similar to: invalid regular expression '[a-Z]'

Displaying 20 results from an estimated 2000 matches similar to: "invalid regular expression '[a-Z]'"

2017 Jun 08
0
regular expression help
Zitat von Ashim Kapoor <ashimkapoor at gmail.com>: > 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.\n Dated: 01.03.2016 Item no.01\n > Present:
2017 Jun 08
2
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.\n Dated: 01.03.2016 Item no.01\n Present: Ms. Sonakshi, the proxy counsel for Ms. Usha Singh, the counsel for ARCIL.\n None for the CDs.\n
2017 Apr 04
2
Bug report: POSIX regular expression doesn't match for somewhat higher values of upper bound
Dear Sirs, while > regexpr('(.{1,2})\\1', 'foo') [1] 2 attr(,"match.length") [1] 2 attr(,"useBytes") [1] TRUE yields the correct match, an incremented upper bound in > regexpr('(.{1,3})\\1', 'foo') [1] -1 attr(,"match.length") [1] -1 attr(,"useBytes") [1] TRUE incorrectly yields no match. R versions tested: 2.11.1
2013 Mar 08
0
DAHDI-Linux and DAHDI-Tools 2.6.2 Now Available
The Asterisk Development Team has announced the release of: DAHDI-Linux 2.6.2 DAHDI-Tools 2.6.2 DAHDI-Linux-Complete 2.6.2+2.6.2 This release is available for immediate download at: http://downloads.asterisk.org/pub/telephony/dahdi-linux http://downloads.asterisk.org/pub/telephony/dahdi-tools http://downloads.asterisk.org/pub/telephony/dahdi-linux-complete 2.6.2 is a bugfix release of which the
2013 Mar 08
0
DAHDI-Linux and DAHDI-Tools 2.6.2 Now Available
The Asterisk Development Team has announced the release of: DAHDI-Linux 2.6.2 DAHDI-Tools 2.6.2 DAHDI-Linux-Complete 2.6.2+2.6.2 This release is available for immediate download at: http://downloads.asterisk.org/pub/telephony/dahdi-linux http://downloads.asterisk.org/pub/telephony/dahdi-tools http://downloads.asterisk.org/pub/telephony/dahdi-linux-complete 2.6.2 is a bugfix release of which the
2013 Jan 07
3
pattern matching
Hi, I have a simple question. Suppose I have a string "x$Expensive". I want to find the position of the $ in this string; i.e., I want a function that returns 2. I tried grep, regexpr, etc with no luck, unless I'm just using them incorrectly. Any suggestions? Thanks, Walt ________________________ Walter R. Paczkowski, Ph.D. Data Analytics Corp. 44 Hamilton Lane Plainsboro,
2006 Jan 27
4
regular expressions, sub
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
2007 Jun 28
3
: regular expressions: escaping a dot
What's really the problem with: > regexpr( '\.odt$', "xxxxYodt", perl=TRUE ) Warning: '\.' is an unrecognized escape in a character string Warning: unrecognized escape removed from "\.odt$" [1] 5 attr(,"match.length") [1] 4 I know that I could use: > regexpr( '[.]odt$', "xxxxYodt", perl=TRUE ) But it seems to me that
2008 Apr 15
1
why does regexpr not work with '.'
Dear R Helpers, I am running R 2.6.2 on a Windows XP machine. I am trying to use regexpr to locate full stops in strings, but, without success. Here an example:- f="a,b.c at d:" #define an arbitrary test string regexpr(',',f) #find the occurrences of ',' in f - should be one at location 2 # and this is what regexpr finds #[1] 2
2017 Apr 05
0
Bug report: POSIX regular expression doesn't match for somewhat higher values of upper bound
>>>>> <dietmar.schindler at manroland-web.com> >>>>> on Tue, 4 Apr 2017 08:45:30 +0000 writes: > Dear Sirs, > while >> regexpr('(.{1,2})\\1', 'foo') > [1] 2 > attr(,"match.length") > [1] 2 > attr(,"useBytes") > [1] TRUE > yields the correct match, an
2008 Feb 04
2
make dataframe from table
Dear R-experts, I have got a dataframe: data ID disease V1 V2 1 p1 1 2 p1 3 3 p3 3 4 p3 5 5 p5 1 From which I extract a usefull table: affect affect 1 3 5 p1 1 1 0 p3 0 1 1 p5 1 0 0 I want to merge this with anotherdataframe: age p1 23 p2 24 p3 23 p4 11 p5 45 If have tried as.data.frame(affect) and other solutions to get the following comment going:
2010 Jun 01
1
regexpr help (match.length=0)
R-help, Sorry if this is more of a regex question than an R question. However, help would be appreciated on my use of the regexpr function. In the first example below, I ask for all characters (a-z) in 'abc123'; regexpr returns a 3-character match beginning at the first character. > regexpr("[[:alpha:]]*", "abc123") [1] 1 attr(,"match.length") [1] 3
2017 Jun 28
1
regexec() bug in R 3.4.0
Hi, In R 3.4.0, the "Pattern Matching and Replacement" documentation that describes regexec(), gregexpr(), etc. states that the "text" argument to regexec is a character vector, "or an object which can be coerced by as.character to a character vector": regexec(pattern, text, ignore.case = FALSE, perl = FALSE, fixed = FALSE, useBytes = FALSE)
2007 Nov 02
1
R timeDate does not allow seconds?
Hello, Sorry if anyone gets this message twice, as my mailserver may not be working. Thanks for your response. Your idea makes a lot of sense to me, but I've been unable to get seconds to work. I ended up with this format finally: "2007-10-31_16:20:22" Problem is I am unable to get it recognized as a date using timeDate(): R>
2006 Nov 09
1
invert argument in grep
Hello, What about an `invert` argument in grep, to return elements that are *not* matching a regular expression : R> grep("pink", colors(), invert = TRUE, value = TRUE) would essentially return the same as : R> colors() [ - grep("pink", colors()) ] I'm attaching the files that I modified (against today's tarball) for that purpose. Cheers, Romain --
2009 Jul 30
1
What changed in the R Windows interface from v2.6.2 to v2.7.0?
Dear R-devel, I would like to launch the Rgui.exe from a Fortran console application (we will call this FortranCon.exe), that I have written myself. To do this, I send the command line "C:\Program Files\R\R-X.X.X\Rgui.exe" (where X.X.X is the version number) as an argument to the Windows API function CreateProcess (see http://msdn.microsoft.com/en-us/library/ms682425(VS.85).aspx)
2000 Jun 21
1
configure: error: Could not find working SSLeay / OpenSSL libraries
Hi, I've built and installed the openssl-0.9.1c package but openssh-2.1.1p1 configure can't apparently 'see' the package. I've tried adding the install directory to LD_LIBRARY_PATH, using CFLAGS, LDFLAGS, LIBS etc. The 'missing' rand.h is present n the included path shown in the config.log output below. What am I missing? Any help would be very much appreciated.
2007 Aug 15
1
installation of packages
Dear All, Have just installed v2.5.1 on Windows XP. Works fine but I had quite a few pakages loaded for 2.5.0 (from contributed) and was wondering how I can get 2.5.1 to recognise them without having to reinstall them all. Is this possible or do I have to reinstall all the packages again? I required 2.5.1 for lme4 and matrix. Many thanks in advance.
2011 Sep 29
2
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/yyyy format and returned the date yyyy.   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 what I did below.  My question is: is there a more efficient way to do this.  Specifically is
2010 May 05
1
extracting a matched string using regexpr
Given a text like I want to be able to extract a matched regular expression from a piece of text. this apparently works, but is pretty ugly # some html test<-"</tr><tr><th>88958</th><th>Abcdsef</th><th>67.8S</th><th>68.9\nW</th><th>26m</th>" # a pattern to extract 5 digits > pattern<-"[0-9]{5}" #