> -----Original Message-----
> From: William Dunlap
> Sent: Thursday, December 04, 2008 9:59 AM
> To: '[EMAIL PROTECTED]'
> Cc: 'R help'
> Subject: Re: [R] How to optimize this codes ?
>
> [R] How to optimize this codes ?
> Daren Tan daren76 at hotmail.com
> Thu Dec 4 17:02:49 CET 2008
>
> How to optimize the for-loop to be reasonably fast for
> sample.size=100000000 ?
> You may want to change sample.size=1000 to have an idea
> what I am achieving.
>
> set.seed(143)
> A <- matrix(sample(0:1, sample.size, TRUE), ncol=10,
> dimnames=list(NULL, LETTERS[1:10]))
>
> B <- list()
> for(i in 1:10) {
> B[[i]] <- apply(combn(LETTERS[1:10], i), 2, function(x)
> { sum(apply(data.frame(A[,x]), 1, all)) })
> }
>
> Instead computing all(A[row,x]) each row of the matrix by looping
> over the rows you could loop over the columns. That is generally
> quicker if there are many more rows than columns. S+ and R have
> functions called pmin and pmax that do this for min and max, but
> no pany or pall functions. In your 0/1 case all is the same as min
> so you can replace
> apply(data.frame(A[,x]), 1, all)
> with
> sum(do.call("pmin", unname(A[,x,drop=FALSE])))
> after first converting A to a data.frame before starting to compute B.
> (The do.call/unname combo is a hack to pass all the columns
> of a data.frame
> as separate arguments to a function. If pmin took a list
> that wouldn't
> be necessary.)
>
> When you do timings, looking at one size of problem often
> isn't helpful,
> as it doesn't tell you how the time depends on the size of
> the input. The
> relationship often is not linear. E.g., I put your method in
> a function
> called computeB0 and my modification of it into computeB1 and
> wrote a makeA
> that makes an A matrix with a given sample.size. Here are
> the times, it looks
> like 1000 is below the linear range for this example:
>
> sample.size time:computeB0 time:computeB1
> 1000 5.36 0.98
> 10000 36.82 2.49
> 100000 381.34 18.97
>
> and here are the functions used
>
> makeA <-
> function(sample.size){
> set.seed(143);
> A<-matrix(sample(0:1,size=sample.size,replace=TRUE),
> ncol=10, dimnames=list(NULL, LETTERS[1:10]));
> A
> }
> computeB0 <-
> function(A){B<-list()
> for(i in 1:10) {
> B[[i]] <- apply(combn(LETTERS[1:10], i), 2, function(x) {
> sum(apply(data.frame(A[,x]), 1, all)) })
> }
> B
> }
> computeB1 <-
> function(A){
> A<-as.data.frame(A)
> B<-list()
> for(i in 1:10) {
> B[[i]] <- apply(combn(LETTERS[1:10], i), 2,
> FUN=function(x) { sum(do.call("pmin",
> unname(A[,x,drop=FALSE]))) }
> )
> }
> B
> }
>
> You could probably same more time by restructuring this as a
> recursive function,
> still operating on columns, that traversed the binary tree of
> column inclusion/exclusion.
> I just wanted to point out the advantage of looping over
> columns instead of looping over
> rows.
A hastily written recursive version is:
computeB3 <-
function(A) {
B <- rep(list(integer(0)), ncol(A))
storage.mode(A) <- "logical"
recurse <- function(thisCol = 1, includedCols = rep(NA, ncol(A)),
allSoFar = rep(TRUE, nrow(A))) {
if (thisCol < ncol(A)) # skip this column
Recall(thisCol+1, replace(includedCols, thisCol, FALSE),
allSoFar = allSoFar)
else if (thisCol > ncol(A))
return()
includedCols[thisCol] <- TRUE
allSoFar <- allSoFar & A[,thisCol]
nIncludedCols <- sum(includedCols[1:thisCol])
# cat(thisCol, ":", nIncludedCols, ":", includedCols[1:thisCol],
sum(allSoFar), "\n")
# Note B<<- in next line uses lexical scoping to change
computeB3:B. Does not work in S+.
B[[nIncludedCols]][length(B[[nIncludedCols]])+1] <<- sum(allSoFar)
Recall(thisCol+1, includedCols, allSoFar = allSoFar)
}
recurse()
B
}
The elements of B[[i]], for each i, are in a different order than they
are in
the other functions, but the histograms are the same.
I added its time to the above table:
sample.size time:computeB0 time:computeB1 time:computeB3
1000 5.36 0.98 0.12
10000 36.82 2.49 0.25
100000 381.34 18.97 1.62
1000000 ? 290.21 16.36
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
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