Re: [R] grep(pattern = each element of a vector) ?
Hi,
res- ddply(.data=df1,
.variables='Taxa',
.fun=transform,
Class=find.class(Taxa))
#Warning messages:
#1: In grep(x, df2$Taxa) :
# argument 'pattern' has length 1 and only the first element will be used
#2: In grep(x, df2$Taxa) :
# argument 'pattern' has length 1 and only the first element will be used
#3: In grep(x, df2$Taxa) :
# argument 'pattern' has length 1 and only the first element will be used
May be it is better to modify the function:
find.class- function(x) df2[grep(unique(x),df2$Taxa),'Class']
res1- ddply(.data=df1,
.variables='Taxa',
.fun=transform,
Class=find.class(Taxa)) #no warnings
#though it doesn't have any effect in the end result.
identical(res,res1)
#[1] TRUE
A.K.
- Original Message -
From: Allen, Joel allen.j...@epa.gov
To: Beaulieu, Jake beaulieu.j...@epa.gov; r-help@r-project.org
r-help@r-project.org
Cc: Farrar, David farrar.da...@epa.gov; Green, Hyatt
green.hy...@epa.gov; McManus, Michael mcmanus.mich...@epa.gov; Wahman,
David wahman.da...@epa.gov
Sent: Thursday, September 12, 2013 2:49 PM
Subject: Re: [R] grep(pattern = each element of a vector) ?
Jake,
You can use the plyr library or some form of apply. If you are on a 64bit
system you can multithread and it goes much faster.
something like this(for 32bit):
require(plyr)
df1 - data.frame(Taxa = c('blue', 'red', NA,'blue', 'red', NA,'blue', 'red',
NA))
df2 - data.frame(Taxa = c( 'blue', 'red', NA), Class = c('Z', 'HI', 'A'))
#function to do the lookup
find.class-function(x)df2[grep(x, df2$Taxa),'Class']
ddply(.data=df1,
.variables='Taxa',
.fun=transform,
Class=find.class(Taxa))
Joel
From: Beaulieu, Jake
Sent: Thursday, September 12, 2013 12:06 PM
To: r-help@r-project.org
Cc: Wahman, David; Farrar, David; Allen, Joel; Green, Hyatt; McManus, Michael
Subject: grep(pattern = each element of a vector) ?
Hi,
I have a large dataframe that contains species names. I have a second
dataframe that contains species names and some additional info, called 'Class',
about each species. I would like match the species name is the first data
frame with the 'Class' information contained in the second. Since the species
names are often formatted differently between the data sets, merge doesn't work
well. grep does the trick, but the function needs to be called separately for
each observation in the first data frame. I put grep into a loop, but this is
too slow. Is there a way to run grep repeatedly without resorting to a loop?
Possibly something in the apply family?
df1 - data.frame(Taxa = c('blue', 'red', NA))
df2 - data.frame(Taxa = c( 'blue', 'red', NA), Class = c('Z', 'HI', 'A'))
index - NULL
for (i in 1:length(df1$Taxa)) {
index[i] - grep(df1$Taxa[1], df2$Taxa)
}
index
sessionInfo()
R version 3.0.1 (2013-05-16)
Platform: i386-w64-mingw32/i386 (32-bit)
==
Jake J. Beaulieu, PhD
US Environmental Protection Agency
National Risk Management Research Lab
26 W. Martin Luther King Drive
Cincinnati, OH 45268
USA
513-569-7842 (desk)
513-487-2511 (fax)
beaulieu.j...@epa.govmailto:beaulieu.j...@epa.gov
[[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.
__
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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] grep(pattern = each element of a vector) ?
Hi,
I have a large dataframe that contains species names. I have a second
dataframe that contains species names and some additional info, called 'Class',
about each species. I would like match the species name is the first data
frame with the 'Class' information contained in the second. Since the species
names are often formatted differently between the data sets, merge doesn't work
well. grep does the trick, but the function needs to be called separately for
each observation in the first data frame. I put grep into a loop, but this is
too slow. Is there a way to run grep repeatedly without resorting to a loop?
Possibly something in the apply family?
df1 - data.frame(Taxa = c('blue', 'red', NA))
df2 - data.frame(Taxa = c( 'blue', 'red', NA), Class = c('Z', 'HI', 'A'))
index - NULL
for (i in 1:length(df1$Taxa)) {
index[i] - grep(df1$Taxa[1], df2$Taxa)
}
index
sessionInfo()
R version 3.0.1 (2013-05-16)
Platform: i386-w64-mingw32/i386 (32-bit)
==
Jake J. Beaulieu, PhD
US Environmental Protection Agency
National Risk Management Research Lab
26 W. Martin Luther King Drive
Cincinnati, OH 45268
USA
513-569-7842 (desk)
513-487-2511 (fax)
beaulieu.j...@epa.gov
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] grep(pattern = each element of a vector) ?
Jake,
You can use the plyr library or some form of apply. If you are on a 64bit
system you can multithread and it goes much faster.
something like this(for 32bit):
require(plyr)
df1 - data.frame(Taxa = c('blue', 'red', NA,'blue', 'red', NA,'blue', 'red',
NA))
df2 - data.frame(Taxa = c( 'blue', 'red', NA), Class = c('Z', 'HI', 'A'))
#function to do the lookup
find.class-function(x)df2[grep(x, df2$Taxa),'Class']
ddply(.data=df1,
.variables='Taxa',
.fun=transform,
Class=find.class(Taxa))
Joel
From: Beaulieu, Jake
Sent: Thursday, September 12, 2013 12:06 PM
To: r-help@r-project.org
Cc: Wahman, David; Farrar, David; