Question

我想创建一个矩阵（A），其中元素是另一个矩阵（B）的每四行的平均值。例如，矩阵A中第1行的元素应该是矩阵B中第1行到第4行的平均值。目前我已经使用了一个循环函数来获得它，但是矩阵的大小非常大，这使得循环很有时间耗时。我想知道是否有更好的方法可以做到这一点。这是一个例子

B = matrix(runif(10000, 0, 10), 100, 100)
A = matrix(0, floor(dim(B)[1]/4), dim(B)[2])
for (im in 1: floor(dim(B)[1]/4)){
    A[im, ] = colMeans(as.matrix(B[c((((im - 1)*4) + 1):(im*4)), ]))
}

Answer 1

您可以使用具有rowsum方法（默认为＆＃39;）的matrix函数轻松地对此进行矢量化，并且可以按组计算总和。然后，除以4以获得均值

grps <- floor(dim(B)[1]/4)
rowsum.default(B[1:(grps*4),], rep(1:grps, each = 4), reorder = FALSE)/4

<强>基准

由于这是一个优化问题，这里有一些基准，所有提出的方法都不是这么大的数据集

library(zoo)
library(microbenchmark)

set.seed(123)
B <- matrix(runif(100, 0, 10), 10000, 100)

OP <- function(B) {
  grps <- floor(dim(B)[1]/4)
  A = matrix(0, grps, dim(B)[2])
  for (im in 1: grps){
    A[im, ] = colMeans(as.matrix(B[c((((im - 1)*4) + 1):(im*4)), ]))
  }
  A
}

DA <- function(B){
  grps <- floor(dim(B)[1]/4)
  rowsum.default(B[1:(grps*4),], rep(1:grps, each = 4), reorder = FALSE)/4
}

JB <- function(B) as.matrix(aggregate(B, list(gl(ceiling(nrow(B)/4), 4, nrow(B))), mean)[, -1])

Thela <- function(B) tapply(B, list((row(B)-1) %/% 4,col(B)), FUN=mean)

RollApply <- function(B) rollapply(B, width = 4, by = 4, FUN = mean, by.column = TRUE)

microbenchmark(OP(B), DA(B), JB(B), RollApply(B), Thela(B), times = 10L)
# Unit: milliseconds
#         expr        min         lq       mean     median         uq        max neval  cld
#        OP(B)   45.57121   48.93491   70.17095   55.77107   65.43564   168.7760    10 a   
#        DA(B)   10.60941   10.87035   11.65232   11.36478   12.07908    14.1551    10 a   
#        JB(B) 1753.39114 1773.83230 1868.60788 1837.47161 1900.38141  2076.5835    10  b  
# RollApply(B) 8946.90359 9009.45160 9380.62408 9294.98441 9450.16426 10922.2595    10    d
#     Thela(B) 4820.36079 4925.70055 5117.22822 5048.89781 5257.58619  5650.2391    10   c

毕竟OP的解决方案不是所以坏了。

Answer 2

您可以使用以下包（zoo）和函数（rollapply）实现此目的。

install.packages("zoo")
require(zoo)

  B <- matrix(runif(100, 0, 10),10, 10)
# with for loop
A = matrix(0,floor(dim(B)[1]/4),dim(B)[2])
for (im in 1 : floor(dim(B)[1]/4)){
+   A[im,] = colMeans(as.matrix(B[c((((im-1)*4)+1):(im*4)),]))}
         [,1]     [,2]     [,3]     [,4]     [,5]     [,6]     [,7]     [,8]     [,9]
[1,] 5.633970 4.092848 3.793473 5.437288 6.316069 4.714015 5.837214 7.150007 4.638332
[2,] 5.445271 2.024052 6.096939 6.165723 3.049140 4.928087 5.433291 5.674594 4.607373
        [,10]
[1,] 5.260153
[2,] 6.589873

# with rowsum @ David
C = grps <- floor(dim(B)[1]/4)
rowsum(B[1:(grps*4),], rep(1:grps, each = 4))/4
      [,1]     [,2]     [,3]     [,4]     [,5]     [,6]     [,7]     [,8]     [,9]
1 5.633970 4.092848 3.793473 5.437288 6.316069 4.714015 5.837214 7.150007 4.638332
2 5.445271 2.024052 6.096939 6.165723 3.049140 4.928087 5.433291 5.674594 4.607373
     [,10]
1 5.260153
2 6.589873

# With rollapply 
D = rollapply(B, width = 4, by = 4, FUN = mean, by.column = T)
D
         [,1]     [,2]     [,3]     [,4]     [,5]     [,6]     [,7]     [,8]     [,9]
[1,] 5.633970 4.092848 3.793473 5.437288 6.316069 4.714015 5.837214 7.150007 4.638332
[2,] 5.445271 2.024052 6.096939 6.165723 3.049140 4.928087 5.433291 5.674594 4.607373
        [,10]
[1,] 5.260153
[2,] 6.589873

Answer 3

aggregate也可以这样做，但需要后续胁迫matrix：

as.matrix(aggregate(B, list(gl(ceiling(nrow(B)/4), 4, nrow(B))), mean)[, -1])

请注意，如果nrow(B)不是4的倍数，则结果将包含最后一行，其中包含最后nrow(B) %% 4行的列平均值。

作为indicated by @thelatemail，tapply可以做一个更简洁的工作：

tapply(B, list((row(B)-1) %/% 4,col(B)), FUN=mean)

如何通过平均R中另一个矩阵的元素来创建矩阵？

3 个答案: