Re: [R] Regular expressions and 2 dots
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.
.* any character (.) repeated zero or more times (*).
$end of string.
All together now.
"\\.{2,}.*$" matches at least 2 dots followed by anything until
the end of the string.
The replacement for this is the empty string "", so what is matched is
removed.
Hope this helps,
Rui Barradas
Às 09:33 de 28/06/19, ptit_b...@yahoo.fr escreveu:
Just a small message to thank you for your fast and working solution.
I now have to understand it ...
Nice day.
_
Sent from http://r.789695.n4.nabble.com
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Re: [R] Regular expressions and 2 dots
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( "colone..xx.","coltwo.ft..rr.","colthree.gh..az.","colfour.DG..lm.")
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[R] Regular expressions and 2 dots
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( "colone","coltwo.ft","colthree.gh","colfour.DG") that means to delete everything after the two dots but to keep strings after intermediate dot like in colfour.DG Thanks in advance for your help (a working solution would be good, explainations with the working solution would be ideal).Nice day,Ptit Bleu. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions, genbank
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)
/gene="REV7"
other text to be ignored...
CDS <4500..4550
/gene="REV7"
other text to be ignored...
CDS complement(join(30708..31700,31931..31984))
/gene="REV7"'))
lines2 <- lines1[grep("CDS",lines1)]
lines3 <- lines2[!grepl("[<>]",lines2)]
indx <- grepl("complement",lines3)*1
mapply(`c`,indx,strapply(lines3,"([0-9]+)",as.numeric))
#[[1]]
#[1] 0 687 3158
#
#[[2]]
#[1] 1 3300 4037
#
#[[3]]
#[1] 1 30708 31700 31931 31984
If you want to have "," as sep:
lapply(mapply(`c`,indx,strapply(lines3,"([0-9]+)",as.numeric)),paste,collapse=",
")
A.K.
For
sure, maybe I could provide a more realistic sample of what I have
rather than the vector. Here is a chunk of the text I'll be processing:
text to be ignored... CDS 687..3158 /gene="AXL2" /note="plasma
membrane glycoprotein"
other text to be ignored...
CDS complement(3300..4037) /gene="REV7"
other text to be ignored...
CDS <4500..4550 /gene="REV7"
other text to be ignored...
CDS complement(join(30708..31700,31931..31984)) /gene="REV7"
and so on ...
processing this text, I want the following output (let's say) in a list called
output with as many elements as there are valid "CDS" (i.e. CDS without "<" or
">"), where the first component of each element of the list is a 0/1 number
that tells if what followed "CDS" included the word "complement" or not. Here
is what I would like to get for the above text:
output:
[[1]] 0, 687, 3158
[[2]] 1, 3300, 4037
[[3]] 1, 30708, 31700, 31931, 31984
Thanks again for the help!
Thank you very much for the response! This is a major improvement on
what I was getting! I need to read and understand what is done as I need to
modify it a little bit. The exact requirement for me is to not only
recognize the numbers that follow "CDS" but also be able to
differentiate between the 4 accepted cases:
"CDS 3300..4037"
or
"CDS complement(3300..4037)"
or
"CDS join(21467..26641,27577..28890)"
or
"CDS complement(join(30708..31700,31931..31984))"
I need to do different things for each for example, when "join"
follows the gap, I need to join the ranges (e.g. in this case have two
intervals [21467 26641] U [27577 28890]) in one set. Many thanks though
for getting me going!
On Thursday, February 6, 2014 2:20 PM, arun wrote:
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 wrote:
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)
as.numeric(unlist(strsplit(str_trim(gsub("\\D+","
",gsub("\\d+<|>\\d+","",vec1)))," ")))
# [1] 3300 4037 3300 4037 4037 21467 26641 27577 28890 30708 31700 31931
#[13] 31984
A.K.
Hi,
I have been using R for the past 1.5 years and usually have
found topics to be relatively easy to learn on your own, but I am
finding the learning curve with the regular expressions to be a little
steep especially since I haven't found any good tutorials. While I
intend to spend more time systematically learning proper ways of making
regular expressions, I have a project that is coming due and can't wait
for that so I was hoping to get some direct help.
I need to extract all the numbers in lines with following formats:
"CDS 3300..4037"
or
"CDS complement(3300..4037)"
or
"CDS join(21467..26641,27577..28890)"
or
"CDS complement(join(30708..31700,31931..31984))"
but not if any of the numbers are preceded by "<" or followed by ">"
Many thanks in advance!
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Re: [R] Regular expressions, genbank
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 wrote:
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)
as.numeric(unlist(strsplit(str_trim(gsub("\\D+","
",gsub("\\d+<|>\\d+","",vec1)))," ")))
# [1] 3300 4037 3300 4037 4037 21467 26641 27577 28890 30708 31700 31931
#[13] 31984
A.K.
Hi,
I have been using R for the past 1.5 years and usually have
found topics to be relatively easy to learn on your own, but I am
finding the learning curve with the regular expressions to be a little
steep especially since I haven't found any good tutorials. While I
intend to spend more time systematically learning proper ways of making
regular expressions, I have a project that is coming due and can't wait
for that so I was hoping to get some direct help.
I need to extract all the numbers in lines with following formats:
"CDS 3300..4037"
or
"CDS complement(3300..4037)"
or
"CDS join(21467..26641,27577..28890)"
or
"CDS complement(join(30708..31700,31931..31984))"
but not if any of the numbers are preceded by "<" or followed by ">"
Many thanks in advance!
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Regular expressions, genbank
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)
as.numeric(unlist(strsplit(str_trim(gsub("\\D+","
",gsub("\\d+<|>\\d+","",vec1)))," ")))
# [1] 3300 4037 3300 4037 4037 21467 26641 27577 28890 30708 31700 31931
#[13] 31984
A.K.
Hi,
I have been using R for the past 1.5 years and usually have
found topics to be relatively easy to learn on your own, but I am
finding the learning curve with the regular expressions to be a little
steep especially since I haven't found any good tutorials. While I
intend to spend more time systematically learning proper ways of making
regular expressions, I have a project that is coming due and can't wait
for that so I was hoping to get some direct help.
I need to extract all the numbers in lines with following formats:
"CDS 3300..4037"
or
"CDS complement(3300..4037)"
or
"CDS join(21467..26641,27577..28890)"
or
"CDS complement(join(30708..31700,31931..31984))"
but not if any of the numbers are preceded by "<" or followed by ">"
Many thanks in advance!
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Regular expressions on filenames
Try sub("\\.[^.]+$", "", basename(FILELIST))
Thanks,
Wojtek
On Wed, 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 the portion before the extension with:
> sub("\\..*$", "", basename(FILELIST))
> then used
> duplicated
> to find duplicates. All was well until I encountered files named:
> FILE.XXX.csv
> FILE.YYY.xls
>
> My regular expression extracted only the FILE portion of the text and
> claimed that the filenames (without the extensions) matched. Can someone
> provide me with the appropriate regular expression to deal with this?
> Thanks.
>
> Dennis
>
>
> Dennis Fisher MD
> P < (The "P Less Than" Company)
> Phone: 1-866-PLessThan (1-866-753-7784)
> Fax: 1-866-PLessThan (1-866-753-7784)
> www.PLessThan.com
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
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Re: [R] Regular expressions on filenames
On Jan 15, 2014, at 4:37 PM, Fisher Dennis wrote:
> R 3.0.2
> OS X
>
> Colleagues
>
> I am writing code to read a large number of files in a particular folder. In
> some situations, there may be two versions of the file with different
> extensions, e.g.:
> FILE.csv
> FILE.xls
> I extracted the portion before the extension with:
> sub("\\..*$", "", basename(FILELIST))
> then used
> duplicated
> to find duplicates. All was well until I encountered files named:
> FILE.XXX.csv
> FILE.YYY.xls
>
> My regular expression extracted only the “FILE” portion of the text and
> claimed that the filenames (without the extensions) matched. Can someone
> provide me with the appropriate regular expression to deal with this? Thanks.
Why not:
sub("\\..{3}$", "", basename(FILELIST))
See ?regex
--
David Winsemius
Alameda, CA, USA
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Re: [R] Regular expressions on filenames
You want to match a period and anything that follows to the end of the string,
as long as what follows has no period in it.
"\\.[^.]*$"
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
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 the portion before the extension with:
> sub("\\..*$", "", basename(FILELIST))
>then used
> duplicated
>to find duplicates. All was well until I encountered files named:
> FILE.XXX.csv
> FILE.YYY.xls
>
>My regular expression extracted only the “FILE” portion of the text and
>claimed that the filenames (without the extensions) matched. Can
>someone provide me with the appropriate regular expression to deal with
>this? Thanks.
>
>Dennis
>
>
>Dennis Fisher MD
>P < (The "P Less Than" Company)
>Phone: 1-866-PLessThan (1-866-753-7784)
>Fax: 1-866-PLessThan (1-866-753-7784)
>www.PLessThan.com
>
>__
>R-help@r-project.org mailing list
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions on filenames
Hi,
Try:
FILELIST <- list.files()
FILELIST
#[1] "FILE.csv" "FILE.XXX.csv" "FILE.YYY.xls"
sub("(.*)\\..*$", "\\1", basename(FILELIST))
#[1] "FILE" "FILE.XXX" "FILE.YYY"
A.K.
On Wednesday, January 15, 2014 7:35 PM, Fisher Dennis
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 the portion before the extension with:
sub("\\..*$", "", basename(FILELIST))
then used
duplicated
to find duplicates. All was well until I encountered files named:
FILE.XXX.csv
FILE.YYY.xls
My regular expression extracted only the “FILE” portion of the text and claimed
that the filenames (without the extensions) matched. Can someone provide me
with the appropriate regular expression to deal with this? Thanks.
Dennis
Dennis Fisher MD
P < (The "P Less Than" Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-866-PLessThan (1-866-753-7784)
www.PLessThan.com
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Regular expressions on filenames
try this:
> x <- c( "FILE.XXX.csv"
+ , "FILE.YYY.xls")
> sub("\\.[^.]*$", "", 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 Wed, Jan 15, 2014 at 7:37 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 the portion before the extension with:
> sub("\\..*$", "", basename(FILELIST))
> then used
> duplicated
> to find duplicates. All was well until I encountered files named:
> FILE.XXX.csv
> FILE.YYY.xls
>
> My regular expression extracted only the “FILE” portion of the text and
> claimed that the filenames (without the extensions) matched. Can someone
> provide me with the appropriate regular expression to deal with this? Thanks.
>
> Dennis
>
>
> Dennis Fisher MD
> P < (The "P Less Than" Company)
> Phone: 1-866-PLessThan (1-866-753-7784)
> Fax: 1-866-PLessThan (1-866-753-7784)
> www.PLessThan.com
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Regular expressions on filenames
R 3.0.2
OS X
Colleagues
I am writing code to read a large number of files in a particular folder. In
some situations, there may be two versions of the file with different
extensions, e.g.:
FILE.csv
FILE.xls
I extracted the portion before the extension with:
sub("\\..*$", "", basename(FILELIST))
then used
duplicated
to find duplicates. All was well until I encountered files named:
FILE.XXX.csv
FILE.YYY.xls
My regular expression extracted only the “FILE” portion of the text and claimed
that the filenames (without the extensions) matched. Can someone provide me
with the appropriate regular expression to deal with this? Thanks.
