r - 如何使用 Reduce 将多个矩阵逆序相乘？

Question

假设我有一个矩阵列表 W，我想同时执行 W[[1]] %*% W[[2]] %*% W[[ 3]] %*% W[[4]] %*% ... 和 W[[4]] %*% W[[3]] %*% W[[2]] %*% W[[1]] %*% ...。看起来 Reduce(`%*%`, W) 和 Reduce(`%*%`, W, right = T) 应该这样做，但事实并非如此似乎不起作用。下面的代码给出了一个例子:

set.seed(90088)
W <- list(
    matrix(sample(1:5, size = 25, replace = T), ncol = 5),
    matrix(sample(1:5, size = 25, replace = T), ncol = 5),
    matrix(sample(1:5, size = 25, replace = T), ncol = 5),
    matrix(sample(1:5, size = 25, replace = T), ncol = 5)
    )
Reduce(`%*%`, W)
Reduce(`%*%`, W, right = T)
W[[1]] %*% W[[2]] %*% W[[3]] %*% W[[4]]
W[[4]] %*% W[[3]] %*% W[[2]] %*% W[[1]]

有谁知道 Reduce(尤其是 right = T 选项)在做什么？

Answer 1

要反转元素相乘的顺序，只需反转您输入到 Reduce() 的列表中元素的顺序即可!

Reduce(`%*%`, rev(W))
#       [,1]  [,2]  [,3] [,4]  [,5]
# [1,] 11583  8978  9082 5034 10443
# [2,] 12759  9950  9967 5604 11565
# [3,]  7269  5686  5702 3176  6575
# [4,] 15459 12094 12050 6857 14060
# [5,]  9834  7684  7691 4314  8916

要查看 right=TRUE 的作用(这很有趣)，请检查对 Reduce() 的更简单调用的中间结果:

Reduce(`*`, 5:1, accumulate=TRUE, right=FALSE)
# [1]   5  20  60 120 120
Reduce(`*`, 5:1, accumulate=TRUE, right=TRUE)
# [1] 120  24   6   2   1

## And a non-commutative function shows even more clearly how right=TRUE works:
Reduce(paste0, letters[1:5], right=TRUE, accumulate=TRUE)
# [1] "abcde" "bcde"  "cde"   "de"    "e"