Re: [R] regular expression, stringr::str_view, grep

This highlights the literal meaning of the last ] in your correct_brackets: aff <- c("affgfk]ing", "fgok", "rafgkah]e","a fgk", "bafghk]") To me, too, the missing_brackets looks more like what was desired, and returns correct results for a PCRE. Perhaps the regular expression should have been

Re: [R] regular expression, stringr::str_view, grep

On 4/28/20 2:29 AM, Sigbert Klinke wrote: Hi, we gave students the task to construct a regular expression selecting some texts. One send us back a program which gives different results on stringr::str_view and grep. The problem is "[^[A-Z]]" / "[^[A-Z]" at the end of the regular

[R] regular expression, stringr::str_view, grep

Hi, we gave students the task to construct a regular expression selecting some texts. One send us back a program which gives different results on stringr::str_view and grep. The problem is "[^[A-Z]]" / "[^[A-Z]" at the end of the regular expression. I would have expected that all four calls

Re: [R] Regular expression help

> On Oct 9, 2017, at 6:08 PM, Georges Monette wrote: > > How about this (I'm showing it as a pipe because it's easier to read that > way): > > library(magrittr) > "f 147/1315/587 2820/1320/587 3624/1321/587 1852/1322/587" %>% > strsplit(' ') %>% > unlist %>% >

Re: [R] Regular expression help

How about this (I'm showing it as a pipe because it's easier to read that way): library(magrittr) "f 147/1315/587 2820/1320/587 3624/1321/587 1852/1322/587" %>%   strsplit(' ') %>%   unlist %>%   sub('^[^/]*/*','',.) %>%   sub('^[^/]*/*','',.) %>%   paste(collapse = ' ') Georges Monette --

Re: [R] Regular expression help

On 09/10/2017 12:06 PM, William Dunlap wrote: "(^| +)([^/ ]*/?){0,2}", with the first "*" replaced by "+" would be a bit better. Thanks! I think I actually need the *, because theoretically the b part of the word could be empty, i.e. "a//c" would be legal and should become "c". Duncan

Re: [R] Regular expression help

On 09/10/2017 11:23 AM, Ulrik Stervbo wrote: Hi Duncan, why not split on / and take the correct elements? It is not as elegant as regex but could do the trick. Thanks for the suggestion. There are likely many thousands of lines of data like the two real examples (which had about 5000 and

Re: [R] Regular expression help

"(^| +)([^/ ]*/?){0,2}", with the first "*" replaced by "+" would be a bit better. Bill Dunlap TIBCO Software wdunlap tibco.com On Mon, Oct 9, 2017 at 8:50 AM, William Dunlap wrote: > > x <- "f 147/1315/587 2820/1320/587 3624/1321/587 1852/1322/587" > > gsub("(^| *)([^/

Re: [R] Regular expression help

> x <- "f 147/1315/587 2820/1320/587 3624/1321/587 1852/1322/587" > gsub("(^| *)([^/ ]*/?){0,2}", "\\1", x) [1] " 587 587 587 587" > y <- "aa aa/ aa/bb aa/bb/ aa/bb/cc aa/bb/cc/ aa/bb/cc/dd aa/bb/cc/dd/" > gsub("(^| *)([^/ ]*/?){0,2}", "\\1", y) [1] "cc cc/ cc/dd cc/dd/" Bill Dunlap TIBCO

Re: [R] Regular expression help

> On 9 Oct 2017, at 17:02 , Duncan Murdoch wrote: > > I have a file containing "words" like > > > a > > a/b > > a/b/c > > where there may be multiple words on a line (separated by spaces). The a, b, > and c strings can contain non-space, non-slash characters.

Re: [R] Regular expression help

Hi Duncan, You can try this: library(readr) f <- function(s) { t <- unlist(readr::tokenize(paste0(gsub(" ",",",s),"\n",collapse=""))) i <- grep("[a-zA-Z0-9]*/[a-zA-Z0-9]*/",t) u <- sub("[a-zA-Z0-9]*/[a-zA-Z0-9]*/","",t[i]) paste0(u,collapse=" ") } f("f 147/1315/587 2820/1320/587

Re: [R] Regular expression help

Hi Duncan, why not split on / and take the correct elements? It is not as elegant as regex but could do the trick. Best, Ulrik On Mon, 9 Oct 2017 at 17:03 Duncan Murdoch wrote: > I have a file containing "words" like > > > a > > a/b > > a/b/c > > where there may be

[R] Regular expression help

I have a file containing "words" like a a/b a/b/c where there may be multiple words on a line (separated by spaces).  The a, b, and c strings can contain non-space, non-slash characters. I'd like to use gsub() to extract the c strings (which should be empty if there are none). A real

Re: [R] regular expression help

Dear Enrico, Many thanks and Best Regards, Ashim. On Thu, Jun 8, 2017 at 5:11 PM, Enrico Schumann wrote: > > Zitat von Ashim Kapoor : > > > Dear All, >> >> My query is: >> >> Do we always need to use perl = TRUE option when doing

Re: [R] regular expression help

Zitat von Ashim Kapoor : Dear All, My query is: Do we always need to use perl = TRUE option when doing ignore.case=TRUE? A small example : my_text = "RECOVERY OFFICER-II\nDEBTS RECOVERY TRIBUNAL-III\n RC No. 162/2015\nSBI VS RAMESH GUPTA.\nDated: 01.03.2016

[R] regular expression help

Dear All, My query is: Do we always need to use perl = TRUE option when doing ignore.case=TRUE? A small example : my_text = "RECOVERY OFFICER-II\nDEBTS RECOVERY TRIBUNAL-III\n RC No. 162/2015\nSBI VS RAMESH GUPTA.\nDated: 01.03.2016 Item no.01\n Present: Ms. Sonakshi,

Re: [R] regular expression question

I know you already have a couple of solutions, but I would like to mention that it can be done in two steps with very simple regular expressions. I would have done: s - c(lngimbintrhofixed,lngimbnointnorhofixed,test, 'rhofixedtest','norhofixedtest') res - gsub('norhofixed$', '',s) res -

Re: [R] regular expression question

On Wed, Jan 14, 2015 at 10:03 AM, MacQueen, Don macque...@llnl.gov wrote: I know you already have a couple of solutions, but I would like to mention that it can be done in two steps with very simple regular expressions. I would have done: s - c(lngimbintrhofixed,lngimbnointnorhofixed,test,

Re: [R] regular expression question

: Wednesday, January 14, 2015 at 8:47 AM To: dh m macque...@llnl.govmailto:macque...@llnl.gov Cc: Mark Leeds marklee...@gmail.commailto:marklee...@gmail.com, r-help-stat.math.ethz.ch r-h...@stat.math.ethz.chmailto:r-h...@stat.math.ethz.ch Subject: Re: [R] regular expression question On Wed, Jan 14, 2015

Re: [R] regular expression question

Hi Mark, Mark Leeds marklee...@gmail.com writes: Hi All: I have a regular expression problem. If a character string ends with rhofixed or norhofixed, I want that part of the string to be removed. If it doesn't end with either of those two endings, then the result should be the same as the

Re: [R] regular expression question

No HTML please. it makes me itchy! grin/ s - c(lngimbintrhofixed,lngimbnointnorhofixed,test) sub('(no)?rhofixed$','',s) [1] lngimbint lngimbnoint test On Mon, Jan 12, 2015 at 1:37 PM, Mark Leeds marklee...@gmail.com wrote: Hi All: I have a regular expression problem. If a character

[R] regular expression question

Hi All: I have a regular expression problem. If a character string ends with rhofixed or norhofixed, I want that part of the string to be removed. If it doesn't end with either of those two endings, then the result should be the same as the original. Below doesn't work for the second case. I know

Re: [R] regular expression help

:26 -0700 Subject: Re: [R] regular expression help To: bac...@hotmail.com CC: dwinsem...@comcast.net; r-help@r-project.org what's the difference between [:space:]+ and[[:space:]]+ ? The pattern '[:space:]' matches any of ':', 's', 'p', 'a', 'c', and 'e' (the second colon is superfluous). I.e

Re: [R] regular expression help

...@tibco.com Date: Fri, 27 Jun 2014 02:35:54 -0700 Subject: Re: [R] regular expression help To: dwinsem...@comcast.net CC: bac...@hotmail.com; r-help@r-project.org You can use parentheses to factor out the common string in David's pattern, as in grep(value=TRUE, (^|//|[[:space:]]+)AARSD1

Re: [R] regular expression help

) grepl('^AARSD1//$|^AARSD1 //$|^//AARSD1$|^AARSD1$', test) [1] FALSE FALSE TRUE TRUE TRUE TRUE -- David. Thanks, Lin From: dulca...@bigpond.com To: bac...@hotmail.com; r-help@r-project.org Subject: RE: [R] regular expression help Date: Fri

Re: [R] regular expression help

From: wdun...@tibco.com Date: Fri, 27 Jun 2014 02:35:54 -0700 Subject: Re: [R] regular expression help To: dwinsem...@comcast.net CC: bac...@hotmail.com; r-help@r-project.org You can use parentheses to factor out the common string in David's pattern, as in grep(value=TRUE

Re: [R] regular expression help

: RE: [R] regular expression help Date: Fri, 27 Jun 2014 10:59:29 +1000 Hi You only have a vector of length 5 and I am not quite sure of the string you are testing so try this grep('[/]*\\AARSD1\\[/]*',test) Duncan Duncan Mackay Department of Agronomy and Soil Science University of New

[R] regular expression help

Dear R users, I need to match a string. It can be followed or preceded by whitespace or // or nothing. How do I code it in R? For example: test - c('AARSD11','AARSD1-','AARSD1//','AARSD1 //','//AARSD1'); grep('AARSD1(\\s*//*)',test); should return 3,4,5 and 6. Thanks in advance for your help.

Re: [R] regular expression help

Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of C Lin Sent: Friday, 27 June 2014 10:05 To: r-help@r-project.org Subject: [R] regular expression help Dear R users, I need to match a string. It can be followed or preceded by whitespace

Re: [R] regular expression help

//','//AARSD1','AARSD1' Thanks, Lin   From: dulca...@bigpond.com To: bac...@hotmail.com; r-help@r-project.org Subject: RE: [R] regular expression help Date: Fri, 27 Jun 2014 10:59:29 +1000 Hi You only have a vector of length 5 and I am not quite sure

Re: [R] regular expression help

From: dulca...@bigpond.com To: bac...@hotmail.com; r-help@r-project.org Subject: RE: [R] regular expression help Date: Fri, 27 Jun 2014 10:59:29 +1000 Hi You only have a vector of length 5 and I am not quite sure of the string you are testing so try

[R] R regular expression and locales

Hi everybody, I am developing some functions that use regular expressions and grepl, to check whether certain strings match a given pattern or not. In the regular expression I use some shortcuts such as [:alnum:]. Reading the documentation for regular expression there is one sentence that is not

[R] why the [^\\d] is not equal to [^0-9] in R regular expression?

gregexpr(pattern=[^0-9],+12345ty) [[1]] [1] 1 7 8 attr(,match.length) [1] 1 1 1 attr(,useBytes) [1] TRUE gregexpr(pattern=[^\\d],+12345ty) [[1]] [1] 1 2 3 4 5 6 7 8 attr(,match.length) [1] 1 1 1 1 1 1 1 1 attr(,useBytes) [1] TRUE why the pattern `[^\\d]` has no same effect as of `[^0-9]` ?

Re: [R] why the [^\\d] is not equal to [^0-9] in R regular expression?

-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of luofeiyu Sent: Sunday, December 29, 2013 3:42 AM To: R-help@r-project.org Subject: [R] why the [^\\d] is not equal to [^0-9] in R regular expression? gregexpr(pattern=[^0-9],+12345ty) [[1]] [1] 1 7 8 attr

[R] Regular Expression returning unexpected results

Hi, So I just took an intro to R programming class and one of the lectures was on Regular Expressions. I've been playing around with various R functions that use Regular Expressions. But this has me stumped. This was part of a quiz and I got it right through understanding the syntax. But when

Re: [R] Regular Expression returning unexpected results

On Tue, Oct 29, 2013 at 1:13 PM, Lopez, Dan lopez...@llnl.gov wrote: grep(^([a-z]+) +\1 +[a-z]+ [0-9],lines) Your expression has a typo: R grep(^([a-z]+) +\\1 +[a-z]+ [0-9],lines) [1] 2 -- Sarah Goslee http://www.functionaldiversity.org __

Re: [R] Regular Expression returning unexpected results

Please read and follow the Posting Guide, in particular re plain text email. You need to keep in mind that the characters in literal strings in R source have to make it into RAM before the regex code can parse it. Since regex needs a single backslash to escape normal parsing and interpret 1 as

Re: [R] Regular Expression returning unexpected results

College Station, TX 77840-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Lopez, Dan Sent: Tuesday, October 29, 2013 12:13 PM To: R help (r-help@r-project.org) Subject: [R] Regular Expression returning unexpected results Hi, So I

Re: [R] Regular Expression returning unexpected results

To: Lopez, Dan; R help (r-help@r-project.org) Subject: Re: [R] Regular Expression returning unexpected results Please read and follow the Posting Guide, in particular re plain text email. You need to keep in mind that the characters in literal strings in R source have to make it into RAM before

[R] regular expression strikes again

Dear experts in regexpr. I have this dput(test[500:510]) c(pH 9,36 2, pH 9,36 3, pH 9,66 1, pH 9,66 2, pH 9,66 3, pH 10,04 1, pH 10,04 2, pH 10,04 3, RGLP 144006 pH 6,13 1, RGLP 144006 pH 6,13 2, RGLP 144006 pH 6,13 3) and I want something like this gsub(^.*([[:digit:]],[[:digit:]]*).*$,

Re: [R] regular expression strikes again

On Jul 9, 2013, at 11:45 , PIKAL Petr wrote: Dear experts in regexpr. I have this dput(test[500:510]) c(pH 9,36 2, pH 9,36 3, pH 9,66 1, pH 9,66 2, pH 9,66 3, pH 10,04 1, pH 10,04 2, pH 10,04 3, RGLP 144006 pH 6,13 1, RGLP 144006 pH 6,13 2, RGLP 144006 pH 6,13 3) and I want

Re: [R] regular expression strikes again

On Tue, Jul 09, 2013 at 09:45:55AM +, PIKAL Petr wrote: Dear experts in regexpr. I have this dput(test[500:510]) c(pH 9,36 2, pH 9,36 3, pH 9,66 1, pH 9,66 2, pH 9,66 3, pH 10,04 1, pH 10,04 2, pH 10,04 3, RGLP 144006 pH 6,13 1, RGLP 144006 pH 6,13 2, RGLP 144006 pH 6,13 3) and

Re: [R] regular expression strikes again

before comma and two digits after comma. Thanks anyway Petr -Original Message- From: peter dalgaard [mailto:pda...@gmail.com] Sent: Tuesday, July 09, 2013 11:59 AM To: PIKAL Petr Cc: r-help Subject: Re: [R] regular expression strikes again On Jul 9, 2013, at 11:45 , PIKAL Petr

Re: [R] regular expression strikes again

On Jul 9, 2013, at 12:19 , PIKAL Petr wrote: Thanks, it works to some extent. The test comes from some file which is not filled propperly. If I use your suggestion I get correct values for those 2 digit numbers before , but I get some other values which do not have space before numbers.

Re: [R] regular expression strikes again

: [R] regular expression strikes again Dear experts in regexpr. I have this dput(test[500:510]) c(pH 9,36 2, pH 9,36 3, pH 9,66 1, pH 9,66 2, pH 9,66 3, pH 10,04 1, pH 10,04 2, pH 10,04 3, RGLP 144006 pH 6,13 1, RGLP 144006 pH 6,13 2, RGLP 144006 pH 6,13 3) and I want something like this gsub

Re: [R] regular expression strikes again

Petr petr.pi...@precheza.cz To: r-help r-help@r-project.org Cc: Sent: Tuesday, July 9, 2013 5:45 AM Subject: [R] regular expression strikes again Dear experts in regexpr. I have this dput(test[500:510]) c(pH 9,36 2, pH 9,36 3, pH 9,66 1, pH 9,66 2, pH 9,66 3, pH 10,04 1, pH 10,04 2, pH

Re: [R] Regular expression

To: r-help Subject: [R] Regular expression Hello again, I am having a problem on Regular expression. Let say I have following code: gsub([',], , 'asd'f) [1] asdf This is perfect. However I am having problem if I include (i.e. the double quote) in the first argument as the pattern

Re: [R] Regular expression

HI, vec1-'asd'f  vec2-'asdf'  gsub([\],,vec2) #[1] asdf  gsub(['],,vec1) #[1] asdf A.K. - Original Message - From: Christofer Bogaso bogaso.christo...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Tuesday, January 15, 2013 4:38 PM Subject: [R] Regular expression Hello again

[R] Regular Expression

Hi-- I have three columns in an input file: MONTH QUARTER YEAR 2012-07 2012-32012 2001-07 2001-32001 2002-01 2002-12002 I want to make output like so: MONTH QUARTER YEAR 07 32012 07 32001 01 1

