hhh - 1 year ago 65

R Question

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

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.

**Trial 1**

`set.seed(10)`

t(matrix(rnorm(16),ncol=4,nrow=4)) -> a

for(i in 1:2){a <- a %*% a}

`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)

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Answer Source

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
```

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 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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