Hi John, As I mentioned in our private exchange, this is well known in R, i.e. vectorized versions are not always faster or more efficient than straight loops. It is a misconception that loops should be avoided at any cost. See John Fox's illuminating article on Rnews (p. 46) on this subject. http://cran.r-project.org/doc/Rnews/Rnews_2008-1.pdf To me, unless the application is way too demanding, I take the vectorized version over the loop any day, as it is much simpler to write, usually a one-liner, and therefore much easier to maintain in the long run. -Christos> -----Original Message----- > From: r-help-bounces at r-project.org > [mailto:r-help-bounces at r-project.org] On Behalf Of john > Sent: Thursday, October 09, 2008 12:38 AM > To: r-help > Subject: Re: [R] vectorized sub, gsub, grep, etc. > > 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-help at r-project.org 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. > >
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
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... at 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. > >