similar to: R Regular Expressions - Metacharacters

Hi folks, Can anyone enlighten me as to why I get the following when I search for ".csv" at the end of a string? > grep("\.csv$","Blah.csv",value=TRUE) [1] "Blah.csv" Warning messages: 1: '\.' is an unrecognized escape in a character string 2: unrecognized escape removed from "[\.]csv$" R reference for regular expressions says

Hi there, I rejoiced when I realized that you can use Perl regex from within R. However, as the FAQ states "Some functions, particularly those involving regular expression matching, themselves use metacharacters, which may need to be escaped by the backslash mechanism. In those cases you may need a quadruple backslash to represent a single literal one. " I was wondering if that is

Hello R-helpers, I am trying to search for string that includes the caret symbol, using the following code: grepl("latitude^2",temp) And R doesn't like that. It gives me: > temp<-c("latitude^2","latitude and latitude^2","longitude^2","longitude and longitude^2") > temp [1] "latitude^2" "latitude and

Hi, I guess the backslash should not be used as the separator for strsplit() in compareVersion(), because the period in [.] is no longer a metacharacter (no need to "escape" it using a backslash): https://github.com/wch/r-source/blob/trunk/src/library/utils/R/packages.R#L866-L867 > compareVersion function (a, b) { .... a <- as.integer(strsplit(a, "[\\.-]")[[1L]])

Hello, when I use list.files with recursive = TRUE and all.files = TRUE, R returns a list of strings/paths. >From all those strings I want to keep only the ones starting with a . I tried using grep to achieve that. However, the problem is that because of the recursive list.files parameter, for some files beginning with a . there is a path attached. I think it is not as simple as it looks

Full_Name: FAN Version: 2.4.0 OS: Windows Submission from: (NULL) (159.50.101.9) These are expected: > grep("[\-|c]", c("a-a","b")) [1] 1 > gsub("[\-|c]", "&", c("a-a","b")) [1] "a&a" "b" but these are strange: > grep("[d|\-|c]", c("a-a","b")) integer(0)

Hello, I have an expression that I want to substitute for something else. myvector<-c("ajkhfkiuwe","Variable(kg/min)") if I use the following code to extract "variable(kg/min)" and substitute it for "va" gsub(myvector[grep("var", myvector, ignore=T)], "va", myvector) grep identifies the element of the vector that matches my

Dear all I'm having some problems with a data set that has parenthesis within the variable names. A example of this kind of variable names is the following: fBodyGyroskewness()Z The case is that R is having a lot of troubles to identify the variable (probably it does understand it like a function). I've tried (among other things) to remove the parenthesis from the name using the

Why sometimes one has to put a double backslash in regular expressions, but often simple backslashes work too? Is only a \ required for giving a metacharacter its usual meaning? --------------------------------------- u=grep('\\{[\\-u]x',a,perl=T) # equivalent to u=grep('\{[\-u]x',a,perl=T) # but u=grep('\w',a,perl=T) # is not correct and requires

I have a path: path = "/nfs/users/nfs_n/ns9/ Phenotype Analysis/Results/Run_AmplRatio_neg BinaryAll trained without akapn+tnik.csv" I wish to replace the spaces with "\ " so that it can be read by a system call to unix. Using gsub I try: > gsub(" ","\\ ",path) [1] "/nfs/users/nfs_n/ns9/Phenotype Analysis/Results/Run_AmplRatio_neg BinaryAll

Commit-ID: afc40b4eb057b08d8cc2eebefdf6cac05849e8ae Gitweb: http://git.kernel.org/?p=libs/klibc/klibc.git;a=commit;h=afc40b4eb057b08d8cc2eebefdf6cac05849e8ae Author: Herbert Xu <herbert at gondor.apana.org.au> AuthorDate: Wed, 28 Mar 2018 18:37:51 +0800 Committer: Ben Hutchings <ben at decadent.org.uk> CommitDate: Fri, 25 Jan 2019 02:57:21 +0000 [klibc] expand: Do not quote

Commit-ID: 6b0cf885180cfb08f7ec5139e67e581bbba5d6be Gitweb: http://git.kernel.org/?p=libs/klibc/klibc.git;a=commit;h=6b0cf885180cfb08f7ec5139e67e581bbba5d6be Author: Herbert Xu <herbert at gondor.apana.org.au> AuthorDate: Wed, 28 Mar 2018 18:37:51 +0800 Committer: Ben Hutchings <ben at decadent.org.uk> CommitDate: Sat, 28 Mar 2020 21:42:55 +0000 [klibc] dash: expand: Do not

Hi, I am trying to use sub, regexpr on expressions like log(D) ~ log(N)+I(log(N)^2)+log(t) being a model specification. The aim is to produce: "ln D ~ ln N + ln^2 N + ln t" The variable names N, t may change, the number of terms too. I succeded only partially, help on regular expressions is hard to understand for me, examples on my case are rare. The help page on R-help

Dear all, I cannot figure out how to solve a small problem (well, not for me), surely somebody can help me in few seconds. I have a series of strings in a vector X of the type "xxx", "yyy", "zzz", "IgA", "IgG", "kkk", "IgM", "aaa". I want to substitute every ENTIRE string beginning with "Ig" with

Greeting R Community, I am a windows user so this problem may be specific to windows. I often want to source files from within R such as: C:\Users\Rinker\Desktop\Research & Law\Data\School Data 09-10. To source this file I need to go through the path and replace all the backslashes (\) with forward slashes (/). I usually do this in MS Word using the replace option, however, I'd like

I am trying to read some file names from an specific directory and such names contains metacharacters. The file names is like V4.35_T01-400720.csv In total I have 14 files for which the value T01 goes up to T14. I need to read the files into a string vector that looks like >names"V4.35_T01-400720.csv" "V4.35_T02-400720.csv" ... "V4.35_T14-400720.csv" So far, I

I'm trying to write a gsub() call that takes a string and escapes all the unescaped quote marks in it. So the string \" would be left unchanged, but \\" would be changed to \\\" because the double backslash doesn't act as an escape for the quote, the first just escapes the second. I have the usual problems of writing regular expressions involving backslashes which

now after this: df_both <- subset(df, grepl("t1", Command) & grepl("t2", Command)) I use factor to apply the subset to df but then the Command level becomes 0 df_both$Command=factor(df_both$Command) str(df_both) $ Protocol : Factor w/ 0 levels: Do you know what is the reason? Thanks for replying On Sunday, April 24, 2016 12:18 PM, jim

'grepl' returns a logical vector; you have to use this to get your subset. You can use: df_tq <- subset(df, grepl("t1", Command)) df_t2 <- subset(df, grepl("t2", Command)) # if you want to also get a subset that has both, use df_both <- subset(df, grepl("t1", Command) & grepl("t2", Command)) Jim Holtman Data Munger Guru What is

How do I remove a "$" character from a string sub() and gsub() with "$" or "\$" as pattern do not work. > sub("$","","ABC$DEF") [1] "ABC$DEF" > sub("\$","","ABC$DEF") Error: '\$' is an unrecognized escape in character string starting "\$" >