Dennis
Dennis Fisher MD
P < (The "P Less Than" Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-866-PLessThan (1-866-753-7784)
www.PLessThan.com
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R Regular Expressions - Metacharacters
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 starting "\w"
>
>> grepl(pattern='\\w', x="what")
> [1] TRUE
>
> I can't find the reason for this on the help pages. Does anyone know why?
The help page for ?regex says near the top ...
"Any metacharacter with special meaning may be quoted by preceding it with a
backslash. The metacharacters in EREs are . \ | ( ) [ { ^ $ * + ?, but note
that whether these have a special meaning depends on the context."
>
> Thanks!
>
> [[alternative HTML version deleted]]
>
> __
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
David Winsemius
Alameda, CA, USA
__
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Re: [R] R Regular Expressions - Metacharacters
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" > grepl(pattern='\\w', x="what") [1] TRUE I can't find the reason for this on the help pages. Does anyone know why? grepl wants a string containing a single backslash. R uses the backslash as an escape character, so you need to double it in your source, so the string ends up containing just one. "\w" is interpreted by R as an escaped w, which doesn't make sense. "\\w" is interpreted by R as a backslash followed by a w, and then the \w is interpreted by grepl the way you want. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R Regular Expressions - Metacharacters
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 can't find the reason for this on the help pages. Does anyone know why? Thanks! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: stuck again...
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 <- gsub("([[:alpha:]]+)_([[:alpha:]]+)\\.([[:alpha:]]+)",
"\"\\1\".\"\\2\".\"\\3\"", x)
--
Noia Raindrops
noia.raindr...@gmail.com
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[R] Regular expressions: stuck again...
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:
SELECT [public_tblFiche].[Fichenr], [public_tblArtnr].[Artnr] FROM
[public_tblFiche], [public_tblArtnr] WHERE [public_tblFiche].[Artnr_ID] =
[public_tblArtnr].[Artnr_ID];
or
SELECT public_tblFiche.Fichenr, public_tblArtnr.Artnr FROM public_tblFiche,
public_tblArtnr WHERE public_tblFiche.Artnr_ID = public_tblArtnr.Artnr_ID;
Should become:
SELECT \"public\".\"tblFiche\".\"Fichenr\",
\"public\".\"tblArtnr\".\"Artnr\" FROM \"public\".\"tblFiche\",
\"public\".\"tblArtnr\" WHERE \"public\".\"tblFiche\".\"Artnr_ID\" =
\"public\".\"tblArtnr\".\"Artnr_ID\";
concrete:
The square backets [ and ] should removed
and xxx_xxx.xxx should become \"xxx\".\"xxx\"\".\"xxx\"
When only queries with square brackets, I used
gsub('[', '\"', x, fixed=TRUE),
gsub(']', '\"', x, fixed=TRUE),
gsub('_', '\"', x, fixed=TRUE),
But to do the trick with regular expressions, I cant get a grip on this
Anyone who can give me some help?
Thanks
Bart
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Re: [R] Regular Expressions in grep - Solution and function to determine significant figures of a number
Am 22.08.2012, 21:46 Uhr, schrieb Dr. Holger van Lishaut
:
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, to work, it must read:
SignifStellen<-function(x){
strx=as.character(x)
intFront <- nchar(regmatches(strx, regexpr("[1-9][0-9]*\\.", strx)))
intEnd <- nchar(regmatches(strx, regexpr("\\.[0-9]*[1-9]", strx)))
intFront+intEnd-2
}
Best regards
H. van Lishaut
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Re: [R] Regular Expressions in grep - Solution and function to determine significant figures of a number
...
On Wed, Aug 22, 2012 at 12:46 PM, Dr. Holger van Lishaut
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
> }
>
> returns the significant figures of a number. Perhaps this can help someone.
except that ?signif already does this, no?
-- Bert
>
> Thanks & best regards
> H. van Lishaut
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
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Re: [R] Regular Expressions in grep - Solution and function to determine significant figures of a number
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 regards
H. van Lishaut
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Re: [R] Regular Expressions in grep
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
To: "r-help@r-project.org"
Cc:
Sent: Tuesday, August 21, 2012 3:24 PM
Subject: [R] Regular Expressions in grep
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 "-01020.909200", exactly a.
Could you please point me to what is wrong?
Thanks and best regards
H. van Lishaut
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Re: [R] Regular Expressions in grep
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 in a[1], all of a[1] is returned (it is a matching element), not just the matching bit: grep(x, value = TRUE) is something like x[grepl(x)] to my mind. I think you want ?regexpr or possibly just substitute out the non-match with gsub. Cheers, Michael On Tue, Aug 21, 2012 at 2:24 PM, Dr. Holger van Lishaut 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" > > 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 "-01020.909200", exactly a. > Could you please point me to what is wrong? > > Thanks and best regards > H. van Lishaut > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular Expressions in grep
'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))
Front
## [1] "1020."
End
## [1] ".9092"
--
Noia Raindrops
noia.raindr...@gmail.com
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Re: [R] Regular Expressions in grep
grep() returns the matches. You want regexpr() and regmatches() -- Bert On Tue, Aug 21, 2012 at 12:24 PM, Dr. Holger van Lishaut 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" > > 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 "-01020.909200", exactly a. > Could you please point me to what is wrong? > > Thanks and best regards > H. van Lishaut > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Regular Expressions in grep
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 "-01020.909200", exactly a. Could you please point me to what is wrong? Thanks and best regards H. van Lishaut __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular Expressions + Matrices
Thanks Bill! Works great! Thanks again guys!
On Fri, Aug 10, 2012 at 2:43 PM, William Dunlap wrote:
> If you think about this as a runs problem you can get a loopless solution
> that I think is easier to read (once the requisite functions are defined).
>
> First define the function to canonicalize the name
>nickname <- function(x) sub(" .*", "", x)
> then define some handy runs functions
> isFirstInRun <- function(x) c(TRUE, x[-1] != x[-length(x)])
> isJustBefore <- function(x) c(x[-1], FALSE) # x should be logical
> then use those functions on your dataset
> > nearDup <- !isFirstInRun(nickname(d$NAME)) & isFirstInRun(d$YEAR)
> > d[ nearDup | isJustBefore(nearDup), ]
> ID NAME YEAR SOURCE
> 1 1New York Mets 1900ESPN
> 2 2 New York Yankees 1920 Cooperstown
> See how it works with triplicates as well
> > dd <- rbind(d, data.frame(ID=6:8,
> NAME=c("Chicago Blacksox", "Chicago Cubs",
> "Chicago Whitesox"),
> YEAR=1701:1703, SOURCE=rep("made up", 3)))
> > nearDup <- !isFirstInRun(nickname(dd$NAME)) & isFirstInRun(dd$YEAR)
> > dd[ nearDup | isJustBefore(nearDup), ]
> ID NAME YEAR SOURCE
> 1 1New York Mets 1900ESPN
> 2 2 New York Yankees 1920 Cooperstown
> 6 6 Chicago Blacksox 1701 made up
> 7 7 Chicago Cubs 1702 made up
> 8 8 Chicago Whitesox 1703 made up
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
>
> > -Original Message-----
> > 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 NAME YEAR SOURCE
> > 1 'New York Mets' 1900 ESPN
> > 2 'New York Yankees' 1920 Cooperstown
> > 3 'Boston Redsox' 1918 ESPN
> > 4 'Washington Nationals' 2010 ESPN
> > 5 'Detroit Tigers' 1990 ESPN
> > "), header=TRUE)
> >
> > d$NAME <- as.character(d$NAME)
> >
> > fun <- function(i, x){
> > if(x[i, "ID"] != x[i + 1, "ID"]){
> > s <- unlist(strsplit(x[i, "NAME"], "[[:space:]]"))[1]
> > if(grepl(s, x[i + 1, "NAME"])) return(TRUE)
> > }
> > FALSE
> > }
> >
> > inx <- sapply(seq_len(nrow(d) - 1), fun, d)
> > inx <- c(inx, FALSE) | c(FALSE, inx)
> > d[inx, ]
> >
> > Hope this helps,
> >
> > Rui Barradas
> > Em 10-08-2012 18:41, Fred G escreveu:
> > > 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$name = first string
> up
> > > to whitespace in row + 1 in inname$name
> > > #AND ID in inname$ID for the top row NOT EQUAL ID in inname$ID for the
> row
> > > below it
> > > #copy these two lines to a new file
> > >
> > > In other words, if the name (up to the first whitespace) in the first
> row
> > > equals the name in the second row (etc for whole file) and the ID in
> the
> > > first row does not equal the ID in the second row, copy both of these
> rows
> > > in full to a new file. Only caveat is that I want a regular
> expression not
> > > to take the full names, but just the first string up to the first
> > > whitespace in the inname$name column (ie if row1 has a name of: New
> York
> > > Mets and row2 has a name of New York Yankees, I would want both of
> these
> > > rows to be copied in full since "New" is the same in both...)
> > >
> > > Here is some example data:
> > > ID NAME YEAR SOURCE NOTES
> > > 1 New York Mets 1900 ESPN
> > > 2 New York Yankees 1920 Cooperstown
> > > 3 Boston Redsox 1918 ESPN
> > > 4 Washington Nationals
Re: [R] Regular Expressions + Matrices
If you think about this as a runs problem you can get a loopless solution
that I think is easier to read (once the requisite functions are defined).
First define the function to canonicalize the name
nickname <- function(x) sub(" .*", "", x)
then define some handy runs functions
isFirstInRun <- function(x) c(TRUE, x[-1] != x[-length(x)])
isJustBefore <- function(x) c(x[-1], FALSE) # x should be logical
then use those functions on your dataset
> nearDup <- !isFirstInRun(nickname(d$NAME)) & isFirstInRun(d$YEAR)
> d[ nearDup | isJustBefore(nearDup), ]
ID NAME YEAR SOURCE
1 1New York Mets 1900ESPN
2 2 New York Yankees 1920 Cooperstown
See how it works with triplicates as well
> dd <- rbind(d, data.frame(ID=6:8,
NAME=c("Chicago Blacksox", "Chicago Cubs", "Chicago
Whitesox"),
YEAR=1701:1703, SOURCE=rep("made up", 3)))
> nearDup <- !isFirstInRun(nickname(dd$NAME)) & isFirstInRun(dd$YEAR)
> dd[ nearDup | isJustBefore(nearDup), ]
ID NAME YEAR SOURCE
1 1New York Mets 1900ESPN
2 2 New York Yankees 1920 Cooperstown
6 6 Chicago Blacksox 1701 made up
7 7 Chicago Cubs 1702 made up
8 8 Chicago Whitesox 1703 made up
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Rui Barradas
> Sent: Friday, August 10, 2012 11:18 AM
> To: Fred G
> Cc: r-help
> Subject: Re: [R] Regular Expressions + Matrices
>
> Hello,
>
> Try the following.
>
>
> d <- read.table(textConnection("
> ID NAME YEAR SOURCE
> 1 'New York Mets' 1900 ESPN
> 2 'New York Yankees' 1920 Cooperstown
> 3 'Boston Redsox' 1918 ESPN
> 4 'Washington Nationals' 2010 ESPN
> 5 'Detroit Tigers' 1990 ESPN
> "), header=TRUE)
>
> d$NAME <- as.character(d$NAME)
>
> fun <- function(i, x){
> if(x[i, "ID"] != x[i + 1, "ID"]){
> s <- unlist(strsplit(x[i, "NAME"], "[[:space:]]"))[1]
> if(grepl(s, x[i + 1, "NAME"])) return(TRUE)
> }
> FALSE
> }
>
> inx <- sapply(seq_len(nrow(d) - 1), fun, d)
> inx <- c(inx, FALSE) | c(FALSE, inx)
> d[inx, ]
>
> Hope this helps,
>
> Rui Barradas
> Em 10-08-2012 18:41, Fred G escreveu:
> > Hi all,
> >
> > My code looks like the following:
> > inname = read.csv("ID_error_checker.csv", as.is=TRUE)
> > outname = read.csv("output.csv", as.is=TRUE)
> >
> > #My algorithm is the following:
> > #for line in inname
> > #if first string up to whitespace in row in inname$name = first string up
> > to whitespace in row + 1 in inname$name
> > #AND ID in inname$ID for the top row NOT EQUAL ID in inname$ID for the row
> > below it
> > #copy these two lines to a new file
> >
> > In other words, if the name (up to the first whitespace) in the first row
> > equals the name in the second row (etc for whole file) and the ID in the
> > first row does not equal the ID in the second row, copy both of these rows
> > in full to a new file. Only caveat is that I want a regular expression not
> > to take the full names, but just the first string up to the first
> > whitespace in the inname$name column (ie if row1 has a name of: New York
> > Mets and row2 has a name of New York Yankees, I would want both of these
> > rows to be copied in full since "New" is the same in both...)
> >
> > Here is some example data:
> > ID NAME YEAR SOURCE NOTES
> > 1 New York Mets 1900 ESPN
> > 2 New York Yankees 1920 Cooperstown
> > 3 Boston Redsox 1918 ESPN
> > 4 Washington Nationals 2010 ESPN
> > 5 Detroit Tigers 1990 ESPN
> >
> > The desired output would be:
> > ID NAMEYEAR SOURCE
> > 1New York Mets1900 ESPN
> > 2New York Yankees 1920 Cooperstown
> >
> > Thanks so much!
> >
> > [[alternative HTML version deleted]]
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular Expressions + Matrices
Thanks so much, and thanks for the clarification. "New York" ---> "New"
should not match "Other New" because "New" is not the first.
Thanks so much, testing it on my data now.
On Fri, Aug 10, 2012 at 2:35 PM, Rui Barradas wrote:
> Hello,
>
> My code doesn't predict a point you've made clear in this post. Inline.
> Em 10-08-2012 19:05, Fred G escreveu:
>
> Thanks Arun. The only issue is that I need the code to be very
>> generalizable, such that the grep() really has to be if the first string
>> up
>> to the whitespace in a row (ie "New", "Boston", "Washington", "Detroit
>> below) is the same as the first string up to the whitespace in the row
>> directly below it
>>
>
> Does this mean that "New York" ---> "New" in one row shouldn't match
> "Other New" in the next row because "New" is not the first string up to the
> whitespace? If this is the case, modify my earlier code to
>
>
>
> fun <- function(i, x){
> if(x[i, "ID"] != x[i + 1, "ID"]){
> s1 <- unlist(strsplit(x[i, "NAME"], "[[:space:]]"))[1] # keep
> first string
> s2 <- unlist(strsplit(x[i + 1, "NAME"], "[[:space:]]"))[1] # keep
> first string
> if(grepl(s1, s2)) return(TRUE)
> }
> FALSE
> }
>
> If it isn't the case, do nothing.
>
> Rui Barradas
>
>
> , AND the ID's are different, then copy. The actual file
>> has thousands of different IDs and names...
>>
>> On Fri, Aug 10, 2012 at 2:01 PM, arun wrote:
>>
>>
>>> Hi,
>>>
>>> Try this:
>>> dat1<-read.table(text="
>>> 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,ESPN
>>> 5,Detroit Tigers,1990,ESPN
>>> ",sep=",",header=TRUE,**stringsAsFactors=FALSE)
>>>
>>> index<-grep("New York.*",dat1$NAME)
>>> dat1[index,]
>>> # ID NAME YEAR SOURCE
>>> #1 1New York Mets 1900ESPN
>>> #2 2 New York Yankees 1920 Cooperstown
>>>
>>> A.K.
>>>
>>>
>>>
>>> - Original Message -
>>> From: Fred G
>>> To: r-help@r-project.org
>>> Cc:
>>> Sent: Friday, August 10, 2012 1:41 PM
>>> Subject: [R] Regular Expressions + Matrices
>>>
>>> Hi all,
>>>
>>> My code looks like the following:
>>> inname = read.csv("ID_error_checker.**csv", as.is=TRUE)
>>> outname = read.csv("output.csv", as.is=TRUE)
>>>
>>> #My algorithm is the following:
>>> #for line in inname
>>> #if first string up to whitespace in row in inname$name = first string up
>>> to whitespace in row + 1 in inname$name
>>> #AND ID in inname$ID for the top row NOT EQUAL ID in inname$ID for the
>>> row
>>> below it
>>> #copy these two lines to a new file
>>>
>>> In other words, if the name (up to the first whitespace) in the first row
>>> equals the name in the second row (etc for whole file) and the ID in the
>>> first row does not equal the ID in the second row, copy both of these
>>> rows
>>> in full to a new file. Only caveat is that I want a regular expression
>>> not
>>> to take the full names, but just the first string up to the first
>>> whitespace in the inname$name column (ie if row1 has a name of: New York
>>> Mets and row2 has a name of New York Yankees, I would want both of these
>>> rows to be copied in full since "New" is the same in both...)
>>>
>>> Here is some example data:
>>> ID NAME YEAR SOURCE NOTES
>>> 1 New York Mets 1900 ESPN
>>> 2 New York Yankees 1920 Cooperstown
>>> 3 Boston Redsox 1918 ESPN
>>> 4 Washington Nationals 2010 ESPN
>>> 5 Detroit Tigers 1990 ESPN
>>>
>>> The desired output would be:
>>> ID NAMEYEAR SOURCE
>>> 1New York Mets1900 ESPN
>>> 2New York Yankees 1920 Cooperstown
>>>
>>> Thanks so much!
>>>
>>> [[alternative HTML version deleted]]
>>>
Re: [R] Regular Expressions + Matrices
Hello,
My code doesn't predict a point you've made clear in this post. Inline.
Em 10-08-2012 19:05, Fred G escreveu:
Thanks Arun. The only issue is that I need the code to be very
generalizable, such that the grep() really has to be if the first string up
to the whitespace in a row (ie "New", "Boston", "Washington", "Detroit
below) is the same as the first string up to the whitespace in the row
directly below it
Does this mean that "New York" ---> "New" in one row shouldn't match
"Other New" in the next row because "New" is not the first string up to
the whitespace? If this is the case, modify my earlier code to
fun <- function(i, x){
if(x[i, "ID"] != x[i + 1, "ID"]){
s1 <- unlist(strsplit(x[i, "NAME"], "[[:space:]]"))[1] #
keep first string
s2 <- unlist(strsplit(x[i + 1, "NAME"], "[[:space:]]"))[1] #
keep first string
if(grepl(s1, s2)) return(TRUE)
}
FALSE
}
If it isn't the case, do nothing.
Rui Barradas
, AND the ID's are different, then copy. The actual file
has thousands of different IDs and names...
On Fri, Aug 10, 2012 at 2:01 PM, arun wrote:
Hi,
Try this:
dat1<-read.table(text="
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,ESPN
5,Detroit Tigers,1990,ESPN
",sep=",",header=TRUE,stringsAsFactors=FALSE)
index<-grep("New York.*",dat1$NAME)
dat1[index,]
# ID NAME YEAR SOURCE
#1 1New York Mets 1900ESPN
#2 2 New York Yankees 1920 Cooperstown
A.K.
- Original Message -
From: Fred G
To: r-help@r-project.org
Cc:
Sent: Friday, August 10, 2012 1:41 PM
Subject: [R] Regular Expressions + Matrices
Hi all,
My code looks like the following:
inname = read.csv("ID_error_checker.csv", as.is=TRUE)
outname = read.csv("output.csv", as.is=TRUE)
#My algorithm is the following:
#for line in inname
#if first string up to whitespace in row in inname$name = first string up
to whitespace in row + 1 in inname$name
#AND ID in inname$ID for the top row NOT EQUAL ID in inname$ID for the row
below it
#copy these two lines to a new file
In other words, if the name (up to the first whitespace) in the first row
equals the name in the second row (etc for whole file) and the ID in the
first row does not equal the ID in the second row, copy both of these rows
in full to a new file. Only caveat is that I want a regular expression not
to take the full names, but just the first string up to the first
whitespace in the inname$name column (ie if row1 has a name of: New York
Mets and row2 has a name of New York Yankees, I would want both of these
rows to be copied in full since "New" is the same in both...)
Here is some example data:
ID NAME YEAR SOURCE NOTES
1 New York Mets 1900 ESPN
2 New York Yankees 1920 Cooperstown
3 Boston Redsox 1918 ESPN
4 Washington Nationals 2010 ESPN
5 Detroit Tigers 1990 ESPN
The desired output would be:
ID NAMEYEAR SOURCE
1New York Mets1900 ESPN
2New York Yankees 1920 Cooperstown
Thanks so much!
[[alternative HTML version deleted]]
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular Expressions + Matrices
Hi,
Try this:
dat1<-read.table(text="
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, ESPN
5, Detroit Tigers, 1990, ESPN
",sep=",",header=TRUE,stringsAsFactors=FALSE)
index<-grep("New York.*",dat1$NAME)
dat1[index,]
# ID NAME YEAR SOURCE
#1 1 New York Mets 1900 ESPN
#2 2 New York Yankees 1920 Cooperstown
A.K.
- Original Message -
From: Fred G
To: r-help@r-project.org
Cc:
Sent: Friday, August 10, 2012 1:41 PM
Subject: [R] Regular Expressions + Matrices
Hi all,
My code looks like the following:
inname = read.csv("ID_error_checker.csv", as.is=TRUE)
outname = read.csv("output.csv", as.is=TRUE)
#My algorithm is the following:
#for line in inname
#if first string up to whitespace in row in inname$name = first string up
to whitespace in row + 1 in inname$name
#AND ID in inname$ID for the top row NOT EQUAL ID in inname$ID for the row
below it
#copy these two lines to a new file
In other words, if the name (up to the first whitespace) in the first row
equals the name in the second row (etc for whole file) and the ID in the
first row does not equal the ID in the second row, copy both of these rows
in full to a new file. Only caveat is that I want a regular expression not
to take the full names, but just the first string up to the first
whitespace in the inname$name column (ie if row1 has a name of: New York
Mets and row2 has a name of New York Yankees, I would want both of these
rows to be copied in full since "New" is the same in both...)
Here is some example data:
ID NAME YEAR SOURCE NOTES
1 New York Mets 1900 ESPN
2 New York Yankees 1920 Cooperstown
3 Boston Redsox 1918 ESPN
4 Washington Nationals 2010 ESPN
5 Detroit Tigers 1990 ESPN
The desired output would be:
ID NAME YEAR SOURCE
1 New York Mets 1900 ESPN
2 New York Yankees 1920 Cooperstown
Thanks so much!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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 'Boston Redsox' 1918 ESPN
4 'Washington Nationals' 2010 ESPN
5 'Detroit Tigers' 1990 ESPN
"), header=TRUE)
d$NAME <- as.character(d$NAME)
fun <- function(i, x){
if(x[i, "ID"] != x[i + 1, "ID"]){
s <- unlist(strsplit(x[i, "NAME"], "[[:space:]]"))[1]
if(grepl(s, x[i + 1, "NAME"])) return(TRUE)
}
FALSE
}
inx <- sapply(seq_len(nrow(d) - 1), fun, d)
inx <- c(inx, FALSE) | c(FALSE, inx)
d[inx, ]
Hope this helps,
Rui Barradas
Em 10-08-2012 18:41, Fred G escreveu:
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$name = first string up
to whitespace in row + 1 in inname$name
#AND ID in inname$ID for the top row NOT EQUAL ID in inname$ID for the row
below it
#copy these two lines to a new file
In other words, if the name (up to the first whitespace) in the first row
equals the name in the second row (etc for whole file) and the ID in the
first row does not equal the ID in the second row, copy both of these rows
in full to a new file. Only caveat is that I want a regular expression not
to take the full names, but just the first string up to the first
whitespace in the inname$name column (ie if row1 has a name of: New York
Mets and row2 has a name of New York Yankees, I would want both of these
rows to be copied in full since "New" is the same in both...)
Here is some example data:
ID NAME YEAR SOURCE NOTES
1 New York Mets 1900 ESPN
2 New York Yankees 1920 Cooperstown
3 Boston Redsox 1918 ESPN
4 Washington Nationals 2010 ESPN
5 Detroit Tigers 1990 ESPN
The desired output would be:
ID NAMEYEAR SOURCE
1New York Mets1900 ESPN
2New York Yankees 1920 Cooperstown
Thanks so much!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular Expressions + Matrices
Thanks Arun. The only issue is that I need the code to be very
generalizable, such that the grep() really has to be if the first string up
to the whitespace in a row (ie "New", "Boston", "Washington", "Detroit
below) is the same as the first string up to the whitespace in the row
directly below it, AND the ID's are different, then copy. The actual file
has thousands of different IDs and names...
On Fri, Aug 10, 2012 at 2:01 PM, arun wrote:
>
>
> Hi,
>
> Try this:
> dat1<-read.table(text="
> ID,NAME,YEAR,SOURCE
> 1,New York Mets,1900,ESPN
> 2,New York Yankees,1920,Cooperstown
> 3,Boston Redsox,1918,ESPN
> 4,Washington Nationals,2010,ESPN
> 5,Detroit Tigers,1990,ESPN
> ",sep=",",header=TRUE,stringsAsFactors=FALSE)
>
> index<-grep("New York.*",dat1$NAME)
> dat1[index,]
> # ID NAME YEAR SOURCE
> #1 1New York Mets 1900ESPN
> #2 2 New York Yankees 1920 Cooperstown
>
> A.K.
>
>
>
> - Original Message -
> From: Fred G
> 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 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$name = first string up
> to whitespace in row + 1 in inname$name
> #AND ID in inname$ID for the top row NOT EQUAL ID in inname$ID for the row
> below it
> #copy these two lines to a new file
>
> In other words, if the name (up to the first whitespace) in the first row
> equals the name in the second row (etc for whole file) and the ID in the
> first row does not equal the ID in the second row, copy both of these rows
> in full to a new file. Only caveat is that I want a regular expression not
> to take the full names, but just the first string up to the first
> whitespace in the inname$name column (ie if row1 has a name of: New York
> Mets and row2 has a name of New York Yankees, I would want both of these
> rows to be copied in full since "New" is the same in both...)
>
> Here is some example data:
> ID NAME YEAR SOURCE NOTES
> 1 New York Mets 1900 ESPN
> 2 New York Yankees 1920 Cooperstown
> 3 Boston Redsox 1918 ESPN
> 4 Washington Nationals 2010 ESPN
> 5 Detroit Tigers 1990 ESPN
>
> The desired output would be:
> ID NAMEYEAR SOURCE
> 1New York Mets1900 ESPN
> 2New York Yankees 1920 Cooperstown
>
> Thanks so much!
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Regular Expressions + Matrices
Hi all,
My code looks like the following:
inname = read.csv("ID_error_checker.csv", as.is=TRUE)
outname = read.csv("output.csv", as.is=TRUE)
#My algorithm is the following:
#for line in inname
#if first string up to whitespace in row in inname$name = first string up
to whitespace in row + 1 in inname$name
#AND ID in inname$ID for the top row NOT EQUAL ID in inname$ID for the row
below it
#copy these two lines to a new file
In other words, if the name (up to the first whitespace) in the first row
equals the name in the second row (etc for whole file) and the ID in the
first row does not equal the ID in the second row, copy both of these rows
in full to a new file. Only caveat is that I want a regular expression not
to take the full names, but just the first string up to the first
whitespace in the inname$name column (ie if row1 has a name of: New York
Mets and row2 has a name of New York Yankees, I would want both of these
rows to be copied in full since "New" is the same in both...)
Here is some example data:
ID NAME YEAR SOURCE NOTES
1 New York Mets 1900 ESPN
2 New York Yankees 1920 Cooperstown
3 Boston Redsox 1918 ESPN
4 Washington Nationals 2010 ESPN
5 Detroit Tigers 1990 ESPN
The desired output would be:
ID NAMEYEAR SOURCE
1New York Mets1900 ESPN
2New York Yankees 1920 Cooperstown
Thanks so much!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] regular expressions in R
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. "." matches any character unless you escape it to mean a 'period'. On Wed, Dec 21, 2011 at 11:11 AM, R. Michael Weylandt wrote: > Do you wish to include .docx files as well or just .doc? > > Michael > > On Wed, Dec 21, 2011 at 10:04 AM, Alaios 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 (is not that regular >> expressions) for doing that. >> >> print(dir(i,full.names=TRUE,pattern=.)) >> >> Could you please help me compose such a pattern? >> >> B.R >> Alex >> [[alternative HTML version deleted]] >> >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- 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. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] regular expressions in R
Do you wish to include .docx files as well or just .doc? Michael On Wed, Dec 21, 2011 at 10:04 AM, Alaios 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 (is not that regular > expressions) for doing that. > > print(dir(i,full.names=TRUE,pattern=.)) > > Could you please help me compose such a pattern? > > B.R > Alex > [[alternative HTML version deleted]] > > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] regular expressions in R
>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" "b.txt" "c.doc" You don't tell us your OS, so I have no idea whether it will work for you. Sarah On Wed, Dec 21, 2011 at 11:04 AM, Alaios 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 (is not that regular > expressions) for doing that. > > print(dir(i,full.names=TRUE,pattern=.)) > > Could you please help me compose such a pattern? > > B.R > Alex > > -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] regular expressions in R
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 a pattern? B.R Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions in R
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 wrote:
> 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 <- c('product + action * mean + CTA + help + mean * product')
> form2 <- c('product+action*mean+CTA+help+mean*product')
>
> ## what I believe your desired output is
> 'product + CTA + help'
> 'product+CTA+help'
>
> gsub("\\s\\+\\s[[:alnum:]]*\\s\\*\\s[[:alnum:]]*", "", form1)
> gsub("\\+[[:alnum:]]*\\*[[:alnum:]]*", "", form2)
>
> ## your code (using gsub() instead of sub())
> gsub("\\+*\\s*[[:alnum:]]*\\s*\\*.[[:alnum:]]", "", form1)
>
>
> Running on r57586 Windows x64
> > gsub("\\s\\+\\s[[:alnum:]]*\\s\\*\\s[[:alnum:]]*", "", form1)
> [1] "product + CTA + help"
> > gsub("\\+[[:alnum:]]*\\*[[:alnum:]]*", "", form2)
> [1] "product+CTA+help"
> >
> > ## your code (using gsub() instead of sub())
> > gsub("\\+*\\s*[[:alnum:]]*\\s*\\*.[[:alnum:]]", "", form1)
> [1] "product ean + CTA + help roduct"
>
> Hope this helps,
>
> Josh
>
> On Tue, Nov 15, 2011 at 9:18 AM, Michael Griffiths
> wrote:
> > 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 +
> mean
> > * product')
> >
> form<-c('~Sentence+LEGAL+Intro+Intro/Intro1+Intro*LEGAL+benefit+benefit/benefit1+product+action*mean+CTA+help+mean*product')
> >
> > I would like to remove the following target strings, either:
> >
> > 1. '+ Intro * LEGAL' which is '+ space name space * space name'
> > 2. '+Intro*LEGAL' which is '+ nospace name nospace * nospace name'
> >
> > Having delved into a variety of sites (e.g.
> > http://www.zytrax.com/tech/web/regex.htm#search) investigating regular
> > expressions I now have a basic grasp, but I am having difficulties
> removing
> > ALL of the instances or 1. or 2.
> >
> > The code below removes just a SINGLE instance of the target string, but I
> > was expecting it to remove all instances as I have \\*.[[allnum]]. I did
> > try \\*.[[allnum]]*, but this did not work.
> >
> > form<-sub("\\+*\\s*[[:alnum:]]*\\s*\\*.[[:alnum:]]", "", form)
> >
> > I am obviously still not understanding something. If the list could offer
> > some guidance I would be most grateful.
