Displaying 20 results from an estimated 3000 matches similar to: "Exact String Compare in R?"
2009 Oct 30
1
exact string match?
Dear R users:
I need to compare character strings stored in 2 separate data frames. I need
an exact match, so finding "a" in "animal" is no good.
I've tried regexpr, match, and grepl, but to no avail.
Anybody know how to accomplish this?
Ben
--
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2009 Sep 01
1
Syntax for crossed random effects in nlme
Hello R users,
I've read the posts on this topic, and had a look at the R documentation for
nlme, but I can't seem to make this work. I'd like to be able to fit a mixed
effects model with crossed random effects, but also be able to specify the
covariance matrix structure for the residuals. Here's the syntax using the
lmer function in lme4 (which doesn't currently allow
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
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
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>
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}"
#
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
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 <-
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]",
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
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 '$' to the to-match string? so that 'M'
won't
2005 Aug 03
2
regexpr and portability issue
Dear all--
I am still forging my first arms with R and I am fighting with regexpr() as
well as portability between unix and windoz. I need to extract barcodes from
filenames (which are located between a double and single underscore) as well
as the directory where the filename is residing. Here is the solution I came
to:
aFileName <-
2017 Jun 08
0
regular expression help
Zitat von Ashim Kapoor <ashimkapoor at gmail.com>:
> Dear All,
>
> My query is:
>
> Do we always need to use perl = TRUE option when doing ignore.case=TRUE?
>
> A small example :
>
> my_text =
> "RECOVERY OFFICER-II\nDEBTS RECOVERY TRIBUNAL-III\n RC No. 162/2015\nSBI
> VS RAMESH GUPTA.\n Dated: 01.03.2016 Item no.01\n
> Present:
2012 Aug 06
5
regexpr with accents
Hello,
I have build a syntax to find out if a given substring is included in a larger string that works like this:
d1$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$V2)>0] <- 9
I have
2003 Aug 13
7
Regexpr with "."
I'm trying to use the regexpr function to locate the decimal in a character
string. Regardless of the position of the decimal, the function returns 1.
For example,
> regexpr(".", "Female.Alabama")
[1] 1
attr(,"match.length")
[1] 1
In trying to figure out what was going on here, I tried the below command:
> gsub(".", ",",
2007 Jun 28
3
: regular expressions: escaping a dot
What's really the problem with:
> regexpr( '\.odt$', "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
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 <-
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