Transform variable length list into matrix in R
2
If I had a list of vectors of variable lengths :
[[1]]
[1] 1 2 3 4
[[2]]
[1] 4 5 6
[[3]]
[1] 1 2 3 4 5 6 7 8 9
[[4]]
[1] 'a' 'b' 'c'
How could I transform this into a data frame / logical matrix with elements of the list represented as columns?
i.e a dataframe like:
1 2 3 4 5 6 7 8 9 'a' 'b' 'c'
[1] 1 1 1 1 0 0 0 0 0 0 0 0
[2] 0 0 0 1 1 1 0 0 0 0 0 0
[3] 1 1 1 1 1 1 1 1 1 0 0 0
[4] 0 0 0 0 0 0 0 0 0 1 1 1
some data:
x <- list(c(1, 2, 3, 4), c(4, 5, 6), c(1, 2, 3, 4, 5, 6, 7, 8, 9), c("a", "b", "c"))
r
edited Nov 16 '18 at 10:58
markus
14.9k11336
asked Nov 16 '18 at 7:02
user124123user124123
46911238
add a comment |
2
If I had a list of vectors of variable lengths :
[[1]]
[1] 1 2 3 4
[[2]]
[1] 4 5 6
[[3]]
[1] 1 2 3 4 5 6 7 8 9
[[4]]
[1] 'a' 'b' 'c'
How could I transform this into a data frame / logical matrix with elements of the list represented as columns?
i.e a dataframe like:
1 2 3 4 5 6 7 8 9 'a' 'b' 'c'
[1] 1 1 1 1 0 0 0 0 0 0 0 0
[2] 0 0 0 1 1 1 0 0 0 0 0 0
[3] 1 1 1 1 1 1 1 1 1 0 0 0
[4] 0 0 0 0 0 0 0 0 0 1 1 1
some data:
x <- list(c(1, 2, 3, 4), c(4, 5, 6), c(1, 2, 3, 4, 5, 6, 7, 8, 9), c("a", "b", "c"))
r
edited Nov 16 '18 at 10:58
markus
14.9k11336
asked Nov 16 '18 at 7:02
user124123user124123
46911238
add a comment |
2
2
2
If I had a list of vectors of variable lengths :
[[1]]
[1] 1 2 3 4
[[2]]
[1] 4 5 6
[[3]]
[1] 1 2 3 4 5 6 7 8 9
[[4]]
[1] 'a' 'b' 'c'
How could I transform this into a data frame / logical matrix with elements of the list represented as columns?
i.e a dataframe like:
1 2 3 4 5 6 7 8 9 'a' 'b' 'c'
[1] 1 1 1 1 0 0 0 0 0 0 0 0
[2] 0 0 0 1 1 1 0 0 0 0 0 0
[3] 1 1 1 1 1 1 1 1 1 0 0 0
[4] 0 0 0 0 0 0 0 0 0 1 1 1
some data:
x <- list(c(1, 2, 3, 4), c(4, 5, 6), c(1, 2, 3, 4, 5, 6, 7, 8, 9), c("a", "b", "c"))
r
edited Nov 16 '18 at 10:58
markus
14.9k11336
asked Nov 16 '18 at 7:02
user124123user124123
46911238
If I had a list of vectors of variable lengths :
[[1]]
[1] 1 2 3 4
[[2]]
[1] 4 5 6
[[3]]
[1] 1 2 3 4 5 6 7 8 9
[[4]]
[1] 'a' 'b' 'c'
How could I transform this into a data frame / logical matrix with elements of the list represented as columns?
i.e a dataframe like:
1 2 3 4 5 6 7 8 9 'a' 'b' 'c'
[1] 1 1 1 1 0 0 0 0 0 0 0 0
[2] 0 0 0 1 1 1 0 0 0 0 0 0
[3] 1 1 1 1 1 1 1 1 1 0 0 0
[4] 0 0 0 0 0 0 0 0 0 1 1 1
some data:
x <- list(c(1, 2, 3, 4), c(4, 5, 6), c(1, 2, 3, 4, 5, 6, 7, 8, 9), c("a", "b", "c"))
r
r
edited Nov 16 '18 at 10:58
markus
14.9k11336
asked Nov 16 '18 at 7:02
user124123user124123
46911238
edited Nov 16 '18 at 10:58
markus
14.9k11336
asked Nov 16 '18 at 7:02
user124123user124123
46911238
edited Nov 16 '18 at 10:58
markus
14.9k11336
edited Nov 16 '18 at 10:58
markus
14.9k11336
edited Nov 16 '18 at 10:58
markus
14.9k11336
14.9k11336
asked Nov 16 '18 at 7:02
user124123user124123
46911238
asked Nov 16 '18 at 7:02
user124123user124123
46911238
asked Nov 16 '18 at 7:02
user124123user124123
46911238
46911238
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
5
Here is a base R option:
# extract unique values from x
uv <- unique(unlist(x))
# Check in each element of lists which values are present and bind everything toegether
out <- do.call(rbind, lapply(x, function(e) as.integer(uv %in% e) ))
# Convert from matrix to data.frame and add column names
out <- setNames(as.data.frame(out), uv)
out
1 2 3 4 5 6 7 8 9 a b c
1 1 1 1 1 0 0 0 0 0 0 0 0
2 0 0 0 1 1 1 0 0 0 0 0 0
3 1 1 1 1 1 1 1 1 1 0 0 0
4 0 0 0 0 0 0 0 0 0 1 1 1
answered Nov 16 '18 at 7:43
sindri_baldursindri_baldur
8,3651033
add a comment |
2
Here is a base R option with stack and table
table(stack(setNames(x, seq_along(x)))[2:1])
# values
#ind 1 2 3 4 5 6 7 8 9 a b c
# 1 1 1 1 1 0 0 0 0 0 0 0 0
# 2 0 0 0 1 1 1 0 0 0 0 0 0
# 3 1 1 1 1 1 1 1 1 1 0 0 0
# 4 0 0 0 0 0 0 0 0 0 1 1 1
answered Nov 16 '18 at 16:31
akrunakrun
418k13206281
add a comment |
1
Something like this?