> >
> > Regards
> >
> > Mike Griffiths
> >
> >
> >
> > --
> >
> > *Michael Griffiths, Ph.D
> > *Statistician
> >
> > *Upstream Systems*
> >
> > 8th Floor
> > Portland House
> > Bressenden Place
> > SW1E 5BH
> >
> > <
> http://www.google.com/url?q=http%3A%2F%2Fwww.upstreamsystems.com%2F&sa=D&sntz=1&usg=AFrqEzfKYfaAalqvahwrpywpJDL9DxUmWw
> >
> >
> > Tel +44 (0) 20 7869 5147
> > Fax +44 207 290 1321
> > Mob +44 789 4944 145
> >
> > www.upstreamsystems.com<
> http://www.google.com/url?q=http%3A%2F%2Fwww.upstreamsystems.com%2F&sa=D&sntz=1&usg=AFrqEzfKYfaAalqvahwrpywpJDL9DxUmWw
> >
> >
> > *griffi...@upstreamsystems.com *
> >
> >
Tel +44 (0) 20 7869 5147
Fax +44 207 290 1321
Mob +44 789 4944 145
www.upstreamsystems.com
*griffi...@upstreamsystems.com *
Re: [R] Regular expressions in R
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 <- c('product + action * mean + CTA + help + mean * product')
form2 <- c('product+action*mean+CTA+help+mean*product')
## what I believe your desired output is
'product + CTA + help'
'product+CTA+help'
gsub("\\s\\+\\s[[:alnum:]]*\\s\\*\\s[[:alnum:]]*", "", form1)
gsub("\\+[[:alnum:]]*\\*[[:alnum:]]*", "", form2)
## your code (using gsub() instead of sub())
gsub("\\+*\\s*[[:alnum:]]*\\s*\\*.[[:alnum:]]", "", form1)
Running on r57586 Windows x64
> gsub("\\s\\+\\s[[:alnum:]]*\\s\\*\\s[[:alnum:]]*", "", form1)
[1] "product + CTA + help"
> gsub("\\+[[:alnum:]]*\\*[[:alnum:]]*", "", form2)
[1] "product+CTA+help"
>
> ## your code (using gsub() instead of sub())
> gsub("\\+*\\s*[[:alnum:]]*\\s*\\*.[[:alnum:]]", "", form1)
[1] "product ean + CTA + help roduct"
Hope this helps,
Josh
On Tue, Nov 15, 2011 at 9:18 AM, Michael Griffiths
wrote:
> 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 + mean
> * product')
> form<-c('~Sentence+LEGAL+Intro+Intro/Intro1+Intro*LEGAL+benefit+benefit/benefit1+product+action*mean+CTA+help+mean*product')
>
> I would like to remove the following target strings, either:
>
> 1. '+ Intro * LEGAL' which is '+ space name space * space name'
> 2. '+Intro*LEGAL' which is '+ nospace name nospace * nospace name'
>
> Having delved into a variety of sites (e.g.
> http://www.zytrax.com/tech/web/regex.htm#search) investigating regular
> expressions I now have a basic grasp, but I am having difficulties removing
> ALL of the instances or 1. or 2.
>
> The code below removes just a SINGLE instance of the target string, but I
> was expecting it to remove all instances as I have \\*.[[allnum]]. I did
> try \\*.[[allnum]]*, but this did not work.
>
> form<-sub("\\+*\\s*[[:alnum:]]*\\s*\\*.[[:alnum:]]", "", form)
>
> I am obviously still not understanding something. If the list could offer
> some guidance I would be most grateful.
>
> Regards
>
> Mike Griffiths
>
>
>
> --
>
> *Michael Griffiths, Ph.D
> *Statistician
>
> *Upstream Systems*
>
> 8th Floor
> Portland House
> Bressenden Place
> SW1E 5BH
>
>
>
> Tel +44 (0) 20 7869 5147
> Fax +44 207 290 1321
> Mob +44 789 4944 145
>
> www.upstreamsystems.com
>
> *griffi...@upstreamsystems.com *
>
>
Re: [R] Regular expressions in R
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 backreferences which are
not defined in ‘pattern’ the result is undefined (but most often
the backreference is taken to be ‘""’).
Sarah
On Tue, Nov 15, 2011 at 12:18 PM, Michael Griffiths
wrote:
> 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 + mean
> * product')
> form<-c('~Sentence+LEGAL+Intro+Intro/Intro1+Intro*LEGAL+benefit+benefit/benefit1+product+action*mean+CTA+help+mean*product')
>
> I would like to remove the following target strings, either:
>
> 1. '+ Intro * LEGAL' which is '+ space name space * space name'
> 2. '+Intro*LEGAL' which is '+ nospace name nospace * nospace name'
>
> Having delved into a variety of sites (e.g.
> http://www.zytrax.com/tech/web/regex.htm#search) investigating regular
> expressions I now have a basic grasp, but I am having difficulties removing
> ALL of the instances or 1. or 2.
>
> The code below removes just a SINGLE instance of the target string, but I
> was expecting it to remove all instances as I have \\*.[[allnum]]. I did
> try \\*.[[allnum]]*, but this did not work.
>
> form<-sub("\\+*\\s*[[:alnum:]]*\\s*\\*.[[:alnum:]]", "", form)
>
> I am obviously still not understanding something. If the list could offer
> some guidance I would be most grateful.
>
> Regards
>
> Mike Griffiths
>
>
>
--
Sarah Goslee
http://www.functionaldiversity.org
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Regular expressions in R
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 + mean
* product')
form<-c('~Sentence+LEGAL+Intro+Intro/Intro1+Intro*LEGAL+benefit+benefit/benefit1+product+action*mean+CTA+help+mean*product')
I would like to remove the following target strings, either:
1. '+ Intro * LEGAL' which is '+ space name space * space name'
2. '+Intro*LEGAL' which is '+ nospace name nospace * nospace name'
Having delved into a variety of sites (e.g.
http://www.zytrax.com/tech/web/regex.htm#search) investigating regular
expressions I now have a basic grasp, but I am having difficulties removing
ALL of the instances or 1. or 2.
The code below removes just a SINGLE instance of the target string, but I
was expecting it to remove all instances as I have \\*.[[allnum]]. I did
try \\*.[[allnum]]*, but this did not work.
form<-sub("\\+*\\s*[[:alnum:]]*\\s*\\*.[[:alnum:]]", "", form)
I am obviously still not understanding something. If the list could offer
some guidance I would be most grateful.
Regards
Mike Griffiths
--
*Michael Griffiths, Ph.D
*Statistician
*Upstream Systems*
8th Floor
Portland House
Bressenden Place
SW1E 5BH
Tel +44 (0) 20 7869 5147
Fax +44 207 290 1321
Mob +44 789 4944 145
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*griffi...@upstreamsystems.com *
Re: [R] Regular Expressions for "Large" Data Set
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
>
> I want to loop through the data and construct a data frame with the zip
> code,
> state abbreviation, and city name in seperate columns. Given the size of
> this
> data set, I was wondering if there was an efficient way to get the desired
> results.
>
> Thanks
> Abraham
Since the original text file is a CSV file (without a header), just use:
> system.time(DF <-
> read.csv("http://www.census.gov/tiger/tms/gazetteer/zips.txt";, header =
> FALSE))
user system elapsed
0.385 0.033 1.832
> str(DF)
'data.frame': 29470 obs. of 8 variables:
$ V1: int 1 1 1 1 1 1 1 1 1 1 ...
$ V2: int 35004 35005 35006 35007 35010 35014 35016 35019 35020 35023 ...
$ V3: Factor w/ 51 levels "AK","AL","AR",..: 2 2 2 2 2 2 2 2 2 2 ...
$ V4: Factor w/ 16698 levels "02821","04465",..: 150 168 180 7710 10434 348
547 812 1250 7044 ...
$ V5: num 86.5 87 87.2 86.8 86 ...
$ V6: num 33.6 33.6 33.4 33.2 32.9 ...
$ V7: int 6055 10616 3205 14218 19942 3062 13650 1781 40549 39677 ...
$ V8: num 0.001499 0.002627 0.000793 0.003519 0.004935 ...
> head(DF)
V1V2 V3 V4 V5 V6V7 V8
1 1 35004 AL ACMAR 86.51557 33.58413 6055 0.001499
2 1 35005 AL ADAMSVILLE 86.95973 33.58844 10616 0.002627
3 1 35006 AL ADGER 87.16746 33.43428 3205 0.000793
4 1 35007 AL KEYSTONE 86.81286 33.23687 14218 0.003519
5 1 35010 AL NEW SITE 85.95109 32.94145 19942 0.004935
6 1 35014 AL ALPINE 86.20893 33.33116 3062 0.000758
HTH,
Marc Schwartz
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[R] Regular Expressions for "Large" Data Set
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 with the zip
code,
state abbreviation, and city name in seperate columns. Given the size of
this
data set, I was wondering if there was an efficient way to get the desired
results.
Thanks
Abraham
WebRep
Overall rating
[[alternative HTML version deleted]]
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Re: [R] Regular Expressions in Column Headings
On Wed, Mar 9, 2011 at 8:52 AM, Matthew DeAngelis 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 matches (so that > the column heading is the given regular expression). My problem is that > read.table seems to be trying to interpret the regular expressions, or has > trouble with the special characters, so that the column headings are not > coming out correctly. For example, a column headed with: \bV\.?A\.?T\.? > will come out as X.bV...A...T... This would not be a problem, since the > regular expressions are still readable, except that I have a number of other > tables that I will need to intersect with these column headings. In some of > those tables, the regular expressions are data, and they are coming in > correctly (although R seems to be doubling "\"s, which is fine so long as it > does this consistently). > > I have also tried importing the column names as a vector and specifying that > vector explicitly using col.names, but R still transforms the provided names > as above. Is it possible to force R to read in regular expressions > completely literally, with no interpretation? Alternately, can I force R to > interpret the column headings in the same way that it interprets data (i.e. > adding the extra slash), so that I can match on these values? > See the read.table check.names argument. -- Statistics & Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Regular Expressions in Column Headings
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 that read.table seems to be trying to interpret the regular expressions, or has trouble with the special characters, so that the column headings are not coming out correctly. For example, a column headed with: \bV\.?A\.?T\.? will come out as X.bV...A...T... This would not be a problem, since the regular expressions are still readable, except that I have a number of other tables that I will need to intersect with these column headings. In some of those tables, the regular expressions are data, and they are coming in correctly (although R seems to be doubling "\"s, which is fine so long as it does this consistently). I have also tried importing the column names as a vector and specifying that vector explicitly using col.names, but R still transforms the provided names as above. Is it possible to force R to read in regular expressions completely literally, with no interpretation? Alternately, can I force R to interpret the column headings in the same way that it interprets data (i.e. adding the extra slash), so that I can match on these values? Thanks, Matt [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular Expressions
2010/11/5 Brian Diggs :
> 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 <- function(pattern, x) {
> s <- sub(pattern, "\\1\034\\2", x)
> matrix(unlist(strsplit(s,"\034")), ncol=2, byrow=TRUE)
> }
>
> both_backrefs(regex, x)
>
> However, putting the parts back together into a string (with a delimiter
> that hopefully won't be in the string otherwise) just to use strsplit to
> pull them apart seems inelegant (as does making multiple calls to sub()).
> sub() (and siblings) surely already have the backreferences as strings at
> some point in the processing, but I don't see a way to return them as a
> vector or matrix, only to substitute using backreferences (sub) or return
> indicies pointing to where the matches start (regexpr) or return the whole
> string matches (grep with value=TRUE).
>
The gsubfn package has gsubfn which is like gsub except it can take a
function in place of the replacement string. The function's arguments
are match or the back references and the function's output replaces
the match.Also it has strapply which will does the same thing
except instead of inserting the function's output it returns the
function's output. See the home page at http://gsubfn.googlecode.com
--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
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Re: [R] Regular Expressions
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 example, given the string: "10 Nov 13.00 (PFE1020K13)"
I want to capture the first to digits and then the month abreviation.
In perl, this would be
/^(\d\d)\s(\w\w\w)\s/
Then I have the variables $1 and $1 assigned to the capturing
parenthesis.
