list 如何将矩阵转换为 R 中的列向量列表？

How to convert a matrix to a list of column-vectors in R?

提问by Joris Meys

Say you want to convert a matrix to a list, where each element of the list contains one column. list()or as.list()obviously won't work, and until now I use a hack using the behaviour of tapply:

假设您要将矩阵转换为列表，其中列表的每个元素都包含一列。list()或者as.list()显然不会工作，直到现在我使用以下行为进行黑客攻击tapply：

x <- matrix(1:10,ncol=2)

tapply(x,rep(1:ncol(x),each=nrow(x)),function(i)i)

I'm not completely happy with this. Anybody knows a cleaner method I'm overlooking?

我对此并不完全满意。有人知道我忽略的更清洁的方法吗？

(for making a list filled with the rows, the code can obviously be changed to :

（为了制作一个充满行的列表，代码显然可以更改为：

tapply(x,rep(1:nrow(x),ncol(x)),function(i)i)

)

)

采纳答案by mdsumner

In the interests of skinning the cat, treat the array as a vector as if it had no dim attribute:

为了给猫剥皮，把数组当作一个向量，就好像它没有暗属性一样：

 split(x, rep(1:ncol(x), each = nrow(x)))

回答by Tommy

Gavin's answer is simple and elegant. But if there are many columns, a much faster solution would be:

加文的回答简单而优雅。但是如果有很多列，一个更快的解决方案是：

lapply(seq_len(ncol(x)), function(i) x[,i])

The speed difference is 6x in the example below:

下例中的速度差为 6 倍：

> x <- matrix(1:1e6, 10)
> system.time( as.list(data.frame(x)) )
   user  system elapsed 
   1.24    0.00    1.22 
> system.time( lapply(seq_len(ncol(x)), function(i) x[,i]) )
   user  system elapsed 
    0.2     0.0     0.2 

回答by Ari B. Friedman

data.frames are stored as lists, I believe. Therefore coercion seems best:

我相信 data.frames 存储为列表。因此强制似乎是最好的：

as.list(as.data.frame(x))
> as.list(as.data.frame(x))
$V1
[1] 1 2 3 4 5

$V2
[1]  6  7  8  9 10

Benchmarking results are interesting. as.data.frame is faster than data.frame, either because data.frame has to create a whole new object, or because keeping track of the column names is somehow costly (witness the c(unname()) vs c() comparison)? The lapply solution provided by @Tommy is faster by an order of magnitude. The as.data.frame() results can be somewhat improved by coercing manually.

基准测试结果很有趣。as.data.frame 比 data.frame 快，要么是因为 data.frame 必须创建一个全新的对象，要么是因为跟踪列名在某种程度上代价高昂（见证 c(unname()) 与 c() 比较）？@Tommy 提供的 lapply 解决方案快了一个数量级。as.data.frame() 结果可以通过手动强制得到一定程度的改善。

manual.coerce <- function(x) {
  x <- as.data.frame(x)
  class(x) <- "list"
  x
}

library(microbenchmark)
x <- matrix(1:10,ncol=2)

microbenchmark(
  tapply(x,rep(1:ncol(x),each=nrow(x)),function(i)i) ,
  as.list(data.frame(x)),
  as.list(as.data.frame(x)),
  lapply(seq_len(ncol(x)), function(i) x[,i]),
  c(unname(as.data.frame(x))),
  c(data.frame(x)),
  manual.coerce(x),
  times=1000
  )

                                                      expr     min      lq
1                                as.list(as.data.frame(x))  176221  183064
2                                   as.list(data.frame(x))  444827  454237
3                                         c(data.frame(x))  434562  443117
4                              c(unname(as.data.frame(x)))  257487  266897
5             lapply(seq_len(ncol(x)), function(i) x[, i])   28231   35929
6                                         manual.coerce(x)  160823  167667
7 tapply(x, rep(1:ncol(x), each = nrow(x)), function(i) i) 1020536 1036790
   median      uq     max
1  186486  190763 2768193
2  460225  471346 2854592
3  449960  460226 2895653
4  271174  277162 2827218
5   36784   37640 1165105
6  171088  176221  457659
7 1052188 1080417 3939286

is.list(manual.coerce(x))
[1] TRUE

回答by Gavin Simpson

Converting to a data frame thence to a list seems to work:

将数据框转换为列表似乎有效：

> as.list(data.frame(x))
$X1
[1] 1 2 3 4 5

$X2
[1]  6  7  8  9 10
> str(as.list(data.frame(x)))
List of 2
 $ X1: int [1:5] 1 2 3 4 5
 $ X2: int [1:5] 6 7 8 9 10

回答by Sacha Epskamp

Using plyrcan be really useful for things like this:

使用plyr对于这样的事情真的很有用：

library("plyr")

alply(x,2)

$`1`
[1] 1 2 3 4 5

$`2`
[1]  6  7  8  9 10

attr(,"class")
[1] "split" "list" 

回答by alfymbohm

I know this is anathema in R, and I don't really have a lot of reputation to back this up, but I'm finding a for loop to be rather more efficient. I'm using the following function to convert matrix mat to a list of its columns:

我知道这是 R 中的诅咒，我真的没有很多声誉来支持这一点，但我发现 for 循环效率更高。我正在使用以下函数将矩阵 mat 转换为其列的列表：

mat2list <- function(mat)
{
    list_length <- ncol(mat)
    out_list <- vector("list", list_length)
    for(i in 1:list_length) out_list[[i]] <- mat[,i]
    out_list
}

Quick benchmark comparing with mdsummer's and the original solution:

与 mdsummer 和原始解决方案相比的快速基准测试：

x <- matrix(1:1e7, ncol=1e6)

system.time(mat2list(x))
   user  system elapsed 
  2.728   0.023   2.720 

system.time(split(x, rep(1:ncol(x), each = nrow(x))))
   user  system elapsed 
  4.812   0.194   4.978 

system.time(tapply(x,rep(1:ncol(x),each=nrow(x)),function(i)i))
   user  system elapsed 
 11.471   0.413  11.817 

回答by nbenn

The new function asplit()will be coming to base R in v3.6. Up until then and in similar spirit to the answer of @mdsumner we can also do

新功能asplit()将在 v3.6 中基于 R。在那之前，本着与@mdsumner 的回答类似的精神，我们也可以这样做

split(x, slice.index(x, MARGIN))

as per the docs of asplit(). As previously shown however, all split()based solutions are much slower than @Tommy's lapply/`[`. This also holds for the new asplit(), at least in its current form.

根据asplit(). 然而，如前所述，所有split()基于解决方案的解决方案都比@Tommy 的lapply/`[`. 这也适用于 new asplit()，至少在其当前形式中。

split_1 <- function(x) asplit(x, 2L)
split_2 <- function(x) split(x, rep(seq_len(ncol(x)), each = nrow(x)))
split_3 <- function(x) split(x, col(x))
split_4 <- function(x) split(x, slice.index(x, 2L))
split_5 <- function(x) lapply(seq_len(ncol(x)), function(i) x[, i])

dat <- matrix(rnorm(n = 1e6), ncol = 100)

#> Unit: milliseconds
#>          expr       min        lq     mean   median        uq        max neval
#>  split_1(dat) 16.250842 17.271092 20.26428 18.18286 20.185513  55.851237   100
#>  split_2(dat) 52.975819 54.600901 60.94911 56.05520 60.249629 105.791117   100
#>  split_3(dat) 32.793112 33.665121 40.98491 34.97580 39.409883  74.406772   100
#>  split_4(dat) 37.998140 39.669480 46.85295 40.82559 45.342010  80.830705   100
#>  split_5(dat)  2.622944  2.841834  3.47998  2.88914  4.422262   8.286883   100

dat <- matrix(rnorm(n = 1e6), ncol = 1e5)

#> Unit: milliseconds
#>          expr       min       lq     mean   median       uq      max neval
#>  split_1(dat) 204.69803 231.3023 261.6907 246.4927 289.5218 413.5386   100
#>  split_2(dat) 229.38132 235.3153 253.3027 242.0433 259.2280 339.0016   100
#>  split_3(dat) 208.29162 216.5506 234.2354 221.7152 235.3539 342.5918   100
#>  split_4(dat) 214.43064 221.9247 240.7921 231.0895 246.2457 323.3709   100
#>  split_5(dat)  89.83764 105.8272 127.1187 114.3563 143.8771 209.0670   100

回答by wjchulme

There's a function array_tree()in the tidyverse's purrrpackage that does this with minimum fuss:

array_tree()tidyverse 的purrr包中有一个函数可以轻松完成此操作：

x <- matrix(1:10,ncol=2)
xlist <- purrr::array_tree(x, margin=2)
xlist

#> [[1]]
#> [1] 1 2 3 4 5
#>  
#> [[2]]
#> [1]  6  7  8  9 10

Use margin=1to list by row instead. Works for n-dimensional arrays. It preserves names by default:

使用margin=1按行列表，而不是。适用于 n 维数组。它默认保留名称：

x <- matrix(1:10,ncol=2)
colnames(x) <- letters[1:2]
xlist <- purrr::array_tree(x, margin=2)
xlist

#> $a
#> [1] 1 2 3 4 5
#>
#> $b
#> [1]  6  7  8  9 10

(this is a near word-for-word copy of my answer to a similar question here)

（这是我回答的近字对字复制到类似的问题在这里）

回答by Daniel Freeman

Use asplitto convert a matrix into a list of vectors

用asplit一个矩阵转换为载体列表

asplit(x, 1) # split into list of row vectors
asplit(x, 2) # split into list of column vectors

回答by Zhilong Jia

convertRowsToList {BBmisc}

convertRowsToList {BBmisc}

Convert rows (columns) of data.frame or matrix to lists.

将 data.frame 或矩阵的行（列）转换为列表。

BBmisc::convertColsToList(x)

ref: http://berndbischl.github.io/BBmisc/man/convertRowsToList.html

参考：http: //berndbischl.github.io/BBmisc/man/convertRowsToList.html