Re: [R] RfW 2.3.1: regular expressions to detect pairs of identical word-final character sequences

From: Greg Snow <Greg.Snow_at_intermountainmail.org>
Date: Wed 26 Jul 2006 - 05:38:50 EST


Before comparing times we should make sure that they functions return the same thing. My original function (f1 below) labels the potential rymes with match numbers as well as finding possible rymes, if you just want the <r> flag then the for loop can be eliminated giving f4 as follows:

 f4 <- function(text) {

	tmp1 <- strsplit(text, ' ')[[1]]
	tmp2 <- nchar(tmp1)
	tmp3 <- substr(tmp1,tmp2-1,tmp2)

	tmp4 <- which(lower.tri(diag(length(tmp3))), arr.ind=TRUE)
	tmp5 <- tmp3[ tmp4[,1] ] == tmp3[ tmp4[,2] ]

	tmp6 <- rep('', length(tmp1))
	tmp6[ unique(c(tmp4[tmp5,])) ] <- '<r>'
	paste( tmp1,tmp6, sep='',collapse=' ') }

The speed of f4 is similar to the speed of f3 (even after correcting f3, the original one just returns the original text string).

But that is on the sample string, what if a longer string is used (more potential for backtracking).

Try the string generated by:

set.seed(1)
text <- paste( sample(c(letters,' ',' ',' '), 1000, replace=T), collapse='')
text <- gsub(" {2,}"," ",text)

Now f4 is much faster than f3. However f3 can be optimized by replacing \\w+ in pat by \\w{2} and that makes it faster than f4 again

It would probably be even faster to use gregexpr to just find the matching endings then create the new regexp based on those endings and do one substitute rather than using multiple gsubs.

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow@intermountainmail.org
(801) 408-8111
 

-----Original Message-----
From: Gabor Grothendieck [mailto:ggrothendieck@gmail.com] 
Sent: Tuesday, July 25, 2006 11:41 AM
To: Greg Snow
Cc: Stefan Th. Gries; r-help@stat.math.ethz.ch
Subject: Re: [R] RfW 2.3.1: regular expressions to detect pairs of
identical word-final character sequences

Regarding having to do a lot of backtracking one can just look at the
relative comparison of speeds and we see that they are comparable in
speed.

In fact the bottleneck is not the backtacking but strapply.
I had coded the regexp version for compactness of code but if we replace
the strapply with custom gsub/strapply code for speed, the new rexexp
version is twice as fast as the for loop version.

Below f1 is the for loop version, f2 is the original regexp version with
strapply and f3 is the revised version using gsub/strsplit instead.

f1 <- function() {
	tmp1 <- strsplit(text, ' ')[[1]]
	tmp2 <- nchar(tmp1)
	tmp3 <- substr(tmp1,tmp2-1,tmp2)

	tmp4 <- which(lower.tri(diag(length(tmp3))), arr.ind=TRUE)
	tmp5 <- tmp3[ tmp4[,1] ] == tmp3[ tmp4[,2] ]

	tmp6 <- rep('', length(tmp1))
	count <- 1
	for( i in which(tmp5) ){
	       tmp6[ tmp4[i,1] ] <- paste(tmp6[ tmp4[i,1] ],
	'<r',count,'>',sep='')
	       tmp6[ tmp4[i,2] ] <- paste(tmp6[ tmp4[i,2] ],
	'<r',count,'>',sep='')
	       count <- count + 1
	}

