随机重新排序(混洗)矩阵的行?

bit*_*ion 5 random r shuffle rows matrix

我想随机重新排序矩阵A的行以生成另一个新矩阵.如何在R中做到这一点?

Rei*_*son 13

用于sample()以(伪)随机顺序生成行索引并使用重新排序矩阵[.

## create a matrix A for illustration
A <- matrix(1:25, ncol = 5)
Run Code Online (Sandbox Code Playgroud)

给予

> A
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    6   11   16   21
[2,]    2    7   12   17   22
[3,]    3    8   13   18   23
[4,]    4    9   14   19   24
[5,]    5   10   15   20   25
Run Code Online (Sandbox Code Playgroud)

接下来,为行生成随机顺序

## generate a random ordering
set.seed(1) ## make reproducible here, but not if generating many random samples
rand <- sample(nrow(A))
rand
Run Code Online (Sandbox Code Playgroud)

这给出了

> rand
[1] 2 5 4 3 1
Run Code Online (Sandbox Code Playgroud)

现在用它来重新排序 A

> A
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    6   11   16   21
[2,]    2    7   12   17   22
[3,]    3    8   13   18   23
[4,]    4    9   14   19   24
[5,]    5   10   15   20   25
> A[rand, ]
     [,1] [,2] [,3] [,4] [,5]
[1,]    2    7   12   17   22
[2,]    5   10   15   20   25
[3,]    4    9   14   19   24
[4,]    3    8   13   18   23
[5,]    1    6   11   16   21
Run Code Online (Sandbox Code Playgroud)


Agi*_*ean 5

使用 tidyverse,您可以使用 one-liner 进行 shuffle:

A %>% sample_n(nrow(.))
Run Code Online (Sandbox Code Playgroud)

这仅适用于数据框或小标题,因此您需要将 A 设为:

A <- tibble(1:25, ncol = 5)
A %>% sample_n(nrow(.))

# A tibble: 25 x 2
   `1:25`  ncol
    <int> <dbl>
 1      9     5
 2      6     5
 3      4     5
 4     15     5
 5     14     5
 6      3     5
 7     23     5
 8     25     5
 9     17     5
10     19     5
# … with 15 more rows
Run Code Online (Sandbox Code Playgroud)