仅计算R中矩阵乘法的对角线 [英] compute only diagonals of matrix multiplication in R

问题描述

我只需要矩阵乘法的对角线元素:

I need only the diagonal elements from a matrix multiplication:

，

inR.由于Z很大，所以我想避免完全相乘....

in R. As Z is huge I want to avoid the full out multiplication....

Z <- matrix(c(1,1,1,2,3,4), ncol = 2)
Z
#     [,1] [,2]
#[1,]    1    2
#[2,]    1    3
#[3,]    1    4

X <- matrix(c(10,-5,-5,20), ncol = 2)
X
#     [,1] [,2]
#[1,]   10   -5
#[2,]   -5   20

Z %*% D %*% t(Z)
#     [,1] [,2] [,3]
#[1,]   70  105  140
#[2,]  105  160  215
#[3,]  140  215  290

diag(Z %*% D %*% t(Z))
#[1]  70 160 290

X始终是一个小的方矩阵(2x2、3x3或4x4)，其中Z的列数等于X的维数.是否有可用的函数来做到这一点?

X is always a small square matrix (2x2 , 3x3 or 4x4), where Z will have the number of columns equal to the dimension of X. Is there a function available to do this?

推荐答案

我认为您不能避免第一个矩阵乘法(即ZX)，但可以避免第二个矩阵乘法，这是昂贵的: /p>

I don't think you can avoid the first matrix multiplication (i.e. ZX), but you can the second one, which is the expensive one:

rowSums((Z %*% X) * Z)
# [1]  70 160 290

第二个乘法是不是矩阵乘法.这要快得多:

The second multiplication is NOT a matrix multiply. This is much faster:

library(microbenchmark)
set.seed(1)
X <- matrix(c(10,-5,-5,20), ncol = 2)
Z <- matrix(sample(1:1000), ncol=2)    # made Z a little bigger    

microbenchmark(
  res.new <- rowSums((Z %*% X) * Z),   # this solution
  res.old <- diag(Z %*% X %*% t(Z))    # original method
)
# Unit: microseconds
#                               expr     min      lq       mean   median        uq       max neval
#  res.new <- rowSums((Z %*% X) * Z)  20.956  23.233   34.77693  29.6150   44.0025    67.852   100
#  res.old <- diag(Z %*% X %*% t(Z)) 571.214 699.247 1761.08885 760.4295 1188.4485 47488.543   100     

all.equal(res.new, res.old)
# [1] TRUE

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