I know this is an old thread, but I wrote a simple FOR loop with vectorized
pattern replacement that is much faster than either of those (it can also
accept outputs differing in length from the patterns):
sub2 <- function(pattern, replacement, x) {
len <- length(x)
y <- character(length=len)
patlen <- length(pattern)
replen <- length(replacement)
if(patlen != replen) stop('Error: Pattern and replacement length do not
match')
for(i in 1:replen) {
y[which(x==pattern[i])] <- replacement[i]
}
return(y)
}
system.time(test <- sub2(patt, repl, XX))
user system elapsed
0 0 0
Cheers,
Adam
On Wednesday, October 8, 2008 at 9:38:01 PM UTC-7, john wrote:
>
> Hello Christos,
> To my surprise, vectorization actually hurt processing speed!
>
> #Example
> X <- c("ab", "cd", "ef")
> patt <- c("b", "cd", "a")
> repl <- c("B", "CD", "A")
>
> sub2 <- function(pattern, replacement, x) {
> len <- length(x)
> if (length(pattern) == 1)
> pattern <- rep(pattern, len)
> if (length(replacement) == 1)
> replacement <- rep(replacement, len)
> FUN <- function(i, ...) {
> sub(pattern[i], replacement[i], x[i], fixed = TRUE)
> }
> idx <- 1:length(x)
> sapply(idx, FUN)
> }
>
> system.time( for(i in 1:10000) sub2(patt, repl, X) )
> user system elapsed
> 1.18 0.07 1.26
>
> system.time( for(i in 1:10000) mapply(function(p, r, x) sub(p, r, x,
> fixed = TRUE), p=patt, r=repl, x=X) )
> user system elapsed
> 1.42 0.05 1.47
>
> So much for avoiding loops.
> John Thaden
>
> ======= At 2008-10-07, 14:58:10 Christos wrote: =======
>
> >John,
> >Try the following:
> >
> > mapply(function(p, r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl, x=X)
> > b cd a
> >"aB" "CD" "ef"
> >
> >-Christos
>
> >> -----My Original Message-----
> >> R pattern-matching and replacement functions are
> >> vectorized: they can operate on vectors of targets.
> >> However, they can only use one pattern and replacement.
> >> Here is code to apply a different pattern and replacement for
> >> every target. My question: can it be done better?
> >>
> >> sub2 <- function(pattern, replacement, x) {
> >> len <- length(x)
> >> if (length(pattern) == 1)
> >> pattern <- rep(pattern, len)
> >> if (length(replacement) == 1)
> >> replacement <- rep(replacement, len)
> >> FUN <- function(i, ...) {
> >> sub(pattern[i], replacement[i], x[i], fixed = TRUE)
> >> }
> >> idx <- 1:length(x)
> >> sapply(idx, FUN)
> >> }
> >>
> >> #Example
> >> X <- c("ab", "cd", "ef")
> >> patt <- c("b", "cd", "a")
> >> repl <- c("B", "CD", "A")
> >> sub2(patt, repl, X)
> >>
> >> -John
>
> ______________________________________________
> r-h...@r-project.org <javascript:> mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
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