[R] efficiently replacing values in a matrix

[Matthew Keller]

Nanmdi,

I think this is simply because a lot of time is taken transforming the
matrix from logical (default when you create it) to numeric (when you
add the number to [1,1]. If you do the same thing again to [1,2], it
is done instantaneously:
> a <- matrix(nrow=10000,ncol=10000)
> system.time(a[1,1] <- 1903908.80385)
   user  system elapsed
 10.000   0.781  10.755
>
> system.time(a[1,2] <- 1903908.80385)
   user  system elapsed
      0       0       0

By the way Rolf - I didn't see your full response last time through. I
do know the difference between a matrix and a data.frame, thank you
very much.


On Fri, Apr 18, 2008 at 11:11 AM, Nnamdi <nnamdii at yahoo.com> wrote:
>
>  Still it is pretty slow when entering values into a large matrix. Case in
>  point:
>
>  > a <- matrix(nrow=10000,ncol=10000)
>  > system.time(a[1,1] <- 1903908.80385)
>    user  system elapsed
>   30.840   6.226  41.416
>  > is.matrix(a)
>  [1] TRUE
>
>  Is there a better way to enter values into large matrices? If I have to
>  spend 41 secs each time I enter into a cell and I have 10000x10000 cells to
>  enter that is impractical!
>
>  --Nnamdi
>
>
>  Rolf Turner-3 wrote:
>  >
>  >
>  > On 17/04/2008, at 9:33 AM, Charles C. Berry wrote:
>  >
>  >       <snip>
>  >
>  >> I'll lay odds that Matthew's 'matrix' is actually a data.frame, and
>  >> I'll not be surprised if the columns are factors.
>  >
>  >       <snip>
>  >
>  > I suspect that you're right.
>  >
>  > ***Why*** can't people distinguish between data frames and matrices?
>  > If they were the same <expletive deleted> thing, there wouldn't be two
>  > different terms for them, would there?
>  >
>  >       cheers,
>  >
>  >               Rolf Turner
>  >
>  > ######################################################################
>
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-- 
Matthew C Keller
Asst. Professor of Psychology
University of Colorado at Boulder
www.matthewckeller.com