	out.text <- paste( tmp1,tmp6, sep='',collapse=' ') }


# places <...> around first occurrences of repeated suffixes
library(gsubfn) f2 <- function() { text <- "And this is the second sentence" pat <- "(\\w+)(?=\\b.+\\1\\b)" # pat <- "(\\w\\w+)(?=\\b.+\\1\\b)" out <- gsub(pat, "\\<\\1\\>", text, perl = TRUE) suff <- strapply(out, "<([^>]+)>", function(x,y)y)[[1]] gsub(paste("(", paste(suff, collapse = "|"), ")\\b", sep = ""), "\\1<r>", text) } f3 <- function() { text <- "And this is the second sentence" pat <- "(\\w+)(?=\\b.+\\1\\b)" # pat <- "(\\w\\w+)(?=\\b.+\\1\\b)" out <- gsub(pat, "\\<\\1\\>", text, perl = TRUE) # redo this strapply by hand for speed purposes # suff <- strapply(out, "<([^>]+)>", function(x,y)y)[[1]] suff <- gsub("[^<>]*<|>[^<>]*<|>[^<>]*$", "<", out) suff <- gsub("^<|<$", "", suff) suff <- strsplit(suff, "<")[[1]] gsub(paste("(", paste(suff, collapse = "|"), ")\\b", sep = ""), "\\1<r>", text) }
# for loop version
system.time(for (i in 1:100) f1()) # 0.32 0.00 0.36 NA NA
# original regexp version with strapply
system.time(for (i in 1:100) f2()) # 0.36 0.00 0.38 NA NA
# regexp version with strapply replaced with gsub/strsplit
system.time(for (i in 1:100) f3()) # 0.15 0.00 0.16 NA NA On 7/25/06, Greg Snow <Greg.Snow@intermountainmail.org> wrote: > Using regular expression matching for this case may be overkill (the > RE engine will be doing a lot of backtracking looking at a lot of > non-matches). Here is an alternative that splits the text into a > vector of words, extracts the last 2 letters of each word (remember if > the last > 3 letters match, then the last 2 have to match, so we only need to > consider the last 2), then looks at all pairwise comparisons for > matches, then pastes everything back together with the marked matches: > > text<-"And this is a second rand sentence" > > tmp1 <- strsplit(text, ' ')[[1]] > tmp2 <- nchar(tmp1) > tmp3 <- substr(tmp1,tmp2-1,tmp2) > > tmp4 <- which(lower.tri(diag(length(tmp3))), arr.ind=TRUE) > tmp5 <- tmp3[ tmp4[,1] ] == tmp3[ tmp4[,2] ] > > tmp6 <- rep('', length(tmp1)) > count <- 1 > for( i in which(tmp5) ){ > tmp6[ tmp4[i,1] ] <- paste(tmp6[ tmp4[i,1] ], > '<r',count,'>',sep='') > tmp6[ tmp4[i,2] ] <- paste(tmp6[ tmp4[i,2] ], > '<r',count,'>',sep='') > count <- count + 1 > } > > out.text <- paste( tmp1,tmp6, sep='',collapse=' ') > > > If you are doing a lot of text processing like this, I would suggest > doing it in Perl rather than R. S Poetry by Dr. Burns has a function > to take a vector of character strings in R and run a Perl script on it > and return the results. > > Hope this helps, > > > > > -- > Gregory (Greg) L. Snow Ph.D. > Statistical Data Center > Intermountain Healthcare > greg.snow@intermountainmail.org > (801) 408-8111 > > > -----Original Message----- > From: r-help-bounces@stat.math.ethz.ch > [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Stefan Th. > Gries > Sent: Saturday, July 22, 2006 7:49 PM > To: r-help@stat.math.ethz.ch > Subject: [R] RfW 2.3.1: regular expressions to detect pairs of > identical word-final character sequences > > Dear all > > I use R for Windows 2.3.1 on a fully updated Windows XP Home SP2 > machine and I have two related regular expression problems. > > platform i386-pc-mingw32 > arch i386 > os mingw32 > system i386, mingw32 > status > major 2 > minor 3.1 > year 2006 > month 06 > day 01 > svn rev 38247 > language R > version.string Version 2.3.1 (2006-06-01) > > > I would like to find cases of words in elements of character vectors > that end in the same character sequences; if I find such cases, I want > to add <r> to both potentially rhyming sequences. An example: > > INPUT:This is my dog. > DESIRED OUTPUT: This<r> is<r> my dog. > > I found a solution for cases where the potentially rhyming words are > adjacent: > > text<-"This is my dog." > gsub("(\\w+?)(\\W\\w+?)\\1(\\W)", "\\1<r>\\2\\1<r>\\3", text, > perl=TRUE) > > However, with another text vector, I came across two problems I cannot > seem to solve and for which I would love to get some input. > > (i) While I know what to do for non-adjacent words in general > > gsub("(\\w+?)(\\W.+?)\\1(\\W)", "\\1<r>\\2\\1<r>\\3", "This not is my > dog", perl=TRUE) # I know this is not proper English ;-) > > this runs into problems with overlapping matches: > > text<-"And this is the second sentence" > gsub("(\\w+?)(\\W.+?)\\1(\\W)", "\\1<r>\\2\\1<r>\\3", text, perl=TRUE) > [1] "And<r> this is the second<r> sentence" > > It finds the "nd" match, but since the "is" match is within the two > "nd"'s, it doesn't get it. Any ideas on how to get all pairwise matches? > > (ii) How would one tell R to match only when there are 2+ characters > matching? If the above expression is applied to another character > string > > text<-"this is an example sentence." > gsub("(\\w+?)(\\W.+?)\\1(\\W)", "\\1<r>\\2\\1<r>\\3", text, perl=TRUE) > > it also matches the "e"'s at the end of example and sentence. It's not > possible to get rid of that by specifying a range such as {2,} > > text<-"this is an example sentence." > gsub("(\\w{2,}?)(\\W.+?)\\1(\\W)", "\\1<r>\\2\\1<r>\\3", text, > perl=TRUE) > > because, as I understand it, this requires the 2+ cases of \\w to be > identical characters: > > text<-"doo yoo see mee?" > gsub("(\\w{2,}?)(\\W.+?)\\1(\\W)", "\\1<r>\\2\\1<r>\\3", text, > perl=TRUE) > > Again, any ideas? > > I'd really appreciate any snippets of codes, pointers, etc. > Thanks so much, > STG > -- > Stefan Th. Gries > ----------------------------------------------- > University of California, Santa Barbara > http://www.linguistics.ucsb.edu/faculty/stgries > > ______________________________________________ > R-help@stat.math.ethz.ch 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. > > ______________________________________________ > R-help@stat.math.ethz.ch 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. > ______________________________________________ R-help@stat.math.ethz.ch 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.
Received on Wed Jul 26 05:45:24 2006

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