开发者

Combining (cbind) vectors of different length

开发者 https://www.devze.com 2023-02-21 14:49 出处:网络
I have several vectors of unequal length and I would like to cbind them. I\'ve put the vectors into a list and I have tried to combine the using do.call(cbind, ...):

I have several vectors of unequal length and I would like to cbind them. I've put the vectors into a list and I have tried to combine the using do.call(cbind, ...):

nm <- list(1:8, 3:8, 1:5)
do.call(cbind, nm)

#      [,1] [,2] [,3]
# [1,]    1    3    1
# [2,]    2    4    2
# [3,]    3    5    3
# [4,]    4    6    4
# [5,]    5    7    5
# [6,]    6    8    1
# [7,]    7    3    2
# [8,]    8    4    3
# Warning message:
#   In (function (..., deparse.level = 1)  :
#         number of rows of result is not a multiple of vector length (arg 2)

As expected, the number of rows in the resulting matrix is the length of the longest vector, and the values of the shorter vectors are recycled to make up for the length.

Instead I'd like to pad the shorter vectors with NA values to obtain the same length as the longest vector. I'd like the matrix to look like this:

#      [,1] [,2] [,3]
# [1,]    1    3    1
# [2,]    2    4    开发者_如何学编程2
# [3,]    3    5    3
# [4,]    4    6    4
# [5,]    5    7    5
# [6,]    6    8    NA
# [7,]    7    NA   NA
# [8,]    8    NA   NA

How can I go about doing this?


You can use indexing, if you index a number beyond the size of the object it returns NA. This works for any arbitrary number of rows defined with foo:

nm <- list(1:8,3:8,1:5)

foo <- 8

sapply(nm, '[', 1:foo)

EDIT:

Or in one line using the largest vector as number of rows:

sapply(nm, '[', seq(max(sapply(nm,length))))

From R 3.2.0 you may use lengths ("get the length of each element of a list") instead of sapply(nm, length):

sapply(nm, '[', seq(max(lengths(nm))))


You should fill vectors with NA before calling do.call.

nm <- list(1:8,3:8,1:5)

max_length <- max(unlist(lapply(nm,length)))
nm_filled <- lapply(nm,function(x) {ans <- rep(NA,length=max_length);
                                    ans[1:length(x)]<- x;
                                    return(ans)})
do.call(cbind,nm_filled)


This is a shorter version of Wojciech's solution.

nm <- list(1:8,3:8,1:5)
max_length <- max(sapply(nm,length))
sapply(nm, function(x){
    c(x, rep(NA, max_length - length(x)))
})


Here is an option using stri_list2matrix from stringi

library(stringi)
out <- stri_list2matrix(nm)
class(out) <- 'numeric'
out
#      [,1] [,2] [,3]
#[1,]    1    3    1
#[2,]    2    4    2
#[3,]    3    5    3
#[4,]    4    6    4
#[5,]    5    7    5
#[6,]    6    8   NA
#[7,]    7   NA   NA
#[8,]    8   NA   NA


Late to the party but you could use cbind.fill from rowr package with fill = NA

library(rowr)
do.call(cbind.fill, c(nm, fill = NA))

#  object object object
#1      1      3      1
#2      2      4      2
#3      3      5      3
#4      4      6      4
#5      5      7      5
#6      6      8     NA
#7      7     NA     NA
#8      8     NA     NA

If you have a named list instead and want to maintain the headers you could use setNames

nm <- list(a = 1:8, b = 3:8, c = 1:5)
setNames(do.call(cbind.fill, c(nm, fill = NA)), names(nm))

#  a  b  c
#1 1  3  1
#2 2  4  2
#3 3  5  3
#4 4  6  4
#5 5  7  5
#6 6  8 NA
#7 7 NA NA
#8 8 NA NA
0

精彩评论

暂无评论...
验证码 换一张
取 消