I've found the grep and sub commands in R, but the docs don't indicate
any way to capture things.
Any suggestions?
Read the the link to ?regexp. It *does* 'indicate the way to capture
things'.
The backreference ‘\N’, where ‘N = 1 ... 9’, matches the substring
previously matched by the Nth parenthesized subexpression of the
regular expression. (This is an extension for extended regular
expressions: POSIX defines them only for basic ones.)
and there is an example on the help page for grep():
## Double all 'a' or 'b's; "\" must be escaped, i.e., 'doubled'
gsub("([ab])", "\\1_\\1_", "abc and ABC")
In your example
x <- "10 Nov 13.00 (PFE1020K13)"
regex <- "(\\d\\d)\\s(\\w\\w\\w).*"
sub(regex, "\\1", x, perl = TRUE)
sub(regex, "\\2", x, perl = TRUE)
A better way to do this would be something like
regex <- "([[:digit:]]{2})\\s([[:alpha:]]{3}).*"
which is also a POSIX extended regexp.
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 <- function(pattern, x) {
s <- sub(pattern, "\\1\034\\2", x)
matrix(unlist(strsplit(s,"\034")), ncol=2, byrow=TRUE)
}
both_backrefs(regex, x)
However, putting the parts back together into a string (with a delimiter
that hopefully won't be in the string otherwise) just to use strsplit to
pull them apart seems inelegant (as does making multiple calls to
sub()). sub() (and siblings) surely already have the backreferences as
strings at some point in the processing, but I don't see a way to return
them as a vector or matrix, only to substitute using backreferences
(sub) or return indicies pointing to where the matches start (regexpr)
or return the whole string matches (grep with value=TRUE).
--
Brian S. Diggs, PhD
Senior Research Associate, Department of Surgery
Oregon Health & Science University
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Re: [R] Regular Expressions
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
an MS in stats. My department is fairly flexible, so I'm taking several
finance courses as part of my work. Currently debating if I want to
graduate with an MS in June, or roll everything into a PhD and be
finished in an extra 1-2 years.
Thanks!
-N
On 11/5/10 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 example, given the string:"10 Nov 13.00 (PFE1020K13)"
>>
>> I want to capture the first to digits and then the month abreviation.
>>
>> In perl, this would be
>>
>> /^(\d\d)\s(\w\w\w)\s/
>>
>> Then I have the variables $1 and $1 assigned to the capturing
>> parenthesis.
>>
>> I've found the grep and sub commands in R, but the docs don't
>> indicate any way to capture things.
>>
>> Any suggestions?
>
> Read the the link to ?regexp. It *does* 'indicate the way to capture
> things'.
>
> The backreference ‘\N’, where ‘N = 1 ... 9’, matches the substring
> previously matched by the Nth parenthesized subexpression of the
> regular expression. (This is an extension for extended regular
> expressions: POSIX defines them only for basic ones.)
>
> and there is an example on the help page for grep():
>
> ## Double all 'a' or 'b's; "\" must be escaped, i.e., 'doubled'
> gsub("([ab])", "\\1_\\1_", "abc and ABC")
>
> In your example
>
> x <- "10 Nov 13.00 (PFE1020K13)"
> regex <- "(\\d\\d)\\s(\\w\\w\\w).*"
> sub(regex, "\\1", x, perl = TRUE)
> sub(regex, "\\2", x, perl = TRUE)
>
> A better way to do this would be something like
>
> regex <- "([[:digit:]]{2})\\s([[:alpha:]]{3}).*"
>
> which is also a POSIX extended regexp.
>
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Re: [R] Regular Expressions
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 (PFE1020K13)"
I want to capture the first to digits and then the month abreviation.
In perl, this would be
/^(\d\d)\s(\w\w\w)\s/
Then I have the variables $1 and $1 assigned to the capturing parenthesis.
I've found the grep and sub commands in R, but the docs don't indicate any
way to capture things.
Any suggestions?
Read the the link to ?regexp. It *does* 'indicate the way to capture
things'.
The backreference ‘\N’, where ‘N = 1 ... 9’, matches the substring
previously matched by the Nth parenthesized subexpression of the
regular expression. (This is an extension for extended regular
expressions: POSIX defines them only for basic ones.)
and there is an example on the help page for grep():
## Double all 'a' or 'b's; "\" must be escaped, i.e., 'doubled'
gsub("([ab])", "\\1_\\1_", "abc and ABC")
In your example
x <- "10 Nov 13.00 (PFE1020K13)"
regex <- "(\\d\\d)\\s(\\w\\w\\w).*"
sub(regex, "\\1", x, perl = TRUE)
sub(regex, "\\2", x, perl = TRUE)
A better way to do this would be something like
regex <- "([[:digit:]]{2})\\s([[:alpha:]]{3}).*"
which is also a POSIX extended regexp.
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
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[R] Regular Expressions
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 and then the month abreviation. In perl, this would be /^(\d\d)\s(\w\w\w)\s/ Then I have the variables $1 and $1 assigned to the capturing parenthesis. I've found the grep and sub commands in R, but the docs don't indicate any way to capture things. Any suggestions? Thanks! -N __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
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 responses! Titus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
On Wed, Sep 29, 2010 at 1:58 PM, Michael Bedward 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 have a skilled C- and R-programmer who could work on it. I'll talk to him. Titus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
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 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.
Michael
On 29 September 2010 20:08, Titus von der Malsburg wrote:
> 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 also the position of the overall match in the original
> string. The ideal solution would provide positions and match lengths
> for the whole pattern and for all groups if desired. Only this would
> solve all related issues. One possibility is to have a subpattern
> argument that accepts a vector of numbers (0 refers to the whole
> pattern):
>
> > gregexpr("a+(b+)", "abcdaabbc", subpattern=c(0,1))
> [[1]]:
> [[1]][[1]]:
> [1] 1 5
> attr(, "match.length"):
> [1] 2 4
> [[1]][[2]]:
> [1] 2 7
> attr(, "match.length"):
> [1] 1 2
>
> A weakness of this solution is that the structure of the return values
> changes if length(subpattern)>1. An alternative is to have a separate
> function, say ggregepxr for group gregexpr, that returns a list of
> lists as in the above example. This function would always return
> positions and match lengths of the whole pattern (group 0) and all
> groups. The original gregexpr could still have the subpattern
> argument but it would only accept single numbers. This way the return
> format of gregexpr remains the same.
>
> Best,
>
> Titus
>
>
> On Wed, Sep 29, 2010 at 2:42 AM, Michael Bedward
> wrote:
>> 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 some consensus on the best
>> way to implement it, ie. as a new R function or by extending existing
>> function(s).
>
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Re: [R] Regular expressions: offsets of groups
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 also the position of the overall match in the original
string. The ideal solution would provide positions and match lengths
for the whole pattern and for all groups if desired. Only this would
solve all related issues. One possibility is to have a subpattern
argument that accepts a vector of numbers (0 refers to the whole
pattern):
> gregexpr("a+(b+)", "abcdaabbc", subpattern=c(0,1))
[[1]]:
[[1]][[1]]:
[1] 1 5
attr(, "match.length"):
[1] 2 4
[[1]][[2]]:
[1] 2 7
attr(, "match.length"):
[1] 1 2
A weakness of this solution is that the structure of the return values
changes if length(subpattern)>1. An alternative is to have a separate
function, say ggregepxr for group gregexpr, that returns a list of
lists as in the above example. This function would always return
positions and match lengths of the whole pattern (group 0) and all
groups. The original gregexpr could still have the subpattern
argument but it would only accept single numbers. This way the return
format of gregexpr remains the same.
Best,
Titus
On Wed, Sep 29, 2010 at 2:42 AM, Michael Bedward
wrote:
> 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 some consensus on the best
> way to implement it, ie. as a new R function or by extending existing
> function(s).
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
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 some consensus on the best
way to implement it, ie. as a new R function or by extending existing
function(s).
Michael
On 29 September 2010 01:46, William Dunlap wrote:
>
> S+ has a subpattern=number argument to regexpr and
> related functions. It means that the text matched
> by the subpattern'th parenthesized expression in the
> pattern will be considered the matched text. E.g.,
> to find runs of b's that come immediately after a's:
>
> > gregexpr("a+(b+)", "abcdaabbc", subpattern=1)
> [[1]]:
> [1] 2 7
> attr(, "match.length"):
> [1] 1 2
>
> or to find bc's that come after 2 or more ab's
> > gregexpr("(ab){2,}bc", "abbcabababbcabcababbc", subpattern=1)
>
> regexpr() and strsplit() have this argument in S+ 8.1 but
> gregexpr() is not yet in a released version of S+.
>
> subpattern=0, the default, means to use the entire
> pattern. regexpr allows subpattern=-1, which means
> to return a list with one element for each subpattern.
> I don't know if the extra complexity is worth it.
> (gregexpr does not allow subpattern=-1.)
>
> The usual C regexec() returns this information.
> Perhaps it would be handy to have it in R.
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
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Re: [R] Regular expressions: offsets of groups
> -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 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 the stringr package (which is nice) but it doesn't
> quite do it either.
S+ has a subpattern=number argument to regexpr and
related functions. It means that the text matched
by the subpattern'th parenthesized expression in the
pattern will be considered the matched text. E.g.,
to find runs of b's that come immediately after a's:
> gregexpr("a+(b+)", "abcdaabbc", subpattern=1)
[[1]]:
[1] 2 7
attr(, "match.length"):
[1] 1 2
or to find bc's that come after 2 or more ab's
> gregexpr("(ab){2,}bc", "abbcabababbcabcababbc", subpattern=1)
regexpr() and strsplit() have this argument in S+ 8.1 but
gregexpr() is not yet in a released version of S+.
subpattern=0, the default, means to use the entire
pattern. regexpr allows subpattern=-1, which means
to return a list with one element for each subpattern.
I don't know if the extra complexity is worth it.
(gregexpr does not allow subpattern=-1.)
The usual C regexec() returns this information.
Perhaps it would be handy to have it in R.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
>
> Michael
>
> On 28 September 2010 01:48, Titus von der Malsburg
> 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 there a way in R
> > to get that?
> >
> > I know about gsubgn and strapply, but they only give me the strings
> > matched by groups not their offsets.
> >
> > I could write something myself that first takes the above matches
> > ("ab" and "aabb") and then searches again using only the group (b+).
> > For this to work, I'd have to parse the regular expression
> and search
> > several times (> 2, for nested groups) instead of just
> once. But I'm
> > sure there is a better way to do this.
> >
> > Thanks for any suggestion!
> >
> > Titus
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
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Re: [R] Regular expressions: offsets of groups
On Tue, Sep 28, 2010 at 6:52 AM, Titus von der Malsburg
wrote:
> On Tue, Sep 28, 2010 at 9:46 AM, Michael Bedward
> wrote:
>> What Titus wants to do is akin to retrieving capturing groups from a
>> Matcher object in Java.
>
> Precisely. Here's the description:
>
> http://download.oracle.com/javase/1.4.2/docs/api/java/util/regex/Matcher.html#start(int)
>
> Gabor's lookbehind trick solves some special cases but it's not the
The only limitation is that in the regular expressions supported by R
you cannot have repitition in the (<=...) portion but none of your
examples -- neither the one you gave nor the one below require that
since if the prior expression ends in X+ you can just use X.Are
you sure it does not cover all your actual situations?
If you truly do have situations where that require repetition a
gregexpr plus gsubfn will do it in one line. Parenthesize the
portion of the regular expression you want to capture and replace
every character in it with X (or some other character that does not
otherwise occur). Then find the positions and lengths of strings of
X.
> gregexpr("X+", gsubfn("a(b+)", ~ gsub(".", "X", x), "abcdaabbcbbb"))
[[1]]
[1] 1 5
attr(,"match.length")
[1] 1 2
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Re: [R] Regular expressions: offsets of groups
On Tue, Sep 28, 2010 at 9:46 AM, Michael Bedward wrote: > What Titus wants to do is akin to retrieving capturing groups from a > Matcher object in Java. Precisely. Here's the description: http://download.oracle.com/javase/1.4.2/docs/api/java/util/regex/Matcher.html#start(int) Gabor's lookbehind trick solves some special cases but it's not the kind of general solution I'm looking for. Let me explain what I'm trying to achieve here. I'm working on a package that provides tools for processing and analyzing eye movements (we're doing reading research). In most situations, eye movements consist of fixations where the eyes are relatively stationary and saccades, quick movements between fixations. A common way to represent eye movements is as strings of symbols, where each symbol corresponds to a fixation on a particular region. AABC means two fixations followed by a fixation on B and then C. When people analyze eye movements it's often necessary to find specific events in the eye movement record like: fixations on the word C preceded by fixations on words D-F and followed by fixations on words A-C. This event can be specified using this regexpr: "[D-F]+(C)[A-C]+" The group (in parenthesis) indicates the substring for which I'd like to know the position in the overall string. Another application is the extraction of subsequences from a sequence of fixations. Note that in some situations people might have to use more groups in their regexprs and that groups can be nested. In this case the user would have to indicate for which group he/she wants to know the offset. I'm not an expert for regexpr engines but I'm pretty sure the necessary information is available in the engine. Gabor, I see you're the author of gsubfn (fantastic package!). Do you see a relatively simple way to expose information about group offsets and their corresponding match lengths? I think this could be useful for other applications as well. At least it seems Michael could use it, too. We can cook up something for ourselves but a general solution would benefit the larger community. Titus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
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 the stringr package (which is nice) but it doesn't
quite do it either.
Michael
On 28 September 2010 01:48, Titus von der Malsburg 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 there a way in R
> to get that?
>
> I know about gsubgn and strapply, but they only give me the strings
> matched by groups not their offsets.
>
> I could write something myself that first takes the above matches
> ("ab" and "aabb") and then searches again using only the group (b+).
> For this to work, I'd have to parse the regular expression and search
> several times (> 2, for nested groups) instead of just once. But I'm
> sure there is a better way to do this.
>
> Thanks for any suggestion!
>
> Titus
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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Re: [R] Regular expressions: offsets of groups
On Mon, Sep 27, 2010 at 1:34 PM, Titus von der Malsburg
wrote:
> On Mon, Sep 27, 2010 at 7:29 PM, Gabor Grothendieck
> 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+", "abcdaabbc", perl = TRUE)
>
> which is wrong if the string is "abcdaabbcbbb".
>
Sorry, try this:
> gregexpr("(?<=a)b+", "abcdaabbcbbb", perl = TRUE)
[[1]]
[1] 2 7
attr(,"match.length")
[1] 1 2
Note that it does not give the same answer as:
> gregexpr("b+", "abcdaabbcbbb", perl = TRUE)
[[1]]
[1] 2 7 10
attr(,"match.length")
[1] 1 2 3
gregexpr("(?<=a)b+", "abcdaabbcbbb", perl = TRUE)
--
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Re: [R] Regular expressions: offsets of groups
You've tried:
gregexpr("b+", "abcdaabbc")
On Mon, Sep 27, 2010 at 12:48 PM, Titus von der Malsburg 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 there a way in R
> to get that?
>
> I know about gsubgn and strapply, but they only give me the strings
> matched by groups not their offsets.
>
> I could write something myself that first takes the above matches
> ("ab" and "aabb") and then searches again using only the group (b+).
> For this to work, I'd have to parse the regular expression and search
> several times (> 2, for nested groups) instead of just once. But I'm
> sure there is a better way to do this.
>
> Thanks for any suggestion!
>
> Titus
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
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25° 25' 40" S 49° 16' 22" O
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Re: [R] Regular expressions: offsets of groups
You could do this:
gregexpr("ab+", "abcdaabbcbb")[[1]] + 1
On Mon, Sep 27, 2010 at 2:25 PM, Titus von der Malsburg
wrote:
> On Mon, Sep 27, 2010 at 7:16 PM, Henrique Dallazuanna
> 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
>
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O
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Re: [R] Regular expressions: offsets of groups
On Mon, Sep 27, 2010 at 7:29 PM, Gabor Grothendieck
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+", "abcdaabbc", perl = TRUE)
which is wrong if the string is "abcdaabbcbbb".
Titus
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Re: [R] Regular expressions: offsets of groups
On Mon, Sep 27, 2010 at 11:48 AM, Titus von der Malsburg
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 there a way in R
> to get that?
>
> I know about gsubgn and strapply, but they only give me the strings
> matched by groups not their offsets.
>
> I could write something myself that first takes the above matches
> ("ab" and "aabb") and then searches again using only the group (b+).
> For this to work, I'd have to parse the regular expression and search
> several times (> 2, for nested groups) instead of just once. But I'm
> sure there is a better way to do this.
>
Try this zero width negative look behind expression:
> gregexpr("(?!a+)(b+)", "abcdaabbc", perl = TRUE)
[[1]]
[1] 2 7
attr(,"match.length")
[1] 1 2
See ?regexp for more info.
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Re: [R] Regular expressions: offsets of groups
On Mon, Sep 27, 2010 at 7:16 PM, Henrique Dallazuanna 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
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Re: [R] Regular expressions: offsets of groups
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 is an interface that exposes this
information.
Titus
On Mon, Sep 27, 2010 at 6:43 PM, jim holtman wrote:
> 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 at 11:48 AM, Titus von der Malsburg
> 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 there a way in R
>> to get that?
>>
>> I know about gsubgn and strapply, but they only give me the strings
>> matched by groups not their offsets.
>>
>> I could write something myself that first takes the above matches
>> ("ab" and "aabb") and then searches again using only the group (b+).
>> For this to work, I'd have to parse the regular expression and search
>> several times (> 2, for nested groups) instead of just once. But I'm
>> sure there is a better way to do this.
>>
>> Thanks for any suggestion!
>>
>> Titus
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem that you are trying to solve?
>
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Re: [R] Regular expressions: offsets of groups
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 at 11:48 AM, Titus von der Malsburg
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 there a way in R
> to get that?
>
> I know about gsubgn and strapply, but they only give me the strings
> matched by groups not their offsets.
>
> I could write something myself that first takes the above matches
> ("ab" and "aabb") and then searches again using only the group (b+).
> For this to work, I'd have to parse the regular expression and search
> several times (> 2, for nested groups) instead of just once. But I'm
> sure there is a better way to do this.
>
> Thanks for any suggestion!
>
> Titus
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
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Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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[R] Regular expressions: offsets of groups
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 strings
matched by groups not their offsets.
I could write something myself that first takes the above matches
("ab" and "aabb") and then searches again using only the group (b+).
For this to work, I'd have to parse the regular expression and search
several times (> 2, for nested groups) instead of just once. But I'm
sure there is a better way to do this.
Thanks for any suggestion!
Titus
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Re: [R] regular expressions
Perfect, thanks!
baptiste
2009/10/26 Gabor Grothendieck :
> 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 = rbind)
>>
>> pat2 <- "([[:alnum:]]+) +[0-9.]+"
>> colnames(s) <- strapply(st2[1], pat2, c, simplify = rbind)
>> s
> text1 text2 text3
> [1,] "1" "2.3" "5"
> [2,] "23.4" "3.1415" "6"
>
> If there are non-START fields that do match pat then grep out the
> START fields first.
>
> On Mon, Oct 26, 2009 at 9:30 AM, baptiste auguie
> wrote:
>> 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 a data.frame in this format:
>>
>> text1 text2
>> 1 2.3
>> 23.4 3.1415
>>
>>
>> All the lines containing "START" have the same number of fields, but
>> this number may vary from file to file.
>>
>> I have managed to get this minimal example work, but I am at a loss as
>> for handling an arbitrary number of couples (text value),
>>
>> library(gsubfn)
>>
>> ( parsed =
>> strapply(st, "^START +([[:alnum:]]+) +([0-9.]+) +([[:alnum:]]+)
>> +([0-9.]+)",c, simplify=rbind,combine=c) )
>>
>> d = data.frame(parsed[ ,c(2,4)])
>> names(d) <- apply(parsed[ ,c(1,3)], 2, unique)
>> d
>>
>> ## 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")
>>
>> Best regards,
>>
>>
>> Baptiste
>>
>> __
>> R-help@r-project.org mailing list
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>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
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Re: [R] regular expressions
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 = rbind)
>
> pat2 <- "([[:alnum:]]+) +[0-9.]+"
> colnames(s) <- strapply(st2[1], pat2, c, simplify = rbind)
> s
text1 text2text3
[1,] "1""2.3""5"
[2,] "23.4" "3.1415" "6"
If there are non-START fields that do match pat then grep out the
START fields first.
On Mon, Oct 26, 2009 at 9:30 AM, baptiste auguie
wrote:
> 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 a data.frame in this format:
>
> text1 text2
> 1 2.3
> 23.4 3.1415
>
>
> All the lines containing "START" have the same number of fields, but
> this number may vary from file to file.
>
> I have managed to get this minimal example work, but I am at a loss as
> for handling an arbitrary number of couples (text value),
>
> library(gsubfn)
>
> ( parsed =
> strapply(st, "^START +([[:alnum:]]+) +([0-9.]+) +([[:alnum:]]+)
> +([0-9.]+)",c, simplify=rbind,combine=c) )
>
> d = data.frame(parsed[ ,c(2,4)])
> names(d) <- apply(parsed[ ,c(1,3)], 2, unique)
> d
>
> ## 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")
>
> Best regards,
>
>
> Baptiste
>
> __
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> and provide commented, minimal, self-contained, reproducible code.
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[R] regular expressions
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 a data.frame in this format:
text1 text2
12.3
23.4 3.1415
All the lines containing "START" have the same number of fields, but
this number may vary from file to file.
I have managed to get this minimal example work, but I am at a loss as
for handling an arbitrary number of couples (text value),
library(gsubfn)
( parsed =
strapply(st, "^START +([[:alnum:]]+) +([0-9.]+) +([[:alnum:]]+)
+([0-9.]+)",c, simplify=rbind,combine=c) )
d = data.frame(parsed[ ,c(2,4)])
names(d) <- apply(parsed[ ,c(1,3)], 2, unique)
d
## 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")
Best regards,
Baptiste
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Re: [R] Regular expressions: bug or misunderstanding?
