Displaying 20 results from an estimated 60000 matches similar to: "why does regexpr not work with '.'"
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
2003 Aug 13
7
Regexpr with "."
I'm trying to use the regexpr function to locate the decimal in a character
string. Regardless of the position of the decimal, the function returns 1.
For example,
> regexpr(".", "Female.Alabama")
[1] 1
attr(,"match.length")
[1] 1
In trying to figure out what was going on here, I tried the below command:
> gsub(".", ",",
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}"
#
2005 Aug 03
2
regexpr and portability issue
Dear all--
I am still forging my first arms with R and I am fighting with regexpr() as
well as portability between unix and windoz. I need to extract barcodes from
filenames (which are located between a double and single underscore) as well
as the directory where the filename is residing. Here is the solution I came
to:
aFileName <-
2012 Aug 06
5
regexpr with accents
Hello,
I have build a syntax to find out if a given substring is included in a larger string that works like this:
d1$V1[regexpr("some text = 9",d1$V2)>0] <- 9
and this works all right till "some text" contains standard ASCII set. However, it does not work when accents are included as the following:
d1$V1[regexpr("some t?xt = 9",d1$V2)>0] <- 9
I have
2010 Sep 22
4
Crash report: regexpr("a{2-}", "")
Each of the following calls crash ("core dumps") R (R --vanilla) on
various versions and OSes:
regexpr("a{2-}", "")
sub("a{2-}", "")
gsub("a{2-}", "")
EXAMPLES:
> sessionInfo()
R version 2.11.1 Patched (2010-09-16 r52949)
Platform: i386-pc-mingw32 (32-bit)
...
> regexpr("a{2-}", "")
Assertion
2010 Sep 22
4
Crash report: regexpr("a{2-}", "")
Each of the following calls crash ("core dumps") R (R --vanilla) on
various versions and OSes:
regexpr("a{2-}", "")
sub("a{2-}", "")
gsub("a{2-}", "")
EXAMPLES:
> sessionInfo()
R version 2.11.1 Patched (2010-09-16 r52949)
Platform: i386-pc-mingw32 (32-bit)
...
> regexpr("a{2-}", "")
Assertion
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
2008 Mar 06
1
invalid regular expression '[a-Z]'
Hi,
just curious, but does anyone know the source/reason of observing the
following error on OSX but not on WinXP and Linux? I've tried with a
few different versions of R (v2.5.1, v2.6.1, v2.6.2, v2.7.0devel).
The locale does not seem to affect the error, i.e. I've tested a few
different and it is still only OSX that gives the error but not the
other two.
> regexpr("[a-Z]",
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
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
2004 Feb 06
3
a grep/regexpr problem
Hi,
I'm trying to parse lines of the form:
dan001.hin (0): fingerprint={256, 411, 426, 947, 973, 976}
What I need is the sequence of number between {}. I'm using grep as
match <- grep("{([0-9,\s]*)}",s,perl=T,value=T)
where s is a character vector.
But all I get is the whole string s. I tried using regexpr in an attempt
to get just the sequence I wanted:
match <-
2007 Jun 29
2
regexpr
Hi,
I 'd like to match each member of a list to a target string, e.g.
------------------------------
mylist=c("MN","NY","FL")
g=regexpr(mylist[1], "Those from MN:")
if (g>0)
{
"On list"
}
------------------------------
My question is:
How to add an end-of-string symbol '$' to the to-match string? so that 'M'
won't
2007 Jun 28
3
: regular expressions: escaping a dot
What's really the problem with:
> regexpr( '\.odt$', "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
2004 Mar 24
1
string problems ( grep and regepxr)
Recently working with strings and data
I have found a small problem.
Windows XP
R 1.8.1
Reading data from a "txt file" with readLine.
finding a specific line with "grep" command, all OK.
but here comes the problem...
After finding the correct line(s) i need to find a substring
inside each string.
In this case "tabs" I think it represented by "\t" in 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,
2007 Aug 02
4
Finding multiple characters in the same string
Hi
I have this problem where I need to find if there is any numbers in a
string, this is no problem if theres only one number per string. I would
then simply use the regexpr() funtion togheter with the substring function
to extract the number. But regexpr only picks one number per string either
from the beginning or the end, but not multiple. Can this be done? And how
for example
My string <-
2009 Feb 25
3
regexp capturing group in R
Hello,
Newbie question: how do you capture groups in a regexp in R?
Let's say I have txt="blah blah start=20080101 end=20090224".
I'd like to get the two dates start and end.
In Perl, one would say:
my ($start,$end) = ($txt =~ /start=(\d{8}).*end=(\d{8})/);
I've tried:
txt <- "blah blah start=20080101 end=20090224"
m <-
2010 Jun 02
2
regexpr mystery can not remove trailing spaces
Dear all
I encountered strange problem with regexpr replacement
I made this character object
str <- "02.06.10 12:40 "
> str(str)
chr "02.06.10 12:40 "
I read in an object which seems to be quite similar
> str(as.character(becva$V1)[1])
chr "02.06.10 12:40 "
However I can not remove trailing spaces from it
> sub(' +$',