Reputation: 43420
How do I make a matrix from a list of vectors in R?
Goal: from a list of vectors of equal length, create a matrix where each vector becomes a row.
Example:
> a <- list()
> for (i in 1:10) a[[i]] <- c(i,1:5)
> a
[[1]]
[1] 1 1 2 3 4 5
[[2]]
[1] 2 1 2 3 4 5
[[3]]
[1] 3 1 2 3 4 5
[[4]]
[1] 4 1 2 3 4 5
[[5]]
[1] 5 1 2 3 4 5
[[6]]
[1] 6 1 2 3 4 5
[[7]]
[1] 7 1 2 3 4 5
[[8]]
[1] 8 1 2 3 4 5
[[9]]
[1] 9 1 2 3 4 5
[[10]]
[1] 10 1 2 3 4 5
I want:
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
Upvotes: 120
Views: 122170
Answers (7)
Reputation: 9486
data.table::transpose(a) can be a useful tool here if you your list elements have unequal size or you actually wanted a data.frame instead.
It efficiently turns a length-n list of length-up-to-p vectors into a length-p list of length-n vectors, padding the missing elements with a value of your choice.
# For list of vectors of unequal size if you want to pad instead of recycle
a <- sapply(1:6, function(i) c(i, seq_len(i)))
a
#> [[1]]
#> [1] 1 1
#>
#> [[2]]
#> [1] 2 1 2
#>
#> [[3]]
#> [1] 3 1 2 3
#>
#> [[4]]
#> [1] 4 1 2 3 4
#>
#> [[5]]
#> [1] 5 1 2 3 4 5
#>
#> [[6]]
#> [1] 6 1 2 3 4 5 6
matrix(unlist(data.table::transpose(a)), nrow=length(a))
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7]
#> [1,] 1 1 NA NA NA NA NA
#> [2,] 2 1 2 NA NA NA NA
#> [3,] 3 1 2 3 NA NA NA
#> [4,] 4 1 2 3 4 NA NA
#> [5,] 5 1 2 3 4 5 NA
#> [6,] 6 1 2 3 4 5 6
#
## neat if you want a data.frame instead
data.table::setDF(data.table::as.data.table(data.table::transpose(a)))[]
#> V1 V2 V3 V4 V5 V6 V7
#> 1 1 1 NA NA NA NA NA
#> 2 2 1 2 NA NA NA NA
#> 3 3 1 2 3 NA NA NA
#> 4 4 1 2 3 4 NA NA
#> 5 5 1 2 3 4 5 NA
#> 6 6 1 2 3 4 5 6
It is almost as fast as the matrix(unlist( ), byrow=TRUE) solution and much faster than the t(sapply( approach that also works for unequal lengths.
a <- sapply(1:6, function(i) c(i, seq_len(i)))
a
bench::mark(
matrix(unlist(data.table::transpose(a)), nrow=length(a)),
t(sapply(a, '[', 1:max(sapply(a, length))))
)
#> # A tibble: 2 × 6
#> expression min median
#> <bch:expr> <bch:tm> <bch:tm>
#> 1 matrix(unlist(data.table::transpose(a)), nrow = length(a)) 6.87µs 8.68µs
#> 2 t(sapply(a, "[", 1:max(sapply(a, length)))) 33.29µs 42.14µs
#> # ℹ 3 more variables: `itr/sec` <dbl>, mem_alloc <bch:byt>, `gc/sec` <dbl>
# small list, equal sizes
a <- sapply(1:6, function(i) c(i, seq_len(5)), simplify = FALSE)
a
#> [[1]]
#> [1] 1 1 2 3 4 5
#>
#> [[2]]
#> [1] 2 1 2 3 4 5
#>
#> [[3]]
#> [1] 3 1 2 3 4 5
#>
#> [[4]]
#> [1] 4 1 2 3 4 5
#>
#> [[5]]
#> [1] 5 1 2 3 4 5
#>
#> [[6]]
#> [1] 6 1 2 3 4 5
bench::mark(
matrix(unlist(data.table::transpose(a)), nrow=length(a)),
t(sapply(a, '[', 1:max(sapply(a, length)))),
do.call(rbind, a),
matrix(unlist(a), byrow=TRUE, nrow=length(a) )
)
#> # A tibble: 4 × 6
#> expression min median
#> <bch:expr> <bch:tm> <bch:tm>
#> 1 matrix(unlist(data.table::transpose(a)), nrow = length(a)) 7.03µs 9.06µs
#> 2 t(sapply(a, "[", 1:max(sapply(a, length)))) 32.99µs 36.18µs
#> 3 do.call(rbind, a) 2.92µs 3.47µs
#> 4 matrix(unlist(a), byrow = TRUE, nrow = length(a)) 2.77µs 3.07µs
#> # ℹ 3 more variables: `itr/sec` <dbl>, mem_alloc <bch:byt>, `gc/sec` <dbl>
# large list, equal sizes
a <- sapply(seq_len(100000), function(i) c(i, seq_len(5)), simplify = FALSE)
bench::mark(
matrix(unlist(data.table::transpose(a)), nrow=length(a)),
t(sapply(a, '[', 1:max(sapply(a, length)))),
do.call(rbind, a),
matrix(unlist(a), byrow=TRUE, nrow=length(a) )
)
#> Warning: Some expressions had a GC in every iteration; so filtering is disabled.
#> # A tibble: 4 × 6
#> expression min median
#> <bch:expr> <bch:tm> <bch:tm>
#> 1 matrix(unlist(data.table::transpose(a)), nrow = length(a)) 11.62ms 12.54ms
#> 2 t(sapply(a, "[", 1:max(sapply(a, length)))) 94.56ms 101.09ms
#> 3 do.call(rbind, a) 59.02ms 70.49ms
#> 4 matrix(unlist(a), byrow = TRUE, nrow = length(a)) 7.02ms 7.82ms
#> # ℹ 3 more variables: `itr/sec` <dbl>, mem_alloc <bch:byt>, `gc/sec` <dbl>
Upvotes: -1
Reputation: 1761
simplify2array is a base function that is fairly intuitive. However, since R's default is to fill in data by columns first, you will need to transpose the output. (sapply uses simplify2array, as documented in help(sapply).)
> t(simplify2array(a))
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
Upvotes: 21
Reputation: 1761
The built-in matrix function has the nice option to enter data byrow. Combine that with an unlist on your source list will give you a matrix. We also need to specify the number of rows so it can break up the unlisted data. That is:
> matrix(unlist(a), byrow=TRUE, nrow=length(a) )
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
Upvotes: 18
Reputation: 589
t(sapply(a, '[', 1:max(sapply(a, length))))
where 'a' is a list. Would work for unequal row size
Upvotes: 8
Reputation: 6649
> library(plyr)
> as.matrix(ldply(a))
V1 V2 V3 V4 V5 V6
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
Upvotes: 3
Reputation: 2815
Not straightforward, but it works:
> t(sapply(a, unlist))
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
Upvotes: 12
Reputation: 43420
One option is to use do.call():
> do.call(rbind, a)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
Upvotes: 142
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