[R] vectorized sub, gsub, grep, etc.
John Thaden
jjthaden at flash.net
Sat Aug 1 08:15:42 CEST 2015
Adam,
You reopened an old thread noting its age, but did you begin at its beginning?
> Subject: vectorized sub, gsub, grep, etc.> Date: Oct 7, 2008
> 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)
If you run that code, you'll see the correct answer is not "" "CD" "", it is
[1] "aB" "CD" "ef"
And the same answer is given by the shorter (but slower) code suggested later that day by Christos
mapply(function(p, r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl, x=X)
b cd a
"aB" "CD" "ef"
By talking instead about simple string matching, I'm afraid you've rather hijacked the thread.
-John
-John
Adam wrote
> I'm not sure I understand your question. Both functions return "" "CD" "" because they
> perform exact string matching. The first demonstrates how string or character replacements
> can be vectorized, while the second merely demonstrates how Rcpp can accelerate this type of operation.
On Jul 30, 2015, at 21:09, John Thaden <jjthaden at flash.net> wrote:
| Can you show what is its solution for the original sample data? Why that discrepancy for you original sub2() function?
| From:"Adam Erickson" <adam.michael.erickson at gmail.com>
Date:Thu, Jul 30, 2015 at 6:11 pm
Subject:Re: [R] vectorized sub, gsub, grep, etc.
Here is a Rcpp version for exact character matching (for example) written in C++ that is substantially faster. Hence, I think this is the way to go where loops may be unavoidable. However, the input vector length has to match the length of the pattern and replacement vectors, as your original code did. That can be changed though.
#include <Rcpp.h>using namespace Rcpp;
// [[Rcpp::export]]CharacterVector subCPP(CharacterVector pattern, CharacterVector replacement, CharacterVector x) { int len = x.size(); CharacterVector y(len); int patlen = pattern.size(); int replen = replacement.size(); if (patlen != replen) Rcout<<"Error: Pattern and replacement length do not match"; for(int i = 0; i < patlen; ++i) { if (*(char*)x[i] == *(char*)pattern[i]) y[x[i] == pattern[i]] = replacement[i]; } return y;}
"" "CD" ""
system.time(for(i in 1:50000) subCPP(patt, repl, X)) user system elapsed 0.16 0.00 0.16
Cheers,
Adam
On Wednesday, July 29, 2015 at 2:42:23 PM UTC-7, Adam Erickson wrote:
Further refining the vectorized (within a loop) exact string match function, I get times below 0.9 seconds while maintaining error checking. This is accomplished by removing which() and replacing 1:length() with seq_along().
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 seq_along(pattern)) { y[x==pattern[i]] <- replacement[i] } return(y) }
system.time(for(i in 1:50000) sub2(patt, repl, X)) user system elapsed 0.86 0.00 0.86
Since the ordered vectors are perfectly aligned, might as well do an exact string match. Hence, I think this is not off-topic.
Cheers,
Adam
On Wednesday, July 29, 2015 at 8:15:52 AM UTC-7, Bert Gunter wrote:
There is confusion here. apply() family functions are **NOT**
vectorization -- they ARE loops (at the interpreter level), just done
in "functionalized" form. Please read background material (John
Chambers's books, MASS, or numerous others) to improve your
understanding and avoid posting erroneous comments.
Cheers,
Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
-- Clifford Stoll
On Tue, Jul 28, 2015 at 3:00 PM, John Thaden <jjth... at flash.net> wrote:
> Adam, The method you propose gives a different result than the prior methods for these example vectors
> X <- c("ab", "cd", "ef")
> patt <- c("b", "cd", "a")
> repl <- c("B", "CD", "A")
>
> Old method 1
>
> mapply(function(p, r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl, x=X)
> gives
> b cd a
> "aB" "CD" "ef"
>
> Old method 2
>
> 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)
> }
> sub2(patt, repl, X)
> gives
> [1] "aB" "CD" "ef"
>
> Your method (I gave it the unique name "sub3")
> sub3 <- 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)}sub3(patt, repl, X)
> gives[1] "" "CD" ""
>
> Granted, whatever it does, it does it faster
> #Old method 1
> system.time(for(i in 1:50000)
> mapply(function(p,r,x) sub(p,r,x, fixed = TRUE),p=patt,r=repl,x=X))
> user system elapsed
> 2.53 0.00 2.52
>
> #Old method 2
> system.time(for(i in 1:50000)sub2(patt, repl, X)) user system elapsed
> 2.32 0.00 2.32
>
> #Your proposed method
> system.time(for(i in 1:50000) sub3(patt, repl, X))
> user system elapsed
> 1.02 0.00 1.01
> but would it still be faster if it actually solved the same problem?
>
> -John Thaden
>
>
>
>
> On Monday, July 27, 2015 11:40 PM, Adam Erickson <adam.micha... at gmail.com> wrote:
>
> 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_________________________ _____________________
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