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( "(? 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 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 make everything I write completely unreadable, so I'm going to change the problem for this post: I will define E to be the escape character, and q to be the quote; the gsub() call would leave Eq unchanged, but would change EEq to EEEq, etc. The expression I have come up with after this change is gsub( "((^|[^E])(EE)*)q", "\\1Eq", x) i.e. "(start of line, or non-escape, followed by an even number of escapes), all of which we call expression 1, followed by a quote, is replaced by expression 1 followed by an escape and a quote". This works sometimes, but not always: > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "Eq") [1] "Eq" > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "EEq") [1] "EEEq" > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qaq") [1] "EqaEq" > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qq") [1] "qEq" Notice that in the final example, the first quote doesn't get escaped. Why not I think (without having done the "experimental diagnostics") that it's because in "qq" the first q mtaches (^|[^E]) because it matches [^E] (i.e. is a "non-escape"); since it is followed by q, it is the second q which gets the escape. Possibly you need to include "^q" as an additional alternative match at the start of the line. Thanks, that sounds right, but now I can't see how to fix it. Is there syntax to say: match A only if it follows B, but don't match the B part? Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: bug or misunderstanding?
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 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 make >>> everything I write completely unreadable, so I'm going to change >>> the problem for this post: I will define E to be the escape >>> character, and q to be the quote; the gsub() call would leave >>> >>> Eq >>> >>> unchanged, but would change >>> >>> EEq >>> >>> to EEEq, etc. >>> >>> The expression I have come up with after this change is >>> >>> gsub( "((^|[^E])(EE)*)q", "\\1Eq", x) >>> >>> i.e. "(start of line, or non-escape, followed by an even number of >>> escapes), all of which we call expression 1, followed by a quote, >>> is replaced by expression 1 followed by an escape and a quote". >>> >>> This works sometimes, but not always: >>> >>> > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "Eq") >>> [1] "Eq" >>> > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "EEq") >>> [1] "EEEq" >>> > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qaq") >>> [1] "EqaEq" >>> > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qq") >>> [1] "qEq" >>> >>> Notice that in the final example, the first quote doesn't get escaped. >>> Why not >> >> I think (without having done the "experimental diagnostics") >> that it's because in "qq" the first q mtaches (^|[^E]) because >> it matches [^E] (i.e. is a "non-escape"); since it is followed >> by q, it is the second q which gets the escape. Possibly you >> need to include "^q" as an additional alternative match at the >> start of the line. > > Thanks, that sounds right, but now I can't see how to fix it. Is there > syntax to say: match A only if it follows B, but don't match the B part? > > Duncan Murdoch > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: bug or misunderstanding?
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 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 make everything I write completely unreadable, so I'm going to change the problem for this post: I will define E to be the escape character, and q to be the quote; the gsub() call would leave Eq unchanged, but would change EEq to EEEq, etc. The expression I have come up with after this change is gsub( "((^|[^E])(EE)*)q", "\\1Eq", x) i.e. "(start of line, or non-escape, followed by an even number of escapes), all of which we call expression 1, followed by a quote, is replaced by expression 1 followed by an escape and a quote". This works sometimes, but not always: > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "Eq") [1] "Eq" > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "EEq") [1] "EEEq" > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qaq") [1] "EqaEq" > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qq") [1] "qEq" Notice that in the final example, the first quote doesn't get escaped. Why not I think (without having done the "experimental diagnostics") that it's because in "qq" the first q mtaches (^|[^E]) because it matches [^E] (i.e. is a "non-escape"); since it is followed by q, it is the second q which gets the escape. Possibly you need to include "^q" as an additional alternative match at the start of the line. Thanks, that sounds right, but now I can't see how to fix it. Is there syntax to say: match A only if it follows B, but don't match the B part? Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: bug or misunderstanding?
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 for the quote, > the first just escapes the second. I have the usual problems of > writing regular expressions involving backslashes which make > everything I write completely unreadable, so I'm going to change > the problem for this post: I will define E to be the escape > character, and q to be the quote; the gsub() call would leave > > Eq > > unchanged, but would change > > EEq > > to EEEq, etc. > > The expression I have come up with after this change is > > gsub( "((^|[^E])(EE)*)q", "\\1Eq", x) > > i.e. "(start of line, or non-escape, followed by an even number of > escapes), all of which we call expression 1, followed by a quote, > is replaced by expression 1 followed by an escape and a quote". > > This works sometimes, but not always: > > > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "Eq") > [1] "Eq" > > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "EEq") > [1] "EEEq" > > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qaq") > [1] "EqaEq" > > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qq") > [1] "qEq" > > Notice that in the final example, the first quote doesn't get escaped. > Why not I think (without having done the "experimental diagnostics") that it's because in "qq" the first q mtaches (^|[^E]) because it matches [^E] (i.e. is a "non-escape"); since it is followed by q, it is the second q which gets the escape. Possibly you need to include "^q" as an additional alternative match at the start of the line. Ted. E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 094 0861 Date: 06-Jul-08 Time: 22:37:10 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: bug or misunderstanding?
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 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 make everything I write completely > unreadable, so I'm going to change the problem for this post: I will define > E to be the escape character, and q to be the quote; the gsub() call would > leave > > Eq > > unchanged, but would change > > EEq > > to EEEq, etc. > > The expression I have come up with after this change is > > gsub( "((^|[^E])(EE)*)q", "\\1Eq", x) > > i.e. "(start of line, or non-escape, followed by an even number of escapes), > all of which we call expression 1, followed by a quote, is replaced by > expression 1 followed by an escape and a quote". > > This works sometimes, but not always: > >> gsub( "((^|[^E])(EE)*)q", "\\1Eq", "Eq") > [1] "Eq" >> gsub( "((^|[^E])(EE)*)q", "\\1Eq", "EEq") > [1] "EEEq" >> gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qaq") > [1] "EqaEq" >> gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qq") > [1] "qEq" > > Notice that in the final example, the first quote doesn't get escaped. Why > not > > Duncan Murdoch > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Regular expressions: bug or misunderstanding?
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 make everything I write completely unreadable, so I'm going to change the problem for this post: I will define E to be the escape character, and q to be the quote; the gsub() call would leave Eq unchanged, but would change EEq to EEEq, etc. The expression I have come up with after this change is gsub( "((^|[^E])(EE)*)q", "\\1Eq", x) i.e. "(start of line, or non-escape, followed by an even number of escapes), all of which we call expression 1, followed by a quote, is replaced by expression 1 followed by an escape and a quote". This works sometimes, but not always: > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "Eq") [1] "Eq" > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "EEq") [1] "EEEq" > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qaq") [1] "EqaEq" > gsub( "((^|[^E])(EE)*)q", "\\1Eq", "qq") [1] "qEq" Notice that in the final example, the first quote doesn't get escaped. Why not Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular Expressions
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 wondering how do I give the regular expression
>
As others have mentioned strsplit is probably easier in this case but it can
be done with a regular expression as shown below where [^_]+ matches a
any string of characters not containing _ :
> re <- "^([^_]+_)*([^_]+)_([^_]+)$"
> gsub(re, "\\2", S)
[1] "is""three" "My"
The strapply function in the gsubfn package can also be used.
out below has the same value as strsplit(S, "_"):
library(gsubfn)
out <- strapply(S, "[^_]+")
sapply(out, function(x) tail(x, 2)[1])
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Re: [R] Regular Expressions
> 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) x[length(x)-1])
You could use regular expressions, but I think it would only be
complicating things.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
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Re: [R] Regular Expressions
try this:
S <- c("World_is_beautiful", "one_two_three_four","My_book")
sapply(strsplit(S, "_"), tail, n = 2)[1, ]
# or
sapply(strsplit(S, "_"), function(x) x[length(x) - 1])
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: "Shubha Vishwanath Karanth" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: 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")
gsub() can do this...but wondering how do I give the regular
expression
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
This e-mail may contain confidential and/or privileged
i...{{dropped:13}}
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Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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[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")
gsub() can do this...but wondering how do I give the regular expression
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
This e-mail may contain confidential and/or privileged i...{{dropped:13}}
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Re: [R] Regular Expressions Help
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")
> [...]
> I have been reading over various post on regular expressions, but
> really haven't made any progress. As far as I can tell there aren't
> standard string functions in R. (Also, as an aside, is there a
> wildcard character in R? I'd want something so if x="Bob" a statment
> of the form x== "B*b" would evaluate true where * is the wildcard.)
I'm not really sure if I understood you correctly.
If you're looking for a way to match St against something like "B*b"
you can make usage of normal regular expressions (like grep()).
For details begin with ?regexp and ?grep
Then you can try for instance:
grep("B.b", St)
to get
[1] 1 3
(the indices of the vector St)
or directly
St[grep("B.b", St)]
Cheers,
--Hans
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[R] Regular Expressions Help
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 column's listed in St
that were in Matrix (which I can do). However, In addition for
elements of the form "Alice:Bob" I want to make a new column in my new
matrix with the product of these columns. I am not sure if a semicolon
is a valid symbol for a column name, if not we can replace it with
something like qqq.
I have been reading over various post on regular expressions, but
really haven't made any progress. As far as I can tell there aren't
standard string functions in R. (Also, as an aside, is there a
wildcard character in R? I'd want something so if x="Bob" a statment
of the form x== "B*b" would evaluate true where * is the wildcard.)
Any help would be appreciated!
(also, thanks again to those who helped on my last question!)
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Re: [R] regular expressions
Try this one:
> gsub("^(plif)([a-z]*)", "\\1ONE", words)
[1] "plifONE" "plafboum" "ploufbang" "plifONE"
-Christos
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of GOUACHE David
> Sent: Wednesday, March 12, 2008 12:15 PM
> To: [EMAIL PROTECTED]
> Subject: [R] 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.
>
> Here is a quick example :
>
> words<-c("plifboum","plafboum","ploufbang","plifplaf")
>
> I want to end up with something like this:
>
> "plifONE","plafboum","ploufbang","plifONE"
>
> All I can produce so far is this :
> gsub("\\bplif.","ONE",words,perl=T)
>
> which turns out :
> "ONEboum","plafboum","ploufbang","ONEplaf"
>
> Thanks in advance for your help.
>
> David Gouache
>
> Arvalis - Institut du Végétal
>
> Station de La Minière
>
> 78280 Guyancourt
>
> Tel: 01.30.12.96.22 / Port: 06.86.08.94.32
>
>
>
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
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[R] 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.
Here is a quick example :
words<-c("plifboum","plafboum","ploufbang","plifplaf")
I want to end up with something like this:
"plifONE","plafboum","ploufbang","plifONE"
All I can produce so far is this :
gsub("\\bplif.","ONE",words,perl=T)
which turns out :
"ONEboum","plafboum","ploufbang","ONEplaf"
Thanks in advance for your help.
David Gouache
Arvalis - Institut du Végétal
Station de La Minière
78280 Guyancourt
Tel: 01.30.12.96.22 / Port: 06.86.08.94.32
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