复制行R的方法 [英] Method to copy down rows R

问题描述

假设我们有一个数据帧或矩阵，其中一列指定整数值N如下（col 5）。
是否有一个向量方法来重新填充对象，使每行被复制N次？

 > y 
 [，1] [，2] [，3] [，4] [，5] 
 [1，] -0.02738267 0.5170621 -0.01644855 0.48830663 1 
 [2，] -0.30076544 1.8136359 0.02319640 -1.59649330 2 
 [3，] 1.73447245 0.4043638 -0.29112385 -0.25102988 3 
 [4，] 0.01025271 -0.4908636 0.80857300 0.08137033 4 
  

结果如下。

  [1，] -0.02738267 0.5170621 -0.01644855 0.48830663 1 
 [2，] -0.30076544 1.8136359 0.02319640 -1.59649330 2 
 [2，] -0.30076544 1.8136359 0.02319640 -1.59649330 2 
 [3，] 1.73447245 0.4043638 -0.29112385 -0.25102988 3 
 [3，] 1.73447245 0.4043638 -0.29112385 -0.25102988 3 
 [3，] 1.73447245 0.4043638 -0.29112385 -0.25102988 3 
 [4，] 0.01025271 -0.4908636 0.80857300 0.08137033 4 
 [4 ，] 0.01025271 -0.4908636 0.80857300 0.08137033 4 
 [4，] 0.01025271 -0.4908636 0.80857300 0.08137033 4 
 [4，] 0.01025271 -0.4908636 0.80857300 0.08137033 4 
  
解决方案
一些化妆数据：
  y<  -  cbind （matrix（runif（16），4，4），1：4）
  
 / p> 
 
 
  z<  -  y [rep（seq_len（nrow（y）），y [，5]），] 
＃[，1] [，2] [，3] [，4] [，5] 
＃[1，] 0.5256007 0.07467979 0.95189484 0.2887943 1 
＃[2，] 0.3083967 0.03518523 0.08380005 0.9168161 2 
＃[3，] 0.3083967 0.03518523 0.08380005 0.9168161 2 
＃[4，] 0.8549639 0.79452728 0.22483537 0.4452553 3 
＃[5，] 0.8549639 0.79452728 0.22483537 0.4452553 3 
＃ ] 0.8549639 0.79452728 0.22483537 0.4452553 3 
＃[7，] 0.5453508 0.47633523 0.51522514 0.3936340 4 
＃[8，] 0.5453508 0.47633523 0.51522514 0.3936340 4 
＃[9，] 0.5453508 0.47633523 0.51522514 0.3936340 4 
＃[10，] 0.5453508 0.47633523 0.51522514 0.3936340 4 
  
我不知道你的意思jitter，但也许
  z < -  z + runif（z）/ 1000 
  
？
 
Suppose we have a dataframe or matrix with one column specifying an integer value N as below (col 5).
Is there a vector approach to repopulate the object such that each row gets copied N times?
> y
            [,1]       [,2]        [,3]        [,4] [,5]
[1,] -0.02738267  0.5170621 -0.01644855  0.48830663    1
[2,] -0.30076544  1.8136359  0.02319640 -1.59649330    2
[3,]  1.73447245  0.4043638 -0.29112385 -0.25102988    3
[4,]  0.01025271 -0.4908636  0.80857300  0.08137033    4
The result would be as follows.
[1,] -0.02738267  0.5170621 -0.01644855  0.48830663    1
[2,] -0.30076544  1.8136359  0.02319640 -1.59649330    2    
[2,] -0.30076544  1.8136359  0.02319640 -1.59649330    2
[3,]  1.73447245  0.4043638 -0.29112385 -0.25102988    3
[3,]  1.73447245  0.4043638 -0.29112385 -0.25102988    3
[3,]  1.73447245  0.4043638 -0.29112385 -0.25102988    3
[4,]  0.01025271 -0.4908636  0.80857300  0.08137033    4
[4,]  0.01025271 -0.4908636  0.80857300  0.08137033    4
[4,]  0.01025271 -0.4908636  0.80857300  0.08137033    4
[4,]  0.01025271 -0.4908636  0.80857300  0.08137033    4
Another question would be how to jitter the newly populated rows, such that there is not compute overlap of the newly copied data.
解决方案
Some made-up data:
y <- cbind(matrix(runif(16), 4, 4), 1:4)
Just do:
z <- y[rep(seq_len(nrow(y)), y[,5]), ]
#            [,1]       [,2]       [,3]      [,4] [,5]
#  [1,] 0.5256007 0.07467979 0.95189484 0.2887943    1
#  [2,] 0.3083967 0.03518523 0.08380005 0.9168161    2
#  [3,] 0.3083967 0.03518523 0.08380005 0.9168161    2
#  [4,] 0.8549639 0.79452728 0.22483537 0.4452553    3
#  [5,] 0.8549639 0.79452728 0.22483537 0.4452553    3
#  [6,] 0.8549639 0.79452728 0.22483537 0.4452553    3
#  [7,] 0.5453508 0.47633523 0.51522514 0.3936340    4
#  [8,] 0.5453508 0.47633523 0.51522514 0.3936340    4
#  [9,] 0.5453508 0.47633523 0.51522514 0.3936340    4
# [10,] 0.5453508 0.47633523 0.51522514 0.3936340    4
And I am not sure what you mean by "jitter", but maybe
z <- z + runif(z) / 1000
?

                        
这篇关于复制行R的方法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！