在矩阵累计总和 [英] Cumulative sum in a matrix

问题描述

我有一个像

A= [ 1 2 4
     2 3 1
     3 1 2 ]

我想通过行和列通过计算其累计总和，就是我想要的结果是

and I would like to calculate its cumulative sum by row and by column, that is, I want the result to be

B = [ 1  3  7 
      3  8  13 
      6  12 19 ]

如何在一个快速的方式使这个R中的任何想法？ （可能使用功能cumsum）
（我有巨大的矩阵）

Any ideas of how to make this in R in a fast way? (Possibly using the function cumsum) (I have huge matrices)

谢谢！

推荐答案

一个单行：

t(apply(apply(A, 2, cumsum)), 1, cumsum))

底层的观察是，你可以先计算在列的累积和，然后这个矩阵在行的累计总和。

The underlying observation is that you can first compute the cumulative sums over the columns and then the cumulative sum of this matrix over the rows.

请注意：在做行，你要转的结果矩阵

Note: When doing the rows, you have to transpose the resulting matrix.

您例如：

> apply(A, 2, cumsum)
     [,1] [,2] [,3]
[1,]    1    2    4
[2,]    3    5    5
[3,]    6    6    7

> t(apply(apply(A, 2, cumsum), 1, cumsum))
     [,1] [,2] [,3]
[1,]    1    3    7
[2,]    3    8   13
[3,]    6   12   19

关于业绩：我现在已经知道这个方法有多好，扩展到大的矩阵。复杂性的角度来看，这应该是接近最优。通常情况下，适用在性能上并不坏为好。

现在我是越来越好奇 - 什么方法是最好的一个？一个简短的风向标：

Now I was getting curious - what approach is the better one? A short benchmark:

> A <- matrix(runif(1000*1000, 1, 500), 1000)
> 
> system.time(
+   B <- t(apply(apply(A, 2, cumsum), 1, cumsum))
+ )
       User      System     elapsed 
      0.082       0.011       0.093 
> 
> system.time(
+   C <- lower.tri(diag(nrow(A)), diag = TRUE) %*% A %*% upper.tri(diag(ncol(A)), diag = TRUE)
+ )
       User      System     elapsed 
      1.519       0.016       1.530 

因此​​：应用由15倍优于矩阵乘法（只是为了对比：MATLAB需要0.10719秒），结果真的不意外，因为适用 - 版本可以在O进行（N ^ 2），而矩阵乘法将需要约为O（n ^ 2.7）计算。因此，如果n是足够大的矩阵乘法提供所有的优化应该会丢失。

Thus: Apply outperforms matrix multiplication by a factor of 15. (Just for comparision: MATLAB needed 0.10719 seconds.) The results do not really surprise, as the apply-version can be done in O(n^2), while the matrix multiplication will need approx. O(n^2.7) computations. Thus, all optimizations that matrix multiplication offers should be lost if n is big enough.

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