英文:
Algorithm for matrix multiplication
问题
我需要编写一个单循环算法来进行矩阵乘法，不使用%*%，而是利用colSums()来完成。
我已经尝试过研究矩阵乘法的特性，但未能找到可以使用colSums()实现的模式。
英文:
I am required to write a one-loop algorithm for matrix multiplication without using %*%, with the help of colSums().
I have tried working through matrix multiplication traits, but failed to find a pattern which can be attained with colSums().
答案1
得分: 1
如果你不允许使用%*%，而必须使用colSums的话，你可以尝试以下代码：
matrix(colSums(t(a)[, rep(1:nrow(a), each = ncol(b))] * b[, rep(1:ncol(b), nrow(a))]),nrow(a),byrow = TRUE)
示例：给定两个如下的矩阵a和b：
> set.seed(0)> (a <- matrix(runif(12), 3))[,1] [,2] [,3] [,4][1,] 0.8966972 0.5728534 0.8983897 0.62911404[2,] 0.2655087 0.9082078 0.9446753 0.06178627[3,] 0.3721239 0.2016819 0.6607978 0.20597457> (b <- matrix(rnorm(8), 4))[,1] [,2][1,] -0.928567035 0.7635935[2,] -0.294720447 -0.7990092[3,] -0.005767173 -1.1476570[4,] 2.404653389 -0.2894616
我们有：
> a %*% b[,1] [,2][1,] 0.50614499 -0.9861506[2,] -0.37108354 -1.6249737[3,] 0.08650475 -0.6949853> matrix(colSums(t(a)[, rep(1:nrow(a), each = ncol(b))] * b[, rep(1:ncol(b), nrow(a))]),nrow(a),byrow = TRUE)[,1] [,2][1,] 0.50614499 -0.9861506[2,] -0.37108354 -1.6249737[3,] 0.08650475 -0.6949853
希望这对你有帮助。
英文:
If you are NOT ALLOWED to use %*% and HAVE TO take colSums anyway, you can try
matrix(colSums(t(a)[, rep(1:nrow(a), each = ncol(b))] * b[, rep(1:ncol(b), nrow(a))]),nrow(a),byrow = TRUE)
Example: Given two matrices a and b like below
> set.seed(0)> (a <- matrix(runif(12), 3))[,1] [,2] [,3] [,4][1,] 0.8966972 0.5728534 0.8983897 0.62911404[2,] 0.2655087 0.9082078 0.9446753 0.06178627[3,] 0.3721239 0.2016819 0.6607978 0.20597457> (b <- matrix(rnorm(8), 4))[,1] [,2][1,] -0.928567035 0.7635935[2,] -0.294720447 -0.7990092[3,] -0.005767173 -1.1476570[4,] 2.404653389 -0.2894616
we have
> a %*% b[,1] [,2][1,] 0.50614499 -0.9861506[2,] -0.37108354 -1.6249737[3,] 0.08650475 -0.6949853> matrix(+ colSums(t(a)[, rep(1:nrow(a), each = ncol(b))] * b[, rep(1:ncol(b), nrow(a))]),+ nrow(a),+ byrow = TRUE+ )[,1] [,2][1,] 0.50614499 -0.9861506[2,] -0.37108354 -1.6249737[3,] 0.08650475 -0.6949853
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