Hello,
Please always cc the list.
To know more about the regular expressions used by r read
help("regex")
The one I used is not very complicated.
\\. match a dot; it is a meta-character so it needs to be escaped.
{2,} repeated at least 2 times, at most an undetermined number of times.
.*
Hello,
Try
s <- c( "colone..xx.","coltwo.ft..rr.","colthree.gh..az.","colfour.DG..lm.")
sub("\\.{2,}.*$", "", s)
#[1] "colone" "coltwo.ft" "colthree.gh" "colfour.DG"
Às 09:00 de 28/06/19, lionel sicot via R-help escreveu:
c(
Hello,
I have files from an equipment with column names including dots.I would like to
simplify these names but all my attempts with sub and regular expressions are
unsuccessful.
I havec(
"colone..xx.","coltwo.ft..rr.","colthree.gh..az.","colfour.DG..lm.")and I would
like to have c(
Hi,
One way would be:
vec1 - c(CDS 3300..4037, CDS
complement(3300..4037), CDS 3300..4037, CDS
join(21467..26641,27577..28890), CDS
complement(join(30708..31700,31931..31984)), CDS 3300..4037)
library(stringr)
You could also try:
library(gsubfn)
strapply(gsub(\\d+|\\d+,,vec1),([0-9]+),as.numeric,simplify=c)
A.K.
On Thursday, February 6, 2014 1:55 PM, arun smartpink...@yahoo.com wrote:
Hi,
One way would be:
vec1 - c(CDS 3300..4037, CDS
complement(3300..4037), CDS
HI,
May be this helps:
lines1 - readLines(textConnection('text to be ignored...
CDS 687..3158
/gene=AXL2
/note=plasma membrane glycoprotein
other text to be ignored...
CDS complement(3300..4037)
R 3.0.2
OS X
Colleagues
I am writing code to read a large number of files in a particular folder. In
some situations, there may be two versions of the file with different
extensions, e.g.:
FILE.csv
FILE.xls
I extracted the portion before the extension with:
sub(\\..*$,
try this:
x - c( FILE.XXX.csv
+ , FILE.YYY.xls)
sub(\\.[^.]*$, , x)
[1] FILE.XXX FILE.YYY
the '[^.]*' says to match anything BUT a period.
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On
Hi,
Try:
FILELIST - list.files()
FILELIST
#[1] FILE.csv FILE.XXX.csv FILE.YYY.xls
sub((.*)\\..*$, \\1, basename(FILELIST))
#[1] FILE FILE.XXX FILE.YYY
A.K.
On Wednesday, January 15, 2014 7:35 PM, Fisher Dennis fis...@plessthan.com
wrote:
R 3.0.2
OS X
Colleagues
I am writing
You want to match a period and anything that follows to the end of the string,
as long as what follows has no period in it.
\\.[^.]*$
---
Jeff NewmillerThe . . Go Live...
On Jan 15, 2014, at 4:37 PM, Fisher Dennis wrote:
R 3.0.2
OS X
Colleagues
I am writing code to read a large number of files in a particular folder. In
some situations, there may be two versions of the file with different
extensions, e.g.:
FILE.csv
FILE.xls
I extracted
Try sub(\\.[^.]+$, , basename(FILELIST))
Thanks,
Wojtek
On Wed, Jan 15, 2014 at 4:37 PM, Fisher Dennis fis...@plessthan.com wrote:
R 3.0.2
OS X
Colleagues
I am writing code to read a large number of files in a particular folder.
In some situations, there may be two versions of the
I thought that I can use metacharacters such as \w to match word characters
with one backslash. But for some reason, I need to include two backslashes.
grepl(pattern='\w', x=what)
Error: '\w' is an unrecognized escape in character string starting \w
grepl(pattern='\\w', x=what)
[1] TRUE
I
On 05/02/2013 12:49 PM, Seth Dickey wrote:
I thought that I can use metacharacters such as \w to match word characters
with one backslash. But for some reason, I need to include two backslashes.
grepl(pattern='\w', x=what)
Error: '\w' is an unrecognized escape in character string starting \w
On Feb 5, 2013, at 9:49 AM, Seth Dickey wrote:
I thought that I can use metacharacters such as \w to match word characters
with one backslash. But for some reason, I need to include two backslashes.
grepl(pattern='\w', x=what)
Error: '\w' is an unrecognized escape in character string
Hi,
I'm currently reworking a report, originating from a MS Access database, but
should be implemented in R.
Now I'm facing the task to convert a lot of queries to postgreSQL.
What I want to do is make a function which takes the MS Access query as an
argument and returns the pgSQL version.
So:
Hello,
try this:
x - c(SELECT [public_tblFiche].[Fichenr], [public_tblArtnr].[Artnr], SELECT
public_tblFiche.Fichenr, public_tblArtnr.Artnr)
# The square backets [ and ] should removed
x - gsub([][], , x)
# and xxx_xxx.xxx should become \xxx\.\xxx\\.\xxx\
x -
Am 22.08.2012, 21:46 Uhr, schrieb Dr. Holger van Lishaut
h.v.lish...@gmx.de:
SignifStellen-function(x){
strx=as.character(x)
nchar(regmatches(strx, regexpr([1-9][0-9]*\\.[0-9]*[1-9],strx)))-1
}
returns the significant figures of a number. Perhaps this can help
someone.
Sorry,
Dear all,
regmatches works.
And, since this has been asked here before:
SignifStellen-function(x){
strx=as.character(x)
nchar(regmatches(strx, regexpr([1-9][0-9]*\\.[0-9]*[1-9],strx)))-1
}
returns the significant figures of a number. Perhaps this can help someone.
Thanks best
...
On Wed, Aug 22, 2012 at 12:46 PM, Dr. Holger van Lishaut
h.v.lish...@gmx.de wrote:
Dear all,
regmatches works.
And, since this has been asked here before:
SignifStellen-function(x){
strx=as.character(x)
nchar(regmatches(strx, regexpr([1-9][0-9]*\\.[0-9]*[1-9],strx)))-1
}
Dear r-help members,
I have a number in the form of a string, say:
a--01020.909200
I'd like to extract 1020. as well as .9092
Front-grep(pattern=[1-9]+[0-9]*\\., value=TRUE, x=a, fixed=FALSE)
End-grep(pattern=\\.[0-9]*[1-9]+, value=TRUE, x=a, fixed=FALSE)
However, both strings give
grep() returns the matches. You want regexpr() and regmatches()
-- Bert
On Tue, Aug 21, 2012 at 12:24 PM, Dr. Holger van Lishaut
h.v.lish...@gmx.de wrote:
Dear r-help members,
I have a number in the form of a string, say:
a--01020.909200
I'd like to extract 1020. as well as .9092
'grep' does not change strings. Use 'gsub' or 'regmatches':
# gsub
Front - gsub(^.*?([1-9][0-9]*\\.).*?$, \\1, a)
End - gsub(^.*?(\\.[0-9]*[1-9]).*?$, \\1, a)
# regexpr and regmatches (R = 2.14.0)
Front - regmatches(a, regexpr([1-9][0-9]*\\., a))
End - regmatches(a, regexpr(\\.[0-9]*[1-9], a))
You're misreading the docs: from grep,
value: if ‘FALSE’, a vector containing the (‘integer’) indices of
the matches determined by ‘grep’ is returned, and if ‘TRUE’,
a vector containing the matching elements themselves is
returned.
Since there's a match somewhere
HI,
Try this:
gsub(^-\\d(\\d{4}.).*,\\1,a)
#[1] 1020.
gsub(^.*(.\\d{5}).,\\1,a)
#[1] .90920
A.K.
- Original Message -
From: Dr. Holger van Lishaut h.v.lish...@gmx.de
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Tuesday, August 21, 2012 3:24 PM
Subject: [R] Regular
Hi all,
My code looks like the following:
inname = read.csv(ID_error_checker.csv, as.is=TRUE)
outname = read.csv(output.csv, as.is=TRUE)
#My algorithm is the following:
#for line in inname
#if first string up to whitespace in row in inname$name = first string up
to whitespace in row + 1 in
New York Mets 1900ESPN
#2 2 New York Yankees 1920 Cooperstown
A.K.
- Original Message -
From: Fred G bayespoker...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Friday, August 10, 2012 1:41 PM
Subject: [R] Regular Expressions + Matrices
Hi all,
My code looks like
Hello,
Try the following.
d - read.table(textConnection(
ID NAME YEAR SOURCE
1 'New York Mets' 1900 ESPN
2 'New York Yankees' 1920 Cooperstown
3 'Boston Redsox' 1918 ESPN
4 'Washington Nationals' 2010
:
Sent: Friday, August 10, 2012 1:41 PM
Subject: [R] Regular Expressions + Matrices
Hi all,
My code looks like the following:
inname = read.csv(ID_error_checker.csv, as.is=TRUE)
outname = read.csv(output.csv, as.is=TRUE)
#My algorithm is the following:
#for line in inname
#if first string up
NAME YEAR SOURCE
#1 1New York Mets 1900ESPN
#2 2 New York Yankees 1920 Cooperstown
A.K.
- Original Message -
From: Fred G bayespoker...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Friday, August 10, 2012 1:41 PM
Subject: [R] Regular Expressions + Matrices
Hi
SOURCE
#1 1New York Mets 1900ESPN
#2 2 New York Yankees 1920 Cooperstown
A.K.
- Original Message -
From: Fred G bayespoker...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Friday, August 10, 2012 1:41 PM
Subject: [R] Regular Expressions + Matrices
Hi all
To: Fred G
Cc: r-help
Subject: Re: [R] Regular Expressions + Matrices
Hello,
Try the following.
d - read.table(textConnection(
ID NAME YEAR SOURCE
1 'New York Mets' 1900 ESPN
2 'New York Yankees' 1920 Cooperstown
3
-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf
Of Rui Barradas
Sent: Friday, August 10, 2012 11:18 AM
To: Fred G
Cc: r-help
Subject: Re: [R] Regular Expressions + Matrices
Hello,
Try the following.
d - read.table(textConnection(
ID
Dear all
I would like to ask from dir function in R (?dir)
to give me only the files that end with .txt or .doc.
The dir functions supports the use of patterns (is not that regular
expressions) for doing that.
print(dir(i,full.names=TRUE,pattern=.))
Could you please help me compose such
From the help for dir:
File naming conventions are platform dependent. The pattern
matching works with the case of file names as returned by the OS
On my linux system, this works:
dir(pattern=*.txt)
[1] a.txt b.txt
dir(pattern=*.doc)
[1] c.doc
dir(pattern=*.doc|*.txt)
[1] a.txt
Do you wish to include .docx files as well or just .doc?
Michael
On Wed, Dec 21, 2011 at 10:04 AM, Alaios ala...@yahoo.com wrote:
Dear all
I would like to ask from dir function in R (?dir)
to give me only the files that end with .txt or .doc.
The dir functions supports the use of patterns
To be correct for the regular expression, it should be:
dir(pattern = \\.(txt|doc)$)
The form
dir(pattern=*.txt)
will match 'txt' appearing anywhere in the name; this looks like the
argument you would have used to Sys.glob which is a UNIX style file
name match and not a regular expression. .
Thanks to everyone who contributed to my questions. As ever, I am extremely
grateful to all those on the R-list who make it what it is.
Regards
Mike Griffiths
On Tue, Nov 15, 2011 at 5:47 PM, Joshua Wiley jwiley.ps...@gmail.comwrote:
Hi Michael,
Your strings were long so I made a bit
Good afternoon list,
I have the following character strings; one with spaces between the maths
operators and variable names, and one without said spaces.
form-c('~ Sentence + LEGAL + Intro + Intro / Intro1 + Intro * LEGAL +
benefit + benefit / benefit1 + product + action * mean + CTA + help +
Hi Michael,
You need to take another look at the examples you were given, and at
the help for ?sub():
The two ‘*sub’ functions differ only in that ‘sub’ replaces only
the first occurrence of a ‘pattern’ whereas ‘gsub’ replaces all
occurrences. If ‘replacement’ contains
Hi Michael,
Your strings were long so I made a bit smaller example. Sarah made
one good point, you want to be using gsub() not sub(), but when I use
your code, I do not think it even works precisely for one instance.
Try this on for size, you were 99% there:
## simplified cases
form1 -
I'm running R 2.13 on Ubuntu 10.10
I have a data set which is comprised of character strings.
site = readLines('http://www.census.gov/tiger/tms/gazetteer/zips.txt')
dat - c(01, 35004, AL, ACMAR, 86.51557, 33.584132, 6055, 0.001499)
dat
I want to loop through the data and construct a data frame
On Jun 7, 2011, at 3:55 PM, Abraham Mathew wrote:
I'm running R 2.13 on Ubuntu 10.10
I have a data set which is comprised of character strings.
site = readLines('http://www.census.gov/tiger/tms/gazetteer/zips.txt')
dat - c(01, 35004, AL, ACMAR, 86.51557, 33.584132, 6055, 0.001499)
dat
Hi all,
I am hoping that someone can help me with a problem I am having with column
headings. I have read a table into R using read.table: the rows are
documents, and the columns are counts of regular expression matches (so that
the column heading is the given regular expression). My problem is
On Wed, Mar 9, 2011 at 8:52 AM, Matthew DeAngelis roni...@gmail.com wrote:
Hi all,
I am hoping that someone can help me with a problem I am having with column
headings. I have read a table into R using read.table: the rows are
documents, and the columns are counts of regular expression
Hi,
I'm trying to figure out how to use capturing parenthesis in regular
expressions in R. (Doing this in Perl, Java, etc. is fairly trivial,
but I can't seem to find the functionality in R.)
For example, given the string:10 Nov 13.00 (PFE1020K13)
I want to capture the first to digits
On Thu, 4 Nov 2010, Noah Silverman wrote:
Hi,
I'm trying to figure out how to use capturing parenthesis in regular
expressions in R. (Doing this in Perl, Java, etc. is fairly trivial, but I
can't seem to find the functionality in R.)
For example, given the string:10 Nov 13.00
That's perfect!
Don't know how I missed that.
I want to start playing with some modeling of financial data and the
only format I can download is rather ugly. So my plan is to use a
series of Regex to extract what I want.
Noticed that you are a Prof. in applied stats. I'm at UCLA working on
On 11/5/2010 12:09 AM, Prof Brian Ripley wrote:
On Thu, 4 Nov 2010, Noah Silverman wrote:
Hi,
I'm trying to figure out how to use capturing parenthesis in regular
expressions in R. (Doing this in Perl, Java, etc. is fairly trivial,
but I can't seem to find the functionality in R.)
For
2010/11/5 Brian Diggs dig...@ohsu.edu:
Is there a standard, built in way to get both (all) backreferences at the
same time with just one call to sub (or the appropriate function)? I can
cobble something together specifically for 2 backreferences (not extensively
tested):
both_backrefs -
Ok, we decided to have a shot at modifying gregexpr. Let's see how it
works out. If anybody is interested in discussing this please contact
me. R-help doesn't seem like the right place for further discussion.
Is there a default place for discussing things like that?
Thanks everybody for your
Bill, Michael,
good to see I'm not the only one who sees potential for improvements
in the regexpr domain. Adding a subpattern argument is certainly a
step in the right direction and would make my life much easier.
However, in my application I need to know not only the position of one
group but
I'd definitely be a customer for it Titus. And it does seem like an
obvious hole in regex processing in R that cries out to be filled.
Um, ggregexpr isn't the sexiest of function names :) Perhaps we can
think of something a little easier ?
How is your C coding ? Bill ? Anyone else ? I could
On Wed, Sep 29, 2010 at 1:58 PM, Michael Bedward
michael.bedw...@gmail.com wrote:
How is your C coding ? Bill ? Anyone else ? I could have a got at
writing some prototype code to test in the next few days, though if
someone else with decent C skills is itching to do it please speak up.
We
What Titus wants to do is akin to retrieving capturing groups from a
Matcher object in Java. I also thought there must be an existing,
elegant solution to this some time ago and searched for it, including
looking at the sources (albeit with not much expertise) but came up
blank.
I also looked at
On Tue, Sep 28, 2010 at 9:46 AM, Michael Bedward
michael.bedw...@gmail.com wrote:
What Titus wants to do is akin to retrieving capturing groups from a
Matcher object in Java.
Precisely. Here's the description:
On Tue, Sep 28, 2010 at 6:52 AM, Titus von der Malsburg
malsb...@gmail.com wrote:
On Tue, Sep 28, 2010 at 9:46 AM, Michael Bedward
michael.bedw...@gmail.com wrote:
What Titus wants to do is akin to retrieving capturing groups from a
Matcher object in Java.
Precisely. Here's the description:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Michael Bedward
Sent: Tuesday, September 28, 2010 12:46 AM
To: Titus von der Malsburg
Cc: r-help@r-project.org
Subject: Re: [R] Regular expressions: offsets of groups
What
Ah, that's interesting - thanks Bill. That's certainly on the right
track for me (Titus, you too ?) especially if the subpattern argument
accepted a vector of multiple group indices.
As you say, this is straightforward in C. I'd be happy to (try to)
make a patch for the R sources if there was
Dear list!
gregexpr(a+(b+), abcdaabbc)
[[1]]
[1] 1 5
attr(,match.length)
[1] 2 4
What I want is the offsets of the matches for the group (b+), i.e. 2
and 7, not the offsets of the complete matches. Is there a way in R
to get that?
I know about gsubgn and strapply, but they only give me the
try this:
x - gregexpr(a+(b+), abcdaabbcaaacaaab)
justA - gregexpr(a+, abcdaabbcaaacaaab)
# find matches in 'x' for 'justA'
indx - which(justA[[1]] %in% x[[1]])
# now determine where 'b' starts
justA[[1]][indx] + attr(justA[[1]], 'match.length')[indx]
[1] 2 7 17
On Mon, Sep 27, 2010
Thank you Jim, but just as the solution that I discussed, your
proposal involves deconstructing the pattern and searching several
times. I'm looking for a general and efficient solution. Internally,
the regexpr engine has all necessary information after one pass
through the string. What I need
On Mon, Sep 27, 2010 at 7:16 PM, Henrique Dallazuanna www...@gmail.com wrote:
You've tried:
gregexpr(b+, abcdaabbc)
But this would match the third occurrence of b+ in abcdaabbcbb. But
in this example I'm only interested in b+ if it's preceded by a+.
Titus
On Mon, Sep 27, 2010 at 11:48 AM, Titus von der Malsburg
malsb...@gmail.com wrote:
Dear list!
gregexpr(a+(b+), abcdaabbc)
[[1]]
[1] 1 5
attr(,match.length)
[1] 2 4
What I want is the offsets of the matches for the group (b+), i.e. 2
and 7, not the offsets of the complete matches. Is
On Mon, Sep 27, 2010 at 7:29 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Try this zero width negative look behind expression:
gregexpr((?!a+)(b+), abcdaabbc, perl = TRUE)
[[1]]
[1] 2 7
attr(,match.length)
[1] 1 2
Thanks Gabor, but this gives me the same result as
gregexpr(b+,
You could do this:
gregexpr(ab+, abcdaabbcbb)[[1]] + 1
On Mon, Sep 27, 2010 at 2:25 PM, Titus von der Malsburg
malsb...@gmail.comwrote:
On Mon, Sep 27, 2010 at 7:16 PM, Henrique Dallazuanna www...@gmail.com
wrote:
You've tried:
gregexpr(b+, abcdaabbc)
But this would match the third
You've tried:
gregexpr(b+, abcdaabbc)
On Mon, Sep 27, 2010 at 12:48 PM, Titus von der Malsburg malsb...@gmail.com
wrote:
Dear list!
gregexpr(a+(b+), abcdaabbc)
[[1]]
[1] 1 5
attr(,match.length)
[1] 2 4
What I want is the offsets of the matches for the group (b+), i.e. 2
and 7, not
On Mon, Sep 27, 2010 at 1:34 PM, Titus von der Malsburg
malsb...@gmail.com wrote:
On Mon, Sep 27, 2010 at 7:29 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Try this zero width negative look behind expression:
gregexpr((?!a+)(b+), abcdaabbc, perl = TRUE)
[[1]]
[1] 2 7
Dear list,
I have the following text to parse (originating from readLines as some
lines have unequal size),
st = c(START text1 1 text2 2.3, whatever intermediate text, START
text1 23.4 text2 3.1415)
from which I'd like to extract the lines starting with START, and
group the subsequent fields in
Assuming only START fields match pat:
## this one has more fields: how do I generalize the regular expression?
st2 = c(START text1 1 text2 2.3 text3 5, whatever intermediate text,
+ START text1 23.4 text2 3.1415 text3 6)
pat - [[:alnum:]]+ +([0-9.]+)
s - strapply(st2, pat, c, simplify =
Perfect, thanks!
baptiste
2009/10/26 Gabor Grothendieck ggrothendi...@gmail.com:
Assuming only START fields match pat:
## this one has more fields: how do I generalize the regular expression?
st2 = c(START text1 1 text2 2.3 text3 5, whatever intermediate text,
+ START text1 23.4 text2
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
Try adding perl = TRUE
On Sun, Jul 6, 2008 at 5:17 PM, Duncan Murdoch [EMAIL PROTECTED] wrote:
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
On 06-Jul-08 21:17:04, Duncan Murdoch wrote:
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
On 06/07/2008 5:37 PM, (Ted Harding) wrote:
On 06-Jul-08 21:17:04, Duncan Murdoch wrote:
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
Look at the discussion of zero width lookahead assertions in ?regex .
Use perl = TRUE as previously indicated.
On Sun, Jul 6, 2008 at 7:29 PM, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 06/07/2008 5:37 PM, (Ted Harding) wrote:
On 06-Jul-08 21:17:04, Duncan Murdoch wrote:
I'm trying to write
On 06/07/2008 7:37 PM, Gabor Grothendieck wrote:
Look at the discussion of zero width lookahead assertions in ?regex .
Use perl = TRUE as previously indicated.
Thanks, this seems to work:
gsub( (?!E)((EE)*)q, \\1Eq, x, perl=TRUE)
Duncan Murdoch
On Sun, Jul 6, 2008 at 7:29 PM, Duncan
Hi R,
Again struck with regular expressions...
Suppose,
S=c(World_is_beautiful, one_two_three_four,My_book)
I need to extract the last but one element of the strings. So, my output should
look like:
Ans=c(is,three,My)
gsub() can do this...but wondering how do I give the
: Tuesday, May 13, 2008 11:02 AM
Subject: [R] Regular Expressions
Hi R,
Again struck with regular expressions...
Suppose,
S=c(World_is_beautiful, one_two_three_four,My_book)
I need to extract the last but one element of the strings. So, my
output should look like:
Ans=c(is,three,My
S=c(World_is_beautiful, one_two_three_four,My_book)
I need to extract the last but one element of the strings. So, my
output should look like:
Ans=c(is,three,My)
gsub() can do this...but wondering how do I give the regular
expression
sapply(strsplit(S, _), function(x)
On Tue, May 13, 2008 at 5:02 AM, Shubha Vishwanath Karanth
[EMAIL PROTECTED] wrote:
Suppose,
S=c(World_is_beautiful, one_two_three_four,My_book)
I need to extract the last but one element of the strings. So, my output
should look like:
Ans=c(is,three,My)
gsub() can do this...but
I am having some trouble learning regular expressions. Let me describe
the general problem I am dealing with. Consider the following setup:
Joe- c(1,2,3)
Bob- c(2,4,6)
Alice - c(9,8,7)
Matrix - cbind(Joe, Bob, Alice)
St - c(Bob, Alice, Alice:Bob)
Now I want to make a new matrix having only the
On 19.04.2008, at 06:46, maud wrote:
I am having some trouble learning regular expressions. Let me describe
the general problem I am dealing with. Consider the following setup:
Joe- c(1,2,3)
Bob- c(2,4,6)
Alice - c(9,8,7)
Matrix - cbind(Joe, Bob, Alice)
St - c(Bob, Alice, Alice:Bob)
Hello all,
Still fighting with regular expressions and such, I am again stuck:
Suppose I have a vector of character chains. In this vector, I wish to identify
which character chains start with a given pattern, and then replace everything
that comes after said pattern.
Here is a quick
] regular expressions
Hello all,
Still fighting with regular expressions and such, I am again stuck:
Suppose I have a vector of character chains. In this vector,
I wish to identify which character chains start with a given
pattern, and then replace everything that comes after said pattern
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