library(tidyverse)
x = list(c(1, 2, 3, 4), c(4, 5, 6), c(1, 2, 3, 4, 5, 6, 7, 8, 9))
y = tibble(column1= map_chr(x, str_flatten, " "))
Where y is this:
# A tibble: 3 x 1
column1
<chr>
1 1 2 3 4
2 4 5 6
3 1 2 3 4 5 6 7 8 9
answered Nov 16 '18 at 7:12
wl1234wl1234
461316
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
Here is a base R option:
# extract unique values from x
uv <- unique(unlist(x))
# Check in each element of lists which values are present and bind everything toegether
out <- do.call(rbind, lapply(x, function(e) as.integer(uv %in% e) ))
# Convert from matrix to data.frame and add column names
out <- setNames(as.data.frame(out), uv)
out
1 2 3 4 5 6 7 8 9 a b c
1 1 1 1 1 0 0 0 0 0 0 0 0
2 0 0 0 1 1 1 0 0 0 0 0 0
3 1 1 1 1 1 1 1 1 1 0 0 0
4 0 0 0 0 0 0 0 0 0 1 1 1
answered Nov 16 '18 at 7:43
sindri_baldursindri_baldur
8,3651033
add a comment |
5
Here is a base R option:
# extract unique values from x
uv <- unique(unlist(x))
# Check in each element of lists which values are present and bind everything toegether
out <- do.call(rbind, lapply(x, function(e) as.integer(uv %in% e) ))
# Convert from matrix to data.frame and add column names
out <- setNames(as.data.frame(out), uv)
out
1 2 3 4 5 6 7 8 9 a b c
1 1 1 1 1 0 0 0 0 0 0 0 0
2 0 0 0 1 1 1 0 0 0 0 0 0
3 1 1 1 1 1 1 1 1 1 0 0 0
4 0 0 0 0 0 0 0 0 0 1 1 1
answered Nov 16 '18 at 7:43
sindri_baldursindri_baldur
8,3651033
add a comment |
5
5
5
Here is a base R option:
# extract unique values from x
uv <- unique(unlist(x))
# Check in each element of lists which values are present and bind everything toegether
out <- do.call(rbind, lapply(x, function(e) as.integer(uv %in% e) ))
# Convert from matrix to data.frame and add column names
out <- setNames(as.data.frame(out), uv)
out
1 2 3 4 5 6 7 8 9 a b c
1 1 1 1 1 0 0 0 0 0 0 0 0
2 0 0 0 1 1 1 0 0 0 0 0 0
3 1 1 1 1 1 1 1 1 1 0 0 0
4 0 0 0 0 0 0 0 0 0 1 1 1
answered Nov 16 '18 at 7:43
sindri_baldursindri_baldur
8,3651033
Here is a base R option:
# extract unique values from x
uv <- unique(unlist(x))
# Check in each element of lists which values are present and bind everything toegether
out <- do.call(rbind, lapply(x, function(e) as.integer(uv %in% e) ))
# Convert from matrix to data.frame and add column names
out <- setNames(as.data.frame(out), uv)
out
1 2 3 4 5 6 7 8 9 a b c
1 1 1 1 1 0 0 0 0 0 0 0 0
2 0 0 0 1 1 1 0 0 0 0 0 0
3 1 1 1 1 1 1 1 1 1 0 0 0
4 0 0 0 0 0 0 0 0 0 1 1 1
answered Nov 16 '18 at 7:43
sindri_baldursindri_baldur
8,3651033
edited Nov 16 '18 at 7:57
answered Nov 16 '18 at 7:43
sindri_baldursindri_baldur
8,3651033
answered Nov 16 '18 at 7:43
sindri_baldursindri_baldur
8,3651033
answered Nov 16 '18 at 7:43
sindri_baldursindri_baldur
8,3651033
8,3651033
add a comment |
add a comment |
2
Here is a base R option with stack and table