Re: [R] Regular Expression

If you want that output. substr() Can help in your task too. I can not help with regular expression, I will learn too. Date: Tue, 24 Jul 2012 13:36:25 -0400 From: bayespoker...@gmail.com To: r-help@r-project.org Subject: [R] Regular Expression Hi-- I have three columns

Re: [R] Regular Expression

Hi, one problem, many solutions, only one of which uses regular expression but work equally well. dat1-read.table(text= MONTH QUARTER YEAR 2012-07 2012-32012 2001-07 2001-32001 2002-01 2002-12002 ,sep=,as.is = TRUE, header=TRUE) # using substr:

Re: [R] Regular Expression

Hi Fred, I'm no regex ninja (and I imagine one will be along shortly to solve your problem) but in your case does it simply suffice to drop the first 5 characters? That might be an easier sub() to write. Best, Michael On Tue, Jul 24, 2012 at 12:36 PM, Fred G bayespoker...@gmail.com wrote: Hi--

Re: [R] Regular Expression

To delete everything from the beginning of the string to and including the hyphen, use sub(^.*-, , tmp) Sarah On Tue, Jul 24, 2012 at 1:36 PM, Fred G bayespoker...@gmail.com wrote: Hi-- I have three columns in an input file: MONTH QUARTER YEAR 2012-07 2012-32012 2001-07

Re: [R] Regular Expression

Is this what you want: x - read.table(text = MONTH QUARTER YEAR + 2012-07 2012-32012 + 2001-07 2001-32001 + 2002-01 2002-12002, header = TRUE, as.is = TRUE) x MONTH QUARTER YEAR 1 2012-07 2012-3 2012 2 2001-07 2001-3 2001 3 2002-01 2002-1 2002 x$MONTH -

Re: [R] Regular Expression

Hello, I believe the following will do it. d - read.table(text= MONTH QUARTER YEAR 2012-07 2012-32012 2001-07 2001-32001 2002-01 2002-12002 , header=TRUE) search - ^.*-([[:digit:]]+)$ sapply(d, function(x) as.integer(sub(search, \\1, x))) Hope this helps,

Re: [R] Regular Expression

-bounces@r- project.org] On Behalf Of Fred G Sent: Tuesday, July 24, 2012 12:36 PM To: r-help@r-project.org Subject: [R] Regular Expression Hi-- I have three columns in an input file: MONTH QUARTER YEAR 2012-07 2012-32012 2001-07 2001-32001 2002-01 2002-1

Re: [R] Regular Expression

On Tue, Jul 24, 2012 at 1:36 PM, Fred G bayespoker...@gmail.com wrote: Hi-- I have three columns in an input file: MONTH QUARTER YEAR 2012-07 2012-32012 2001-07 2001-32001 2002-01 2002-12002 I want to make output like so: MONTH QUARTER YEAR 07

Re: [R] Regular Expression

Thank you! :) On Tue, Jul 24, 2012 at 1:42 PM, Sarah Goslee sarah.gos...@gmail.comwrote: To delete everything from the beginning of the string to and including the hyphen, use sub(^.*-, , tmp) Sarah On Tue, Jul 24, 2012 at 1:36 PM, Fred G bayespoker...@gmail.com wrote: Hi-- I have

Re: [R] Regular Expression

-help@r-project.org Cc: Sent: Tuesday, July 24, 2012 1:36 PM Subject: [R] Regular Expression Hi-- I have three columns in an input file: MONTH  QUARTER  YEAR 2012-07  2012-3        2012 2001-07  2001-3        2001 2002-01  2002-1        2002 I want to make output like so: MONTH  QUARTER  YEAR

[R] regular expression and R

Dear R People: Are there any courses which describe how to use regular expressions in R, please? Or any books, please? I know a little bit (very little) but would like to know more. Thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University

Re: [R] regular expression and R

On Jun 4, 2012, at 3:48 PM, Erin Hodgess wrote: Dear R People: Are there any courses which describe how to use regular expressions in R, please? Or any books, please? I know a little bit (very little) but would like to know more. Thanks, Erin Hi Erin, The two places that I

Re: [R] regular expression and R

On Mon, Jun 4, 2012 at 4:48 PM, Erin Hodgess erinm.hodg...@gmail.com wrote: Dear R People: Are there any courses which describe how to use regular expressions in R, please?  Or any books, please? I know a little bit (very little) but would like to know more. You might want to go through

