2

I have 3 lists, each with n elements of length fixed (10 in the example):

my_list <- list(1:10,11:20,21:30)
my_list2 <- list(31:40,41:50,51:60)
my_list3 <- list(rep("a", 10), rep("b", 10), rep("c", 10))

I'd like to put them in a list of data.frames, combining "vertically" each list element:

final <- list(
  tb1 = data.frame(v1=1:10, v2=31:40, v3=rep("a",10)),
  tb2 = data.frame(v1=11:20, v2=41:50, v3=rep("b", 10)),
  tb3 = data.frame(v1=21:30, v2=51:60, v3=rep("c", 10))
)
# map(final, head, 2)
# $`tb1`
# v1 v2 v3
# 1  1 31  a
# 2  2 32  a
# 
# $tb2
# v1 v2 v3
# 1 11 41  b
# 2 12 42  b
# 
# $tb3
# v1 v2 v3
# 1 21 51  c
# 2 22 52  c

With c and rbind at least I combined them in the right order, but I can't think of a way to put them in a data.frame "every 3 elements".

ab <- c(rbind(my_list, my_list2, my_list3))

# head(ab,3)
# [[1]]
# [1]  1  2  3  4  5  6  7  8  9 10
# 
# [[2]]
# [1] 31 32 33 34 35 36 37 38 39 40
# 
# [[3]]
# [1] "a" "a" "a" "a" "a" "a" "a" "a" "a" "a"
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0

3 Answers 3

Reset to default
3

You could something hackish like this:

n_lists <- 3 # length(ls(pattern = "my_list"))
n_cols_each <- 4 # length(get(ls(pattern = "my_list")[[1]]))
temp <- do.call(data.frame, mget(ls(pattern = "my_list"))) 
final <- lapply(
  1:n_lists,
  function(x) setNames(temp[x + (0:(n_cols_each-1))*3], paste0("v", 1:n_cols_each))
)
lapply(final, head, 2)
[[1]]
  v1 v2 v3
1  1 31  a
2  2 32  a

[[2]]
  v1 v2 v3
1 11 41  b
2 12 42  b

[[3]]
  v1 v2 v3
1 21 51  c
2 22 52  c

EDITED

Made more generalisable with predefined, n variables.

EDIT2: Benchmark

bench::mark(
  snoram = {
    n_lists <- 3
    n_cols_each <- 3
    temp <- do.call(data.frame, list(my_list, my_list2, my_list3))
    final <- lapply(
      1:n_lists,
      function(x) setNames(temp[x + (0:(n_cols_each-1))*3], paste0("v", 1:3))
    )
    NULL  
  },
  Jimbou = {
    res <- list()
    for(i in 1:3) {
      res[[i]] <- data.frame(V1=my_list[[i]], V2=my_list2[[i]], V3=my_list3[[i]])
    }    
  },
  ELrico = {
    ll <- list(my_list,my_list2,my_list3)

    final <-
    lapply(seq_along(ll), function(x){ structure(as.data.frame(
            sapply(seq_along(ll[[1]]), function(u) { ll[[u]][x] })
            ), .Names = paste0("v",seq_along(ll))) })
    NULL
  }
)

# A tibble: 3 x 14
  expression    min   mean median    max `itr/sec` mem_alloc  n_gc n_itr
  <chr>      <bch:> <bch:> <bch:> <bch:>     <dbl> <bch:byt> <dbl> <int>
1 snoram     1.13ms 1.49ms 1.33ms 2.68ms      673.    34.9KB     6   297
2 Jimbou     3.87ms 4.37ms 4.25ms 8.46ms      229.    20.3KB     7    86
3 ELrico     1.17ms 1.38ms  1.3ms  2.9ms      725.    26.7KB     7   315
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4 Comments

I was just running it, thank you for the update. It seems that yours is the fastest so far. I'll wait if someone has another solution before accepting yours.
@AndreElrico just did - our solutions are comparable on this small dataset.
lapply(purrr::transpose(mget(ls(pattern = 'my_list'))),function(x)setNames(data.frame(x,stringsAsFactors = F),paste0("V",1:length(x))))
@Onyambu, interesting, probably the best solution so far unless OP wants to stick to base R.
3

There are various solutions:

library(tidyverse)

map(transpose(mget(ls(pattern = 'my_list'))),~set_names(as.tibble(.x),paste0("V",1:length(.x))))

You can also do

split.default(map_dfc(mget(ls(pattern='my_list')),~.x),f=row(diag(3)))

Also you can decide to form a grouped dataframe:

 map_df(set_names(transpose(mget(ls(pattern = 'my_list'))),1:3),~.x)
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Comments

2

You can try also this

res <- list()
for(i in 1:3) {
  res[[i]] <- data.frame(V1=my_list[[i]], V2=my_list2[[i]], V3=my_list3[[i]])
}
map(res, head, 2)
[[1]]
  V1 V2 V3
1  1 31  a
2  2 32  a

[[2]]
  V1 V2 V3
1 11 31  a
2 12 32  a

[[3]]
  V1 V2 V3
1 21 31  a
2 22 32  a
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