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speed up this algorithm (apply-fuction / 4D array)

3 messages · Martin Batholdy, William Dunlap, Claudia Beleites

Martin Batholdy
# 
Hi,


I have this sample-code (see above) and I was wondering wether it is possible to speed things up.



What this code does is the following:

x is 4D array (you can imagine it as x, y, z-coordinates and a time-coordinate).

So x contains 50x50x50 data-arrays for 91 time-points.

Now I want to reduce the 91 time-points.
I want to merge three consecutive time points to one time-points by calculating the mean of this three time-points for every x,y,z coordinate.

The reduce-sequence defines which time-points should get merged.
And the apply-function in the for-loop calculates the mean of the three 3D-Arrays and puts them into a new 4D array (data_reduced).



The problem is that even in this example it takes really long.
I thought apply would already vectorize, rather than loop over every coordinate.

But for my actual data-set it takes a really long time ? So I would be really grateful for any suggestions how to speed this up.




x <- array(rnorm(50 * 50 * 50 * 90, 0, 2), dim=c(50, 50, 50, 91))



data_reduced <- array(0, dim=c(50, 50, 50, 90/3))

reduce <- seq(1,90, 3)



for( i in 1:length(reduce) ) {

	data_reduced[ , , , i]    <-    apply(x[ , , , reduce[i] : (reduce[i]+3) ], 1:3, mean) 
}
William Dunlap
# 
I corrected your code a bit and put it into a function, f0, to
make testing easier.  I also made a small dataset to make
testing easier.  Then I made a new function f1 which does
what f0 does in a vectorized manner:

  x <- array(rnorm(50 * 50 * 50 * 91, 0, 2), dim=c(50, 50, 50, 91))
  xsmall <- array(log(seq_len(2 * 2 * 2 * 91)), dim=c(2, 2, 2, 91))

  f0 <- function(x) {
      data_reduced <- array(0, dim=c(dim(x)[1:3], trunc(dim(x)[4]/3)))
      reduce <- seq(1, dim(x)[4]-1, by=3)
      for( i in 1:length(reduce) ) {
          data_reduced[ , , , i] <- apply(x[ , , , reduce[i] : (reduce[i]+2) ], 1:3, mean)
     }
     data_reduced
  }

  f1 <- function(x) {
     reduce <- seq(1, dim(x)[4]-1, by=3)
     data_reduced <- (x[, , , reduce] + x[, , , reduce+1] + x[, , , reduce+2]) / 3
     data_reduced
  }

The results were:

  > system.time(v1 <- f1(x))
     user  system elapsed
    0.280   0.040   0.323
  > system.time(v0 <- f0(x))
     user  system elapsed
   73.760   0.060  73.867
  > all.equal(v0, v1)
  [1] TRUE
No, you have that backwards.  Use *apply functions when you cannot figure
out how to vectorize.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Claudia Beleites
# 
here's another one - which is easier to generalize:

x <- array(rnorm(50 * 50 * 50 * 91, 0, 2), dim=c(50, 50, 50, 91))
y <- x [,,,1:90] # decide yourself what to do with slice 91, but
                  # 91 is not divisible by 3
system.time ({
dim (y) <- c (50, 50, 50, 3, 90 %/% 3)
y <- aperm (y, c (4, 1:3, 5))
v2 <- colMeans (y)
})
        User      System verstrichen
        0.32        0.08        0.40

(my computer is a bit slower than Bill's:)
 > system.time (v1 <- f1 (x))
        User      System verstrichen
       0.360       0.030       0.396

Claudia


Am 05.10.2011 20:24, schrieb William Dunlap: