Re: [R] string concatenation

Hi, Thanks for clarifying, there is no quote in the result. The quote in the output is just R's way to tell you that the variable got printed is a string. If you add quotes around a string, it will get printed like "\"Alice, Bob, Charles\"" I hope this helps. Best, Jiefei On Thu, Nov 5, 2020

Re: [R] string concatenation

On 05/11/2020 3:20 a.m., Edjabou Vincent wrote: Following John request, I am wondering if it is possible to get this result: Alice, Bob, Charles (without bracket ) I think you mean "without the quotes". Use noquote(): noquote(paste0(x,collapse = ", ")) will print without quotes.

Re: [R] string concatenation

Following John request, I am wondering if it is possible to get this result: Alice, Bob, Charles (without bracket ) Thank you for your help Med venlig hilsen/ Best regards Vincent Edjabou Mobile: +45 31 95 99 33 linkedin.com/vincent

Re: [R] string concatenation

Hi John, Try paste0(x,collapse = ", ") Best, Jiefei On Thu, Nov 5, 2020 at 1:16 PM John wrote: > Hi, > > I have a sequence of characters: > > x <- c("Alice", "Bob", "Charles") > > How can I produce the following results: > > "Alice, Bob, Charles" > > ? > > paste? merge? > > Thank you very

[R] string concatenation

Hi, I have a sequence of characters: x <- c("Alice", "Bob", "Charles") How can I produce the following results: "Alice, Bob, Charles" ? paste? merge? Thank you very much! [[alternative HTML version deleted]] __ R-help@r-project.org

Re: [R] String replace

On Wed, 3 Apr 2019 15:01:37 +0100 Graham Leask via R-help wrote: Suppose that `BHC$Date` contains a string "M_24". You do: > BHC <-BHC %>% mutate ( Date = stringr :: str_replace ( Date , "M_2" , > "01-04-2017")) before you have a chance to do: > BHC <-BHC %>% mutate ( Date = stringr ::

Re: [R] String replace

Try reversing the order, or extending the to-replace bit. "M_1" is finding and replacing "M_10", "M_11", etc. "M_2" is finding and replacing "M_20", "M_21", etc. So M_14 becomes "01-03-20174" In the future, a toy dataset would help with the reproducible example aspect, and make your question

[R] String replace

I’m attempting to replace a string variable that normally works fine. However when trying to do this with a string in the form of a date the output becomes corrupted. See below: BHC <-BHC %>% mutate ( Date = stringr :: str_replace ( Date , "M_1" , "01-03-2017")) BHC <-BHC %>% mutate ( Date =

Re: [R] string pattern matching

Oh, I see the utility of that now. Definitely will be using "term.labels" (which I was not aware of). Thanks! On Thu, Mar 23, 2017 at 9:48 AM, William Dunlap wrote: > If you are trying to see if one model nested in another then I think > looking at the 'term.labels' attribute

Re: [R] string pattern matching

If you are trying to see if one model nested in another then I think looking at the 'term.labels' attribute of terms(formula) is the way to go. Most formula-based modelling functions store the output of terms(formula) in their output and many supply a method for the terms function that extracts

Re: [R] string pattern matching

Thanks for the additional response, Bill. I did not want to bog down the question with the full context of the function. Briefly, given a set of nested and non-nested regression models, I want to compare AIC (bigger model - smaller model) and do an LRT for all the nested models that differ by a

Re: [R] string pattern matching

You did not describe the goal of your pattern matching. Were you trying to match any string that could be interpreted as an R expression containing X1 and X3 as additive terms? If so, you could turn the string into a one-sided formula and use the terms() function. E.g., f <- function(string)

Re: [R] string pattern matching

Wow. Thanks to everyone (Jim, Ng Bo Lin, Bert, David, and Ulrik) for all the quick and helpful responses. They have given me a better understanding of regular expressions, and certainly answered my question. Joe On Wed, Mar 22, 2017 at 12:22 AM, Ulrik Stervbo wrote: >

Re: [R] string pattern matching

Hi Joe, you could also rethink your pattern: grep("x1 \\+ x2", test, value = TRUE) grep("x1 \\+ x", test, value = TRUE) grep("x1 \\+ x[0-9]", test, value = TRUE) HTH Ulrik On Wed, 22 Mar 2017 at 02:10 Jim Lemon wrote: > Hi Joe, > This may help you: > > test <- c("x1",

Re: [R] string pattern matching

> On Mar 21, 2017, at 6:31 PM, Bert Gunter wrote: > > It is not clear to me what you mean, but: > >> grep ("x1 \\+.* \\+ x3",test, value = TRUE) > [1] "x1 + x2 + x3" > > ## This will miss "x1 + x3" though. So then this might be acceptable: grep ("x1\\ \\+.* x3",

Re: [R] string pattern matching

It is not clear to me what you mean, but: > grep ("x1 \\+.* \\+ x3",test, value = TRUE) [1] "x1 + x2 + x3" ## This will miss "x1 + x3" though. seems to do what you want, maybe. Perhaps you need to read up about regular expressions and/or clarify what you want to do. Cheers, Bert Bert Gunter

Re: [R] string pattern matching

Hi Joe, This may help you: test <- c("x1", "x2", "x3", "x1 + x2 + x3") multigrep<-function(x1,x2) { xbits<-unlist(strsplit(x1," ")) nbits<-length(xbits) xans<-rep(FALSE,nbits) for(i in 1:nbits) if(length(grep(xbits[i],x2))) xans[i]<-TRUE return(all(xans)) } multigrep("x1 + x3","x1 + x2 +

[R] string pattern matching

Hi Folks, Is there a way to find "x1 + x2 + x3" given "x1 + x3" as the pattern? Or is that a ridiculous question, since I'm trying to find something based on a pattern that doesn't exist? test <- c("x1", "x2", "x3", "x1 + x2 + x3") test [1] "x1" "x2" "x3" "x1 + x2 +

Re: [R] String match

Please (re)-read the Help file for pmatch, which says: ... " (A partial match occurs if the whole of the element of x matches the beginning of the element of table.) " This is clearly not your situation. One approach (there may be others depending on what your detailed situation actually is) is

Re: [R] String match

try this: > WhereToLook = c("ultracemco.bo.openam" , "ultracemco.bo.higham" + ,"ultracemco.bo.lowam" , "ultracemco.bo.closeam" , + "ultracemco.bo.volumeam" ,"ultracemco.bo.adjustedam") > > WhatToLook = c("volume", "close") > > # create pattern match > pat <- paste(WhatToLook,

[R] String match

Hi again, I am facing trouble to match a vector strings. Let say I have below full string vector WhereToLook = c("ultracemco.bo.openam" , "ultracemco.bo.higham" ,"ultracemco.bo.lowam" , "ultracemco.bo.closeam" , "ultracemco.bo.volumeam" ,"ultracemco.bo.adjustedam") WhatToLook =

Re: [R] String Matching

>> > > or the last six? > >> substr(id, nchar(id)-6, nchar(id))=="432.rds" > [1] TRUE >> > > Cheers > Petr > > >> -Original Message- >> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Glenn >&

Re: [R] String Matching

gt; -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Glenn > Schultz > Sent: Friday, January 29, 2016 6:02 PM > To: R Help R > Subject: [R] String Matching > > All, > > I have a file named as so 313929BL4FNMA2432.rds the user may

[R] String Matching

All, I have a file named as so 313929BL4FNMA2432.rds  the user may pass either the first 9 character or the last six characters.  I need to match the remainder of the file name using either the first nine or last six.  I have read the help files for Regular Expression as used in R and I think

Re: [R] String Matching

On 29/01/2016 12:02 PM, Glenn Schultz wrote: All, I have a file named as so 313929BL4FNMA2432.rds the user may pass either the first 9 character or the last six characters. I need to match the remainder of the file name using either the first nine or last six. I have read the help files

Re: [R] string split problem

Hi Marc/Bill Thanks for reply. That's exactly what I am looking for. Jun On Fri, Oct 23, 2015 at 3:53 PM, William Dunlap wrote: > > test <- c('aaa.bb.cc','aaa.dd', 'aaa', 'aaa.', '.eee', '') > > sub("([^.]*)(.*)", "\\1", test) > [1] "aaa" "aaa" "aaa" "aaa" """" > >

[R] string split problem

Dear list, Say I have a vector that has two different types of string test <- c('aaa.bb.cc','aaa.dd') I want to extract the first part of the string (aaa) as a name and save the rest of the string as another name. I was thinking something like sub('(.*)\\.(.*)','\\1',test) but doesn't give me

Re: [R] string split problem

> test <- c('aaa.bb.cc','aaa.dd', 'aaa', 'aaa.', '.eee', '') > sub("([^.]*)(.*)", "\\1", test) [1] "aaa" "aaa" "aaa" "aaa" """" > sub("([^.]*)(.*)", "\\2", test) [1] ".bb.cc" ".dd""" "." ".eee" "" Bill Dunlap TIBCO Software wdunlap tibco.com On Fri, Oct 23, 2015 at 12:17 PM,

Re: [R] string split problem

> On Oct 23, 2015, at 2:17 PM, Jun Shen wrote: > > Dear list, > > Say I have a vector that has two different types of string > > test <- c('aaa.bb.cc','aaa.dd') > > I want to extract the first part of the string (aaa) as a name and save the > rest of the string as

Re: [R] String Matching

I haven't tested this, but what about: df - data.frame(mtch=c(matchString, string1, string2)) grep(searchString, df$mtch, ignore.case=FALSE) Depending on what your next step is, you might prefer grepl. Sometimes using fixed=TRUE in grep() helps. -Don -- Don MacQueen Lawrence Livermore

[R] String Matching

Hey everyone,    I have been having an issue trying to find a specific string of text in a log of system messages.  I have tried to use pmatch, match, and some regular expressions but all to no avail.   I have a matrix / data.frame (either one, the file outputs a tens of thousands of rows

Re: [R] String Matching

Hi Kevin, It's not totally clear to me what the desired output is. grep(searchString, matchString, ignore.case=FALSE) told you that searchString is in the first element of matchString. Isn't that what you want to know? If not, perhaps you can be more specific about what the desired result is.

[R] String manipulation

I want to do the following: if a string does not contain a colon (:), no change is needed; if it contains one or more colons, break the string into multiple strings using the colon as a separator. For example, happy: becomes happy : :sad turns to : sad and happy:sad changes to happy : sad

Re: [R] String manipulation

strsplit(split=:) does almost what you want, but it omits the colons from the output. You can use perl zero-length look-ahead and look-behind operators in the split argument to get the colons as well: strsplit(c(:sad, happy:, happy:sad), split=(?=:)|(?=:), perl=TRUE) [[1]] [1] : sad [[2]]

Re: [R] String manipulation

Actually, the zero-length look-ahead expression is enough to get the job done: strsplit(c(:sad, happy:, happy:sad, :happy:sad:subdued:), split=(?=:), perl=TRUE) [[1]] [1] : sad [[2]] [1] happy : [[3]] [1] happy : sad [[4]] [1] : happy : sad : subdued : Bill

Re: [R] String comparison, trailing blanks make a difference.

On Fri, Jul 18, 2014 at 11:17 AM, John McKown john.archie.mck...@gmail.com wrote: Well, this was a shock to me. And I don't really see any documentation about it, but perhaps I just can't see it. abc == abc [1] FALSE I guess that I thought of strings in R like I do is some other languages

Re: [R] String comparison, trailing blanks make a difference.

If you have unicode strings, you may need to do even more because there are often multiple ways of representing the same glyph. I made a little demo at http://rpubs.com/hadley/unicode-normalisation, since any unicode characters are likely to get mangled by email. Hadley On Fri, Jul 18, 2014 at

[R] String comparison, trailing blanks make a difference.

Well, this was a shock to me. And I don't really see any documentation about it, but perhaps I just can't see it. abc == abc [1] FALSE I guess that I thought of strings in R like I do is some other languages where the shorter value is padded with blanks to the length of the longer value, then

Re: [R] String comparison, trailing blanks make a difference.

abc == abc [1] FALSE R does no interpretation of strings when doing comparisons so you do have do your own canonicalization. That may involve removing trailing, leading, or all white space or punctuation, converting to lower or upper case, mapping nicknames to official names, trimming to a

Re: [R] String comparison, trailing blanks make a difference.

Hi John, On 07/18/2014 09:17 AM, John McKown wrote: Well, this was a shock to me. And I don't really see any documentation about it, but perhaps I just can't see it. abc == abc [1] FALSE I guess that I thought of strings in R like I do is some other languages where the shorter value is

Re: [R] String substitution

Wonderful! Thank you Arun! Irene   Irene Ruberto Da: arun kirshna [via R] ml-node+s789695n4677558...@n4.nabble.com Inviato: Giovedì 3 Ottobre 2013 22:51 Oggetto: Re: String substitution Hi, Try: dat$y- as.character(dat$y) dat1- dat dat2- dat

Re: [R] String substitution

Hi, Try: dat$y- as.character(dat$y) dat1- dat dat2- dat library(stringr)  dat$y[NET]- substr(word(dat$y[NET],2),1,1)  dat$y #[1] n n house n tree #or for(i in 1:length(NET)){dat1$y[NET[i]]- n}  dat1$y #[1] n n house n tree #or dat2$y[NET]- gsub(.*(n).*,\\1,dat2$y[NET])  

[R] string processing(regular expressions)

I have a variable that is course # nCourse - as.factor(c(002A,002B,002C,007A,007B,007C,101,118A,118B,118C)) And I would like to get rid of the leading zeros, and have the following set (2A,2B,2C,7A,7B,7C,101,118A,118B,118C) to paste() together with the department, B,P,C (bio, phys, chem etc) I

Re: [R] string processing(regular expressions)

Hello, Try the following. gsub(^0+, , as.character(nCourse)) Hope this helps, Rui Barradas Em 01-09-2013 21:41, Robert Lynch escreveu: I have a variable that is course # nCourse - as.factor(c(002A,002B,002C,007A,007B,007C,101,118A,118B,118C)) And I would like to get rid of the leading

Re: [R] string processing(regular expressions)

),c(B,P,C)) head(res) #    B   P   C #1 2AB 2AP 2AC #2 2BB 2BP 2BC #3 2CB 2CP 2CC #4 7AB 7AP 7AC #5 7BB 7BP 7BC #6 7CB 7CP 7CC A.K. - Original Message - From: Robert Lynch robert.b.ly...@gmail.com To: R help r-help@r-project.org Cc: Sent: Sunday, September 1, 2013 4:41 PM Subject: [R

[R] String based chemical name identification

The problem is the following: I have two big databases one look like this: 2-Methyl-4-trimethylsilyloxyoct-5-yne Benzoic acid, methyl ester Benzoic acid, 2-methyl-, methyl ester Acetic acid, phenylmethyl ester 2,7-Dimethyl-4-trimethylsilyloxyoct-7-en-5-yne etc. The second one looks

Re: [R] String based chemical name identification

-project.org] On Behalf Of Zsurzsa Laszlo Sent: Wednesday, July 03, 2013 7:28 AM To: r-help@r-project.org Subject: [R] String based chemical name identification The problem is the following: I have two big databases one look like this: 2-Methyl-4-trimethylsilyloxyoct-5-yne Benzoic acid, methyl

[R] string size limits in RCurl

the string into R using either con - file(paste(.project.path, data/stream.json, sep = ), r) string - readLines(con) or directly to list as tmp - fromJSON(file = paste(.project.path, data/stream.json, sep = )) Any thoughts are very much appreciated. Note that I posted this same question

[R] string split at xth position

Hi, I have a vector of strings like: c(a1b1,a2b2,a1b2) which I want to spilt into two parts like: c(a1,a2,a2) and c(b1,b2,b2). So there is always a first part with a+number and a second part with b+number. Unfortunately there is no separator I could use to directly split the vectors.. Any idea

Re: [R] string split at xth position

Dear Johannes, May not be the best way, but this looks like what you described: x - c(a1b1,a2b2,a1b2) x [1] a1b1 a2b2 a1b2 substr(x, 1, 2) [1] a1 a2 a1 substr(x, 3, 4) [1] b1 b2 b2 HTH, Jorge.- On Wed, Mar 13, 2013 at 7:37 PM, Johannes Radinger wrote: Hi, I have a vector of strings

Re: [R] string split at xth position

Thank you Jorge! thats working perfectly... /johannes On Wed, Mar 13, 2013 at 9:45 AM, Jorge I Velez jorgeivanve...@gmail.com wrote: Dear Johannes, May not be the best way, but this looks like what you described: x - c(a1b1,a2b2,a1b2) x [1] a1b1 a2b2 a1b2 substr(x, 1, 2) [1] a1 a2 a1

Re: [R] string split at xth position

, March 13, 2013 4:37 AM Subject: [R] string split at xth position Hi, I have a vector of strings like: c(a1b1,a2b2,a1b2) which I want to spilt into two parts like: c(a1,a2,a2) and c(b1,b2,b2). So there is always a first part with a+number and a second part with b+number. Unfortunately

[R] string split at xth position

Here is another way require(stringr) aaa-paste0(a, 1:20) bbb-paste0(b, 101:120) ab-paste0(aaa,bbb) ab ptrn-([ab][[:digit:]]*) unlist(str_extract_all(ab, ptrn)) Hi, I have a vector of strings like: c(a1b1,a2b2,a1b2) which I want to spilt into two parts like: c(a1,a2,a2) and

Re: [R] string split at xth position

\\2,x1), )),let1) #$a #[1] a1   a10  a2   a140 #$b #[1] b11 b2  b2  b31 A.K. - Original Message - From: arun smartpink...@yahoo.com To: Johannes Radinger johannesradin...@gmail.com Cc: Sent: Wednesday, March 13, 2013 8:32 AM Subject: Re: [R] string split at xth position Also, You could

Re: [R] String Handling() for Split a word by a letter

On 08/27/2012 07:29 PM, Rantony wrote: Hi, here im unable to run a string handle function called unpaste(). for eg:- a- 12345_mydata Actually my requirement what is i need to get , only 12345. Means that , i need the all letter as a word which is before of first _ - symbol of a. i tried

Re: [R] String Handling() for Split a word by a letter

Here I have a variable called Variable_1. Variable_1 - MyDataFrame Here I want to create another variable, by assigning the value of Variable_1 . So, it will come like, Assign(Variable_1,data.frame(read.csv(c:\\Mydata.csv))) ---[this was the 1st requirement, now I got the solution]

[R] String Handling() for Split a word by a letter

Hi, here im unable to run a string handle function called unpaste(). for eg:- a- 12345_mydata Actually my requirement what is i need to get , only 12345. Means that , i need the all letter as a word which is before of first _ - symbol of a. i tried to do with unpaste, it says function not

Re: [R] String Handling() for Split a word by a letter

Is this what you want: a- 12345_mydata sub(_.*, , a) [1] 12345 On Mon, Aug 27, 2012 at 5:29 AM, Rantony antony.akk...@ge.com wrote: Hi, here im unable to run a string handle function called unpaste(). for eg:- a- 12345_mydata Actually my requirement what is i need to get , only 12345.

Re: [R] String Handling() for Split a word by a letter

Hello, If the op says he has tried strsplit, maybe a simple subsetting afterwards would solve it. a - 12345_mydata strsplit(a, _)[[1]][1] Hope this helps, Rui Barradas Em 27-08-2012 15:22, jim holtman escreveu: Is this what you want: a- 12345_mydata sub(_.*, , a) [1] 12345 On Mon,

[R] String Manipulation in R

Hi , Is there any inbuilt functions to check whether a substring is present in a string and give the result as boolean Thanks -- View this message in context: http://r.789695.n4.nabble.com/String-Manipulation-in-R-tp4633104.html Sent from the R help mailing list archive at Nabble.com.

Re: [R] String Manipulation in R

grepl Michael On Tue, Jun 12, 2012 at 8:51 AM, anjali jeevi...@gmail.com wrote: Hi , Is there any inbuilt functions  to check whether a substring is present in a string and give the result as boolean Thanks -- View this message in context:

Re: [R] String Manipulation in R

?grepl Note that this function uses regular expressions, in which certain characters have special meanings, so depending on what string you are looking for you may have to know something about regex patterns to get it to work.

Re: [R] String Manipulation in R

Hello, Yes, there is. See ?grepl or help('grepl'). Hope this helps, Rui Barradas Em 12-06-2012 14:51, anjali escreveu: Hi , Is there any inbuilt functions to check whether a substring is present in a string and give the result as boolean Thanks -- View this message in context:

Re: [R] String Manipulation in R

Or use 'fixed=TRUE' as an argument to grepl to avoid the regular expression matching (but learning regular expressions will be a useful tool in the long run). On Tue, Jun 12, 2012 at 9:15 AM, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: ?grepl Note that this function uses regular

Re: [R] string substitution for argument in function

Me again, what I really dont understand is the behaivour of R here. notice, elem is not written in brackets, why it is interpreted as string and not as a variable? I would expect an error elem not found because it is not defined as variable, but in contrary it is accepted as column name, see

Re: [R] string substitution for argument in function

HI, thanks Weidong, Henrique, this works for the example (may be a bad example, but it should be as simple as possible), This is however only a workaround to name the columns of a dataframe, but is not addressing the problem itself, that is, how can I substitute a string (used as argument in

Re: [R] string substitution for argument in function

'elem' is an attribute (the name) of the 'data' variable (bad name for a variable), not something that is independently accessible. By very (exceptionally!) loose analogy, think of it as being something like a field/property of an object -- you can't get to it directly (and that's a good thing)

[R] string substitution for argument in function

hello, I want to iterate through a list of names and use each element as an argument in a function. For instance: a = c('one','two','three') data= c() for(elem in a){data=cbind(elem = 2,data)} data elem elem elem [1,]222 instead I want 'elem' to be substituted by the string

Re: [R] string substitution for argument in function

Hi, This may help a = c('one','two','three') data.frame(eval(substitute(rbind(var,2),list(var=a ?substitute ?eval Weidong Gu On Sun, Mar 25, 2012 at 5:22 PM, Pedro Martinez pedro.marti...@senckenberg.de wrote: hello, I want to iterate through a list of names and use each element as an

Re: [R] string substitution for argument in function

try this: `names-`(rep(2, 3), a) On Sun, Mar 25, 2012 at 6:22 PM, Pedro Martinez pedro.marti...@senckenberg.de wrote: hello, I want to iterate through a list of names and use each element as an argument in a function. For instance: a = c('one','two','three') data= c() for(elem in

[R] String position character replacement

Hi, Is there a way to efficiently replace specified indices in a string with another character? For example, if I had a vector of strings such as [1] hellohowareyoudoing [2] imgoodhowareyou [3] goodandyou [4] yesimgoodijusttoldyou [5] ohyesthatsright and had a list of positions that I want to

Re: [R] String position character replacement

Hi Joy, Perhaps not the easiest way, but the following seems to work: x - c(hellohowareyoudoing, imgoodhowareyou, goodandyou, yesimgoodijusttoldyou, ohyesthatsright) pos - list(c(3, 9), c(3,4), c(4,7), 5:9, c(2, 5, 7, 12)) sapply(1:length(pos), function(i){ xx - strsplit(x, )[[i]]

Re: [R] String position character replacement

And here's an alternative solution: subchar - function(string, pos, char=-) { for(i in pos) { string - gsub(paste(^(.{, i-1, })., sep=), \\1-, string) } string } subchar(hellohowareyoudoing, 3) [1] he-lohowareyoudoing subchar(hellohowareyoudoing, c(3, 9)) [1]

Re: [R] String position character replacement

On Wed, Feb 8, 2012 at 12:33 PM, Yang, Joy (NIH/NHGRI) [F] joy.y...@nih.gov wrote: Hi, Is there a way to efficiently replace specified indices in a string with another character? For example, if I had a vector of strings such as [1] hellohowareyoudoing [2] imgoodhowareyou [3] goodandyou

Re: [R] String position character replacement

, Joy (NIH/NHGRI) [F] Cc: r-help@R-project.org Subject: Re: [R] String position character replacement And here's an alternative solution: subchar - function(string, pos, char=-) { for(i in pos) { string - gsub(paste(^(.{, i-1, })., sep=), \\1-, string) } string

Re: [R] String position character replacement

On Wed, Feb 08, 2012 at 01:30:55PM -0500, Sarah Goslee wrote: And here's an alternative solution: subchar - function(string, pos, char=-) { for(i in pos) { string - gsub(paste(^(.{, i-1, })., sep=), \\1-, string) } string } Hi. Try the following modification.

Re: [R] String position character replacement

Try this: sapply(mapply(replace, x = strsplit(avec, NULL), list = alist, MoreArgs = list(values = -)), paste, collapse = ) On Wed, Feb 8, 2012 at 3:33 PM, Yang, Joy (NIH/NHGRI) [F] joy.y...@nih.govwrote: Hi, Is there a way to efficiently replace specified indices in a string with another

Re: [R] String position character replacement

Oh cool! I didn't realize you could assign a different value to a substring like that. Thanks! Joy From: Petr Savicky [savi...@cs.cas.cz] Sent: Wednesday, February 08, 2012 3:26 PM To: r-help@r-project.org Subject: Re: [R] String position character

[R] string to list()

I can get an array of strings for the data that I want using 'paste()' as follows: paste('ma', 1:am$arma[2], '=', coef(am)[1:am$arma[2] + am$arma[1]], sep='') This results in a vector of strings like: [1] ma1=1.17760133668255 ma2=0.649795570407939 ma3=0.329456750858276 What I

Re: [R] string to list()

Do it in 2 steps: z - as.list( coef(am)[1:am$arma[2] + am$arma[1]]) names(z) - paste(ma,seq_along(z), sep=) -- Bert On Mon, Nov 14, 2011 at 10:40 AM, Kevin Burton rkevinbur...@charter.netwrote: I can get an array of strings for the data that I want using 'paste()' as follows: paste('ma',

Re: [R] String manipulation with regexpr, got to be a better way

Hi Chris, why not using routines for dates dates - c(09/10/2003, 10/22/2005) format(strptime(dates,format=%m/%d/%Y),%Y) or take just the last 4 chars from dates gsub(.*([0-9]{4})$,\\1,dates) cheers Am 29.09.2011 16:23, schrieb Chris Conner: Help-Rs, I'm doing some string manipulation in a

[R] String manipulation with regexpr, got to be a better way

Help-Rs,   I'm doing some string manipulation in a file where I converted a string date in mm/dd/ format and returned the date .   I've used regexpr (hat tip to Gabor G for a very nice earlier post on this function) in steps (I've un-nested the code and provided it and an example of

Re: [R] String manipulation with regexpr, got to be a better way

Chris Conner wrote on 09/29/2011 09:23:02 AM: Help-Rs, I'm doing some string manipulation in a file where I converted a string date in mm/dd/ format and returned the date . I've used regexpr (hat tip to Gabor G for a very nice earlier post on this function) in steps (I've

Re: [R] string manipulation

Message- From: Steven Kennedy [mailto:stevenkennedy2...@gmail.com] Sent: Friday, 26 August 2011 11:31 AM To: Henrique Dallazuanna Cc: Lorenzo Cattarino; r-help@r-project.org Subject: Re: [R] string manipulation You can split your string, and then only take the first 4 digits after that (this is only

Re: [R] string manipulation

On Thu, Aug 25, 2011 at 9:51 PM, Lorenzo Cattarino l.cattar...@uq.edu.au wrote: Apologies for confusion. What I meant was the following: mytext - I want the number 2000, not the number two thousand and the problem is to select 2000 as the first four digits after the word number. The

Re: [R] string manipulation

.* is greedy... might want regex number[^0-9]*([0-9] {4}) to avoid getting 1999 from I want the number 2000, not the number 1999. --- Jeff Newmiller The . . Go Live... DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#.

Re: [R] string manipulation

On Fri, Aug 26, 2011 at 7:27 AM, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: .* is greedy... might want regex number[^0-9]*([0-9] {4}) to avoid getting 1999 from I want the number 2000, not the number 1999. If such inputs are possible we could also do this where we have added a ? after the *

[R] string manipulation

I R-users, I am trying to find the way to manipulate a character string to select a 4 digit number after some specific word/s. Example: mytext - I do not want the first number 1234, but the second number 5678 Is there any function that allows you to select a certain number of digits (in this

Re: [R] string manipulation

Try this: gsub(.*second number , , mytext) On Thu, Aug 25, 2011 at 8:00 PM, Lorenzo Cattarino l.cattar...@uq.edu.au wrote: I R-users, I am trying to find the way to manipulate a character string to select a 4 digit number after some specific word/s. Example: mytext - I do not want the

Re: [R] string manipulation

You can split your string, and then only take the first 4 digits after that (this is only an improvement if your numbers might not be at the end of mytext): mytext - I do not want the first number 1234, but the second number 5678 sstr-strsplit(mytext,split=second number )[[1]][2]

Re: [R] string manipulation

To be on the safe side in case there are other characters at the end of the string, use: mytext - I do not want the first number 1234, but the second number 5678sadfsadffdsa # make sure you get 4 digits sub(^.*second number[^[0-9]]*([0-9]{4}).*, \\1, mytext) [1] 5678 On Thu, Aug 25, 2011

Re: [R] string manipulation

- From: Steven Kennedy [mailto:stevenkennedy2...@gmail.com] Sent: Friday, 26 August 2011 11:31 AM To: Henrique Dallazuanna Cc: Lorenzo Cattarino; r-help@r-project.org Subject: Re: [R] string manipulation You can split your string, and then only take the first 4 digits after that (this is only

[R] String manipulation

Dear all, I have following kind of character vector: Vec - c(344426, dwjjsgcj, 123sgdc, aagha123, sdh343asgh, 123jhd51) Now I want to split each element of this vector according to numeric and string element. For example in the 1st element of that vector, there is no string element. Therefore

Re: [R] String manipulation

On Sun, Jun 26, 2011 at 10:54 AM, Megh Dal megh700...@yahoo.com wrote: Dear all, I have following kind of character vector: Vec - c(344426, dwjjsgcj, 123sgdc, aagha123, sdh343asgh, 123jhd51) Now I want to split each element of this vector according to numeric and string element. For

Re: [R] String manipulation

On Jun 26, 2011, at 10:54 AM, Megh Dal wrote: Dear all, I have following kind of character vector: Vec - c(344426, dwjjsgcj, 123sgdc, aagha123, sdh343asgh, 123jhd51) Now I want to split each element of this vector according to numeric and string element. For example in the 1st element

Re: [R] String manipulation

On Sun, Jun 26, 2011 at 11:00 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote: On Sun, Jun 26, 2011 at 10:54 AM, Megh Dal megh700...@yahoo.com wrote: Dear all, I have following kind of character vector: Vec - c(344426, dwjjsgcj, 123sgdc, aagha123, sdh343asgh, 123jhd51) Now I want to

Re: [R] R string functions

? For example, the letter A appeared 6 times, letter T appeared 5 times, how can I use a string function to get the these number? thanks, karena -- View this message in context: http://r.789695.n4.nabble.com/R-string-functions-tp3600484p3600568.html Sent from the R help mailing list archive

[R] R string functions

use a string function to get the these number? thanks, karena -- View this message in context: http://r.789695.n4.nabble.com/R-string-functions-tp3600484p3600484.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org

Re: [R] R string functions

] R string functions Hi, I have a string GGCCCAATCGCAATTCCAATT What I want to do is to count the percentage of each letter in the string, what string functions can I use to count the number of each letter appearing in the string? For example, the letter A appeared 6 times, letter

Re: [R] R string functions

Hi, On Wed, Jun 15, 2011 at 4:37 PM, karena dr.jz...@gmail.com wrote: Hi, I have a string GGCCCAATCGCAATTCCAATT What I want to do is to count the percentage of each letter in the string, what string functions can I use to count the number of each letter appearing in the string? For

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