[R] regular expression

Computer Friends, with the following example lines: [107] 98-610: Cell type: S; Surv(months): 6; STATUS(0=alive, 1=dead): 1 [108] 99-625: Cell type: S; Surv(months): 21; STATUS(0=alive, 1=dead): 1 i want to be able to isolate the number of months of survival for each row. is there a regular

Re: [R] regular expression

On Feb 29, 2012, at 2:24 PM, Fred G wrote: Computer Friends, with the following example lines: Modified to be correct R code. Please emulate my example in the future. inp -c( 98-610: Cell type: S; Surv(months): 6; STATUS(0=alive, 1=dead): 1, 99-625: Cell type: S; Surv(months): 21;

Re: [R] regular expression

On Wed, Feb 29, 2012 at 2:24 PM, Fred G bayespoker...@gmail.com wrote: Computer Friends, with the following example lines: [107] 98-610: Cell type: S; Surv(months): 6; STATUS(0=alive, 1=dead): 1 [108] 99-625: Cell type: S; Surv(months): 21; STATUS(0=alive, 1=dead): 1 i want to be able to

Re: [R] regular expression

gsub('.+; (.+);.+','\\1',x) or if you just want the value out: gsub('.+; Surv\\(months\\): ([0-9]+);.+','\\1',x) You can also look at strsplit: strsplit(x,';') [[1]] [1] 99-625: Cell type: S Surv(months): 21 STATUS(0=alive, 1=dead): 1 lapply(strsplit(x,';'),'[',2) [[1]]

[R] regular expression for selection

Dear all I am again (as usual) lost in regular expression use for selection. Here are my data: dput(mena) c(138516_10g_50ml_50c_250utes1_m53.00-_s1.imp, 138516_10g_50ml_50c_250utes1_m54.00_s1.imp, 138516_10g_50ml_50c_250utes1_m55.00_s1.imp, 138516_10g_50ml_50c_250utes1_m56.00_s1.imp,

Re: [R] regular expression for selection

Hi, Try grepl instead of sub, mena[grepl(m5., mena)] HTH, baptiste On 14 November 2011 21:45, Petr PIKAL petr.pi...@precheza.cz wrote: Dear all I am again (as usual) lost in regular expression use for selection. Here are my data: dput(mena)

Re: [R] regular expression for selection

On 11/14/2011 07:45 PM, Petr PIKAL wrote: Dear all I am again (as usual) lost in regular expression use for selection. Here are my data: dput(mena) c(138516_10g_50ml_50c_250utes1_m53.00-_s1.imp, 138516_10g_50ml_50c_250utes1_m54.00_s1.imp, 138516_10g_50ml_50c_250utes1_m55.00_s1.imp,

Re: [R] regular expression for selection

Hi Hi, Try grepl instead of sub, mena[grepl(m5., mena)] It does not select those m5? strings from those character vectors. I need as an output a vector m53, m54, m55, m56, m57, m58, m59 Regards Petr HTH, baptiste On 14 November 2011 21:45, Petr PIKAL petr.pi...@precheza.cz

Re: [R] regular expression for selection

Hi On 11/14/2011 07:45 PM, Petr PIKAL wrote: Dear all I am again (as usual) lost in regular expression use for selection. Here are my data: dput(mena) c(138516_10g_50ml_50c_250utes1_m53.00-_s1.imp, 138516_10g_50ml_50c_250utes1_m54.00_s1.imp,

Re: [R] regular expression for selection

On 14.11.2011 10:22, Petr PIKAL wrote: Hi On 11/14/2011 07:45 PM, Petr PIKAL wrote: Dear all I am again (as usual) lost in regular expression use for selection. Here are my data: dput(mena) c(138516_10g_50ml_50c_250utes1_m53.00-_s1.imp, 138516_10g_50ml_50c_250utes1_m54.00_s1.imp,

Re: [R] regular expression for selection

Does library( stringr ) str_extract( mena, m5[0-9] ) achieve what you are looking for? Rgds, Rainer On Monday 14 November 2011 10:22:09 Petr PIKAL wrote: Hi On 11/14/2011 07:45 PM, Petr PIKAL wrote: Dear all I am again (as usual) lost in regular expression use for selection.

