R右矩阵除法 [英] R right matrix division

问题描述

在R中进行左右矩阵除法的最简洁，最快，最数值稳定，最R惯用的方法是什么?我知道通常用solve(a,b)进行左除法inv(A)*B，但是B*inv(A)怎么样?真的是计算t(solve(t(A),t(B)))的最佳方法吗?

What's the most succinct, fastest, most numerically stable, most R-idiomatic way to do left and right matrix division in R? I understand left division inv(A)*B is usually done with solve(a,b), but how about B*inv(A)? Is the best way really to compute t(solve(t(A),t(B)))?

推荐答案

我没有比B %*% solve(A)更好的解决方案，但是我确实想指出，通常solve(A,B)比[c5]更快，并且在数值上更稳定solve(A) %*% B.

I don't have a solution better than B %*% solve(A), but I did want to point out that in general solve(A,B) is faster and more numerically stable than solve(A) %*% B.

> A = matrix(rnorm(10000),100,100)
> B = matrix(rnorm(10000),100,100)
> microbenchmark(solve(A,B), solve(A) %*% B, t(solve(t(B),t(A))), B %*% solve(A))
Unit: microseconds
             expr     min       lq      mean   median       uq       max neval
      solve(A, B)     481.695 604.2435  722.2512 677.2455  761.735  1280.888   100
   solve(A) %*% B     628.243 830.2095 1056.3947 927.0130 1204.682  5275.030   100
t(solve(t(B), t(A)))  603.855 792.1360 1164.7210 924.0895 1122.184 10351.307   100
   B %*% solve(A)     645.119 784.1990 1070.4751 927.9400 1097.601  7866.591   100

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