R中矩阵乘法的A ^ k? [英] A^k for matrix multiplication in R?

问题描述

假设 A 是一些正方形矩阵.如何在R中轻松对这个矩阵求幂?

Suppose A is some square matrix. How can I easily exponentiate this matrix in R?

我已经尝试了两种方法:带有for-loop hack的Trial 1和Trial 2更加优雅，但是与 A ^k 简单性相去甚远.

I tried two ways already: Trial 1 with a for-loop hack and Trial 2 a bit more elegantly but it is still a far cry from A^k simplicity.

试验1

set.seed(10)
t(matrix(rnorm(16),ncol=4,nrow=4)) -> a 
for(i in 1:2){a <- a %*% a}

审判2

a <- t(matrix(c(0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0),nrow=4))
i <- diag(4) 
(function(n) {if (n<=1) a else (i+a) %*% Recall(n-1)})(10)

推荐答案

尽管Reduce更为优雅，但是for循环解决方案速度更快，并且似乎与expm ::％^％

Although Reduce is more elegant, a for-loop solution is faster and seems to be as fast as expm::%^%

m1 <- matrix(1:9, 3)
m2 <- matrix(1:9, 3)
m3 <- matrix(1:9, 3)
system.time(replicate(1000, Reduce("%*%" , list(m1,m1,m1) ) ) )
#   user  system elapsed 
#  0.026   0.000   0.037 
mlist <- list(m1,m2,m3)
m0 <- diag(1, nrow=3,ncol=3)
system.time(replicate(1000, for (i in 1:3 ) {m0 <- m0 %*% m1 } ) )
#   user  system elapsed 
#  0.013   0.000   0.014 

library(expm)  # and I think this may be imported with pkg:Matrix
system.time(replicate(1000, m0%^%3))
# user  system elapsed 
#0.011   0.000   0.017 

另一方面，matrix.power解决方案要慢得多:

On the other hand the matrix.power solution is much, much slower:

system.time(replicate(1000, matrix.power(m1, 4)) )
   user  system elapsed 
  0.677   0.013   1.037 

@BenBolker是正确的(再次).随着指数的增加，for循环在时间上呈现线性关系，而expm ::％^％函数似乎甚至比log(exponent)更好.

@BenBolker is correct (yet again). The for-loop appears linear in time as the exponent rises whereas the expm::%^% function appears to be even better than log(exponent).

> m0 <- diag(1, nrow=3,ncol=3)
> system.time(replicate(1000, for (i in 1:400 ) {m0 <- m0 %*% m1 } ) )
   user  system elapsed 
  0.678   0.037   0.708 
> system.time(replicate(1000, m0%^%400))
   user  system elapsed 
  0.006   0.000   0.006 

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