Re: [R] regular expression for selection

Hi Thank you. It is a pure magic, something taught in Unseen University. this is what I got as a help for selecting only letters from set of character vector. vzor [1] 61A 62C/27 65A/27 66C/29 69A/29 70C/31 73A/31 [8] 74C/33 77A/33 81A/35 82C/37 85A/37 86C/39 89A/39 [15]

Re: [R] regular expression for selection

On 14.11.2011 11:27, Petr PIKAL wrote: Hi Thank you. It is a pure magic, something taught in Unseen University. this is what I got as a help for selecting only letters from set of character vector. vzor [1] 61A 62C/27 65A/27 66C/29 69A/29 70C/31 73A/31 [8] 74C/33 77A/33

[R] regular expression in gsub() for strings with leading backslash

Hello, Can anyone help on gsub() in R? I have a string like something below, and wanted to delete all the strings with leading backslash, including \xa0On, \023, \xab, and many others. How should I write a regular expression pattern in gsub()? I don't care how many characters following

Re: [R] regular expression in gsub() for strings with leading backslash

On 29/04/2011 7:41 PM, Miao wrote: Hello, Can anyone help on gsub() in R? I have a string like something below, and wanted to delete all the strings with leading backslash, including \xa0On, \023, \xab, and many others. How should I write a regular expression pattern in gsub()? I don't care

Re: [R] regular expression in gsub() for strings with leading backslash

Thanks Duncan for clarifying this. I'm pretty a newbie to such type of characters and special characters. In R's gsub() what regular expressions shall I use to handle all these situations? On Fri, Apr 29, 2011 at 6:07 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote: On 29/04/2011 7:41 PM,

Re: [R] regular expression in gsub() for strings with leading backslash

On 29/04/2011 9:34 PM, Miao wrote: Thanks Duncan for clarifying this. I'm pretty a newbie to such type of characters and special characters. In R's gsub() what regular expressions shall I use to handle all these situations? I don't know. This might work: gsub([\x01-\x1f\x7f-\xff], , x)

Re: [R] regular expression in gsub() for strings with leading backslash

On Fri, 29 Apr 2011, Duncan Murdoch wrote: On 29/04/2011 7:41 PM, Miao wrote: Can anyone help on gsub() in R? I have a string like something below, and wanted to delete all the strings with leading backslash, including \xa0On, \023, \xab, and many others. How should I write a regular

Re: [R] regular expression in gsub() for strings with leading backslash

That works like a charm! Thanks so much Duncan. On Fri, Apr 29, 2011 at 6:37 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote: On 29/04/2011 9:34 PM, Miao wrote: Thanks Duncan for clarifying this. I'm pretty a newbie to such type of characters and special characters. In R's gsub() what

[R] regular expression for nth character in a string

Hi, I have a string InTrouble and want to extract, say, the first two characters: In or the last three: blee or the 3rd, 4th, and 5th: Trou Is there an easy way of doing this quickly with regular expressions in gsub, grep or similar? Thank you for any help. Gonçalo

Re: [R] regular expression for nth character in a string

will this do it: x - InTrouble sub(^(..).*, \\1, x) # first two [1] In sub(.*(...)$, \\1, x) # last three [1] ble sub(^..(...).*, \\1, x) # 3rd,4th,5th char [1] Tro 2011/4/25 Gonçalo Ferraz gferra...@gmail.com: Hi, I have a string InTrouble and want to extract, say, the first two

Re: [R] regular expression for nth character in a string

On 04/25/2011 08:17 PM, Gonçalo Ferraz wrote: Hi, I have a string InTrouble and want to extract, say, the first two characters: In or the last three: blee or the 3rd, 4th, and 5th: Trou Is there an easy way of doing this quickly with regular expressions in gsub, grep or similar? Hi

Re: [R] regular expression for nth character in a string

On Apr 25, 2011, at 6:17 AM, Gonçalo Ferraz wrote: Hi, I have a string InTrouble and want to extract, say, the first two characters: In or the last three: blee or the 3rd, 4th, and 5th: Trou Is there an easy way of doing this quickly with regular expressions in gsub, grep or similar?

Re: [R] regular expression for nth character in a string

2011/4/25 Gonçalo Ferraz gferra...@gmail.com: Hi, I have a string InTrouble and want to extract, say, the first two characters: In or the last three: blee or the 3rd, 4th, and 5th: Trou Is there an easy way of doing this quickly with regular expressions in gsub, grep or similar?

