[R] Matrix mulitplication

[Spencer Graves]

      One can also use "crossprod" AND use "solve" to actually "solve" 
the system of linear equations rather than just get the inverse, which 
is later multiplied by t(BA)%*%D.  However, the difference seems very 
small: 

 > set.seed(1)
 > n <- 500
 > A <- array(rnorm(n^2), dim=c(n,n))
 > B <- array(rnorm(n^2), dim=c(n,n))
 > C. <- array(rnorm(n^2), dim=c(n,n))
 > D <- array(rnorm(n^2), dim=c(n,n))
 >
 > BA <- B%*%A
 >
 > start.time <- proc.time()
 > A1 <- A%*%solve(t(BA)%*%BA+C.)%*%BA%*%D
 > proc.time()-start.time
[1] 4.75 0.03 5.13   NA   NA
 >
 > start.time <- proc.time()
 > A2 <- A%*%solve(crossprod(BA)+C., crossprod(t(BA), D))
 > proc.time()-start.time
[1] 4.19 0.01 4.49   NA   NA
 > all.equal(A1, A2)
[1] TRUE

      This was in R 1.8.1 under Windows 2000 on an IBM Thinkpad T30 with 
a Mobile Intel Pentium 4-M, 1.8Ghz, 1Gbyte RAM.  The same script under 
S-Plus 6.2 produced the following elapsed times: 

[1] 3.325 0.121 3.815 0.000 0.000

[1] 2.934 0.070 3.355 0.000 0.000

      Thus, roughly, using "crossprod" plus "solving with solve" gave a 
10% speed improvement and S-Plus 6.2 gave a 25% speed improvement in 
this one small benchmark.  Using smaller matrices did not show as big a 
difference just because the time was too short to measure accurately, I 
think. 

      Enough trivia for now. 
      Best Wishes,
      spencer graves

(Ted Harding) wrote:

>On 16-Feb-04 Peter Dalgaard wrote:
>  
>
>>(Ted Harding) <Ted.Harding at nessie.mcc.ac.uk> writes:
>>
>>    
>>
>>>Sorry! Missed a trick here:
>>>
>>>    At <- t(A)
>>>    Bt <- t(B)
>>>    E  <- B%*%A
>>>    Et <- t(E)
>>>    A%*%solve(Et%*%E + C)%*%Et%*%D
>>>
>>>(saves 2 multiplications at the relatively cheap cost of 1 transpose)
>>>      
>>>
>>Well, you might consider getting rid of the first two transposes since
>>you're not actually using them for anything....
>>    
>>
>
>Touché ... !
>Ted.
>
>
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>Date: 16-Feb-04                                       Time: 23:41:57
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