尽管 r 中有独特的列标题，但仍通过位置模式堆叠列

Question

I have a series of columns in my data frame (299 of them) which have the same information type repeated every 3 columns. I need to stack them so each set of 3 retain their referential integrity with each other, and the preceding 22 columns. Normally I'd use melt() to go from wide to long form, but that would involve one or two categories with common names. Each of these columns has a unique name (eg "x1", "y1", "z1","x2","y2","z2").

For the sake of example, lets take example data mtcars. These data contain the 11 columns, each with unique names: mpg, cyl, disp, hp, drat, wt, qsec, vs, am, gear, carb. Lets say I want to keep columns 1:5 as identifiers, then stack the data columns 6:8 on top of 9:11. In this simple example I could use stack or melt and say which columns to use, but when we extrapolate out to hundreds of columns this gets to be a pretty horrid task very quickly... thoughts?

Answer 1

Given that your column names all share a prefix, tidyr::pivot_longer makes this easy. First I'll create some sample data based on your description:

## semi-realistic sample data
nr = 3
n_group = 4
n_per_group = 3
nc = n_group * n_per_group
df = as.data.frame(matrix(rnorm(nr * nc), nrow = nr))
names(df) = outer(letters[1:n_group], 1:n_per_group, paste, sep = ".")
df = cbind(id = 1:nr, df)
#   id        a.1         b.1         c.1         d.1        a.2       b.2        c.2        d.2
# 1  1  0.7898394 -2.34853091 -0.15575050 -1.18205111 -0.6348410  2.545300  0.8265652  0.8817947
# 2  2  0.5717691 -1.46586503  0.01206218  0.83525523  1.6629318  1.602706 -1.5781837  1.1245715
# 3  3 -1.5727813  0.04925075 -0.82081511 -0.06905755 -0.4137177 -1.734805 -0.3742163 -1.0701552
#         a.3        b.3         c.3         d.3
# 1 0.3238154  2.2401514 -0.70958661 -0.09062723
# 2 0.7016303 -0.9547929  0.23150225  0.39295502
# 3 0.7229709 -1.7137527  0.09044422 -0.59431548

With tidyr :

## convert to long format
library(tidyr)
pivot_longer(df, cols = -id, names_sep = "\\.", names_to = c(".value", "iteration"))
# # A tibble: 9 x 6
#      id iteration      a       b       c       d
#   <int> <chr>      <dbl>   <dbl>   <dbl>   <dbl>
# 1     1 1          0.790 -2.35   -0.156  -1.18  
# 2     1 2         -0.635  2.55    0.827   0.882 
# 3     1 3          0.324  2.24   -0.710  -0.0906
# 4     2 1          0.572 -1.47    0.0121  0.835 
# 5     2 2          1.66   1.60   -1.58    1.12  
# 6     2 3          0.702 -0.955   0.232   0.393 
# 7     3 1         -1.57   0.0493 -0.821  -0.0691
# 8     3 2         -0.414 -1.73   -0.374  -1.07  
# 9     3 3          0.723 -1.71    0.0904 -0.594 

Using the same sample data, here's a data.table::melt approach based on column numbers:

## note that `nc` is the number of non-ID columns.
## The `1:nc + 1` is a way to get the column numbers of the stack columns
## (+ 1 because there is 1 id column in my example)
dt = as.data.table(df)
melt(dt, id.vars = "id", measure.vars = split(1:nc + 1, rep(1:n_group, times = n_per_group)))
#    id variable          1           2           3           4
# 1:  1        1  0.7898394 -2.34853091 -0.15575050 -1.18205111
# 2:  2        1  0.5717691 -1.46586503  0.01206218  0.83525523
# 3:  3        1 -1.5727813  0.04925075 -0.82081511 -0.06905755
# 4:  1        2 -0.6348410  2.54530008  0.82656523  0.88179474
# 5:  2        2  1.6629318  1.60270621 -1.57818365  1.12457150
# 6:  3        2 -0.4137177 -1.73480524 -0.37421630 -1.07015525
# 7:  1        3  0.3238154  2.24015140 -0.70958661 -0.09062723
# 8:  2        3  0.7016303 -0.95479285  0.23150225  0.39295502
# 9:  3        3  0.7229709 -1.71375273  0.09044422 -0.59431548

(The results look different at a glance, but it's just a different order of the same rows.)