df1[, 1:2],setnames(df1[, 3:4],names(df1[,1:2])))
)

The results are:

Unit: microseconds
expr     min         lq       mean         median     uq         max neval      cld
Map        214.724   232.9380  246.2936    244.1240   255.9240   343.611    100  bc  
Reshape    716.962   739.8940  778.7912    749.7550   767.6725   2834.192   100     e
Purr       309.559   324.6545  339.2973    334.0440   343.4290   551.746    100    d 
DataTable  98.228    111.6080  122.7753    119.2320   129.2640   189.614    100 a    
Mapply     233.577   258.2605  271.1881    270.7895   281.6305   339.291    100   c  
Rbind      206.001   221.1515  228.5956    226.6850   235.2670   283.957    100  b  

Answer 4

A classical split-apply-combine approach will scale flexibly:

as.data.frame(lapply(split(unclass(df), 
                           names(df)[seq(ncol(df) / 2)]), 
                     unlist, use.names = FALSE))
##   A  B
## 1 1  4
## 2 2  5
## 3 3  6
## 4 7 10
## 5 8 11
## 6 9 12

or with a hint of purrr,

library(purrr)

df %>% unclass() %>%    # convert to non-data.frame list
    split(names(.)[seq(length(.) / 2)]) %>%     # split columns by indexed names
    map_df(simplify)    # simplify each split to vector, coerce back to data.frame
## # A tibble: 6 × 2
##       A     B
##   <int> <int>
## 1     1     4
## 2     2     5
## 3     3     6
## 4     7    10
## 5     8    11
## 6     9    12

Answer 5

reshape to the rescue:

reshape(df, direction="long", varying=split(names(df), rep(seq_len(ncol(df)/2), 2)))

#    time A  B id
#1.1    1 1  4  1
#2.1    1 2  5  2
#3.1    1 3  6  3
#1.2    2 7 10  1
#2.2    2 8 11  2
#3.2    2 9 12  3