Question

我想按以下方式重新排列矩阵b：

> b <- matrix(1:24, 6, 4)
> b
      [,1] [,2] [,3] [,4]
[1,]    1    7   13   19
[2,]    2    8   14   20
[3,]    3    9   15   21
[4,]    4   10   16   22
[5,]    5   11   17   23
[6,]    6   12   18   24

> cbind(b[1:2, ], b[3:4, ], b[5:6, ])
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,]    1    7   13   19    3    9   15   21    5    11    17    23
[2,]    2    8   14   20    4   10   16   22    6    12    18    24

行数（此处为2）总是相同，但矩阵b将非常大，大小约为1M x 100.有一些快速的方法我可以做到这一点，没有环？感谢。

Answer 1

首先，使用b将lapply拆分成两行的块，然后使用cbind将子组划分为do.call。

do.call(cbind, lapply(seq(1, NROW(b), 2), function(i) b[i:(i+1),]))
#     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#[1,]    1    7   13   19    3    9   15   21    5    11    17    23
#[2,]    2    8   14   20    4   10   16   22    6    12    18    24

另一种方法是将奇数行和偶数行的元素分成两个向量，rbind将它们

rbind(do.call(c, data.frame(t(b[(1:NROW(b) %% 2) == 1,]))),
      do.call(c, data.frame(t(b[(1:NROW(b) %% 2) == 0,]))))
#     X11 X12 X13 X14 X21 X22 X23 X24 X31 X32 X33 X34
#[1,]   1   7  13  19   3   9  15  21   5  11  17  23
#[2,]   2   8  14  20   4  10  16  22   6  12  18  24

另一种方式，在很大程度上，类似于第二种方法

do.call(rbind, lapply(split(data.frame(b), 1:2), function(x) do.call(c, data.frame(t(x)))))
#  X11 X12 X13 X14 X31 X32 X33 X34 X51 X52 X53 X54
#1   1   7  13  19   3   9  15  21   5  11  17  23
#2   2   8  14  20   4  10  16  22   6  12  18  24

Answer 2

您可以对数组进行整形，使用aperm对其进行转置并重新整形：

array(aperm(array(b, c(2,3,4)), c(1,3,2)), c(2,12))

#     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#[1,]    1    7   13   19    3    9   15   21    5    11    17    23
#[2,]    2    8   14   20    4   10   16   22    6    12    18    24

另一种选择：

tb <- t(b)
rbind(c(tb[,c(T,F)]), c(tb[,c(F,T)]))
#     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
#[1,]    1    7   13   19    3    9   15   21    5    11    17    23
#[2,]    2    8   14   20    4   10   16   22    6    12    18    24

第一步，将其重塑为2, 3, 4：

array(b, c(2,3,4))
, , 1

     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6

, , 2

     [,1] [,2] [,3]
[1,]    7    9   11
[2,]    8   10   12

, , 3

     [,1] [,2] [,3]
[1,]   13   15   17
[2,]   14   16   18

, , 4

     [,1] [,2] [,3]
[1,]   19   21   23
[2,]   20   22   24

第二步，切换轴/转置2和3：

aperm(array(b, c(2,3,4)), c(1,3,2))

, , 1

     [,1] [,2] [,3] [,4]
[1,]    1    7   13   19
[2,]    2    8   14   20

, , 2

     [,1] [,2] [,3] [,4]
[1,]    3    9   15   21
[2,]    4   10   16   22

, , 3

     [,1] [,2] [,3] [,4]
[1,]    5   11   17   23
[2,]    6   12   18   24

第三步，将其重塑为所需的尺寸：

array(aperm(array(b, dim = c(2,3,4)), c(1,3,2)), dim = c(2,12))
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,]    1    7   13   19    3    9   15   21    5    11    17    23
[2,]    2    8   14   20    4   10   16   22    6    12    18    24

Answer 3

使用split对矩阵进行分块，然后循环添加维度cbind：

# number of rows
n <- 2

do.call(cbind, lapply(split(b, (seq(nrow(b))-1) %/% n), matrix, nrow = n))

#      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
# [1,]    1    7   13   19    3    9   15   21    5    11    17    23
# [2,]    2    8   14   20    4   10   16   22    6    12    18    24

在R中用cbind组合相同大小的行

3 个答案: