Move several chunks of columns dynamically to another position-CodePudding

My data is:

df <- data.frame(a   = 1:2,
                 x   = 1:2,
                 b   = 1:2,
                 y   = 3:4,
                 x_2 = 1:2,
                 y_2 = 3:4,
                 c   = 1:2,
                 x_3 = 5:6,
                 y_3 = 1:2)

I now want to put together the x vars, and the y vars so that the order of columns would be:

a, x, x_2, x_3, b, y, y_2, y_3, c

I thought, I could use tidyverse's relocate function in combination with lapply or map or reduce (?), but it doesn't work out.

E.g. if I do:

move_names <- c("x", "y")

library(tidyverse)
moved_data <- lapply(as.list(move_names), function(x)
{
  df <- df |> 
    relocate(!!!syms(paste0(x, "_", 2:3)),
             .after = all_of(x))
}
)

It does the moving for x and y separately, but it creates separate list, but I want to have just my original df with relocated columns.

Update:

I should have been clear that my real data frame has ~500 columns where the to-be-moved columns are all over the place. So providing the full vector of desired column name order won't be feasible.

What I instead have: I have the names of my original columns, i.e. x and y, and I have the names of the to-be-moved columns, i.e. x_2, x_3, y_2, y_3.

CodePudding user response：

In base R:

df[match(c('a', 'x', 'x_2', 'x_3', 'b', 'y', 'y_2', 'y_3', 'c'), names(df))]
#>   a x x_2 x_3 b y y_2 y_3 c
#> 1 1 1   1   5 1 3   3   1 1
#> 2 2 2   2   6 2 4   4   2 2

CodePudding user response：

Not sure if it's what you want.

Vector with order of column names

Let's say you have a vector relocate_name that contains the order of your columns:

library(tidyverse)

relocate_name <- c("a", "x", "x_2", "x_3", "b", "y", "y_2", "y_3", "c")

df %>% relocate(any_of(relocate_name))

Vector with prefix of column names

Or if you only have the prefix of the order, let's call it relocate_name2:

relocate_name2 <- c("a", "x", "b", "y", "c")

df %>% relocate(starts_with(relocate_name2))

Group x and y together

Or if you only want to "group" x and y together:

df %>% 
  relocate(starts_with("x"), .after = "x") %>% 
  relocate(starts_with("y"), .after = "y")

Output

All of the above output is the same.

  a x x_2 x_3 b y y_2 y_3 c
1 1 1   1   5 1 3   3   1 1
2 2 2   2   6 2 4   4   2 2

CodePudding user response：

library(rlist)
# split based in colname-part before _
L <- split.default(df, f = gsub("(.*)_.*", "\\1", names(df)))

# remove names with an underscore
# this is the new order, it should match the names of list L !!
neworder <- names(df)[!grepl("_", names(df))]
# [1] "a" "x" "b" "y" "c"

# cbind list elements together
ans <- rlist::list.cbind(L[neworder])
# a x.x x.x_2 x.x_3 b y.y y.y_2 y.y_3 c
# 1 1   1     1     5 1   3     3     1 1
# 2 2   2     2     6 2   4     4     2 2

# create tidy names again
names(ans) <- gsub(".*\\.(.*)", "\\1", names(ans))
#   a x x_2 x_3 b y y_2 y_3 c
# 1 1 1   1   5 1 3   3   1 1
# 2 2 2   2   6 2 4   4   2 2

CodePudding user response：

Ok, this is probably the worst workaround ever and I don't really understand what exactly I'm doing (especially with the <<-), but it is does the trick.

My general idea after realizing the problem a bit more with the help of you guys here was to "loop" through both of my x and y names, remove these new _2 and _3 columns from the vector of column names and re-append them after their "base" x and y columns.

search_names <- c("x", "y")

df_names <- names(df)

new_names <- lapply(search_names, function(x)
{
  start <- which(df_names == x)

  without_new_names <- setdiff(df_names, paste0(x, "_", 2:3))
  
  df_names <<- append(without_new_names, values = paste0(x, "_", 2:3), after = start)
})[[length(search_names)]]

df |> 
  relocate(any_of(new_names))

  a x x_2 x_3 b y y_2 y_3 c
1 1 1   1   5 1 3   3   1 1
2 2 2   2   6 2 4   4   2 2