Re: [R] regular expression question

On Mon, Apr 11, 2011 at 10:49 PM, Erin Hodgess erinm.hodg...@gmail.com wrote: Dear R People: I have a data frame with the following column names: names(funky)  [1] UHD.1   UHD.2   UHD.3   UHD.4   L..W..1 L..W..2 L..W..3  [8] L..W..4 B..W..1 B..W..2 B..W..3 B..W..4 W..B..1 W..B..2 [15]

Re: [R] regular expression question

Hi Erin, Please read ?grep. It is clearly not the function you want (neither is strsplit() either really). This does what you want and you can modify for upper/lower case if you need it. Also note that regular expressions exist separate from R, so while : may have seemed natural to select a

[R] regular expression question

Dear R People: I have a data frame with the following column names: names(funky) [1] UHD.1 UHD.2 UHD.3 UHD.4 L..W..1 L..W..2 L..W..3 [8] L..W..4 B..W..1 B..W..2 B..W..3 B..W..4 W..B..1 W..B..2 [15] W..B..3 W..B..4 B..G..1 B..G..2 B..G..3 B..G..4 I would like to extract the letters

Re: [R] regular expression

Great. thank you Bernd! Learned a new thing here. John From: Bernd Weiss bernd.we...@uni-koeln.de Cc: r-help@r-project.org Sent: Thu, March 31, 2011 6:19:25 PM Subject: Re: [R] regular expression Am 31.03.2011 21:06, schrieb array chip: Ok then this code

Re: [R] regular expression

From: Bernd Weiss bernd.we...@uni-koeln.de Cc: r-help@r-project.org Sent: Thu, March 31, 2011 6:19:25 PM Subject: Re: [R] regular expression Am 31.03.2011 21:06, schrieb array chip: Ok then this code didn't do what I wanted. I want not including 'arg' before

[R] regular expression

Hi, I am stuck on this: how to specify a match pattern that means not to include abc? I tried: grep(^(abc), hello, value=T) should return hello. while grep(^(abc), hello abcd foo, value=T) should return character(0). But both returned character(0). Thanks John [[alternative

Re: [R] regular expression

Am 31.03.2011 19:31, schrieb array chip: Hi, I am stuck on this: how to specify a match pattern that means not to include abc? I tried: grep(^(abc), hello, value=T) should return hello. grep([^(abc)], hello, value=T) [1] hello HTH, Bernd __

Re: [R] regular expression

, but the 2nd example didn't. What was wrong here? Thanks John From: Bernd Weiss bernd.we...@uni-koeln.de Sent: Thu, March 31, 2011 5:32:25 PM Subject: Re: [R] regular expression Am 31.03.2011 19:31, schrieb array chip: Hi, I am stuck on this: how to specify

Re: [R] regular expression

On Thu, Mar 31, 2011 at 5:49 PM, array chip arrayprof...@yahoo.com wrote: Thanks Bernd! I tried your approach with my real example, sometimes it worked, sometimes it didn't. For example grep('[^(arg)]\\.symptom',stomach.symptom,value=T) [1] stomach.symptom

Re: [R] regular expression

, does include .symptom. Any other suggestions would be appreciated John From: Peter Langfelder peter.langfel...@gmail.com Cc: Bernd Weiss bernd.we...@uni-koeln.de; r-help@r-project.org Sent: Thu, March 31, 2011 5:55:26 PM Subject: Re: [R] regular expression

Re: [R] regular expression

Am 31.03.2011 21:06, schrieb array chip: Ok then this code didn't do what I wanted. I want not including 'arg' before '.symptom', not individual letters of arg, but rather as a word. Bill Dunlap suggested using invert=T, it works for single 1 condition, but not for 2 conditions here: not

[R] Regular Expression

Hi R users,   Thanks in advance.   I am using R-2.12.1 on Windows XP.   I am looking for some good literature on Regular Expression. May I request you to assist me please. Once again, thank you very much for the time you have given.   Regards,   Deb   [[alternative HTML version

Re: [R] Regular Expression

On Mon, Feb 14, 2011 at 4:13 AM, Deb Midya debmi...@yahoo.com wrote: Hi R users, Thanks in advance. I am using R-2.12.1 on Windows XP. I am looking for some good literature on Regular Expression. May I request you to assist me please. There are regular expression links on the gsubfn home

[R] Regular expression to find value between brackets

Hi, this should be an easy one, but I can't figure it out. I have a vector of tests, with their units between brackets (if they have units). eg tests - c(pH, Assay (%), Impurity A(%), content (mg/ml)) Now I would like to hava a function where I use a test as input, and which returns the units

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