table(stack(setNames(x, seq_along(x)))[2:1])
# values
#ind 1 2 3 4 5 6 7 8 9 a b c
# 1 1 1 1 1 0 0 0 0 0 0 0 0
# 2 0 0 0 1 1 1 0 0 0 0 0 0
# 3 1 1 1 1 1 1 1 1 1 0 0 0
# 4 0 0 0 0 0 0 0 0 0 1 1 1
answered Nov 16 '18 at 16:31
akrunakrun
418k13206281
add a comment |
2
Here is a base R option with stack and table
table(stack(setNames(x, seq_along(x)))[2:1])
# values
#ind 1 2 3 4 5 6 7 8 9 a b c
# 1 1 1 1 1 0 0 0 0 0 0 0 0
# 2 0 0 0 1 1 1 0 0 0 0 0 0
# 3 1 1 1 1 1 1 1 1 1 0 0 0
# 4 0 0 0 0 0 0 0 0 0 1 1 1
answered Nov 16 '18 at 16:31
akrunakrun
418k13206281
add a comment |
2
2
2
Here is a base R option with stack and table
table(stack(setNames(x, seq_along(x)))[2:1])
# values
#ind 1 2 3 4 5 6 7 8 9 a b c
# 1 1 1 1 1 0 0 0 0 0 0 0 0
# 2 0 0 0 1 1 1 0 0 0 0 0 0
# 3 1 1 1 1 1 1 1 1 1 0 0 0
# 4 0 0 0 0 0 0 0 0 0 1 1 1
answered Nov 16 '18 at 16:31
akrunakrun
418k13206281
Here is a base R option with stack and table
table(stack(setNames(x, seq_along(x)))[2:1])
# values
#ind 1 2 3 4 5 6 7 8 9 a b c
# 1 1 1 1 1 0 0 0 0 0 0 0 0
# 2 0 0 0 1 1 1 0 0 0 0 0 0
# 3 1 1 1 1 1 1 1 1 1 0 0 0
# 4 0 0 0 0 0 0 0 0 0 1 1 1
answered Nov 16 '18 at 16:31
akrunakrun
418k13206281
answered Nov 16 '18 at 16:31
akrunakrun
418k13206281
answered Nov 16 '18 at 16:31
akrunakrun
418k13206281
answered Nov 16 '18 at 16:31
akrunakrun
418k13206281
418k13206281
add a comment |
add a comment |
1
Something like this?
library(tidyverse)
x = list(c(1, 2, 3, 4), c(4, 5, 6), c(1, 2, 3, 4, 5, 6, 7, 8, 9))
y = tibble(column1= map_chr(x, str_flatten, " "))
Where y is this:
# A tibble: 3 x 1
column1
<chr>
1 1 2 3 4
2 4 5 6
3 1 2 3 4 5 6 7 8 9
answered Nov 16 '18 at 7:12
wl1234wl1234
461316
add a comment |
1
Something like this?
library(tidyverse)
x = list(c(1, 2, 3, 4), c(4, 5, 6), c(1, 2, 3, 4, 5, 6, 7, 8, 9))
y = tibble(column1= map_chr(x, str_flatten, " "))
Where y is this:
# A tibble: 3 x 1
column1
<chr>
1 1 2 3 4
2 4 5 6
3 1 2 3 4 5 6 7 8 9
answered Nov 16 '18 at 7:12
wl1234wl1234
461316
add a comment |
1
1
1
Something like this?
library(tidyverse)
x = list(c(1, 2, 3, 4), c(4, 5, 6), c(1, 2, 3, 4, 5, 6, 7, 8, 9))
y = tibble(column1= map_chr(x, str_flatten, " "))
Where y is this:
# A tibble: 3 x 1
column1
<chr>
1 1 2 3 4
2 4 5 6
3 1 2 3 4 5 6 7 8 9
answered Nov 16 '18 at 7:12
wl1234wl1234
461316
Something like this?
library(tidyverse)
x = list(c(1, 2, 3, 4), c(4, 5, 6), c(1, 2, 3, 4, 5, 6, 7, 8, 9))
y = tibble(column1= map_chr(x, str_flatten, " "))
Where y is this:
# A tibble: 3 x 1
column1
<chr>
1 1 2 3 4
2 4 5 6
3 1 2 3 4 5 6 7 8 9
answered Nov 16 '18 at 7:12
wl1234wl1234
461316
answered Nov 16 '18 at 7:12
wl1234wl1234
461316
answered Nov 16 '18 at 7:12
wl1234wl1234
461316
answered Nov 16 '18 at 7:12
wl1234wl1234
461316
461316
add a comment |
add a comment |
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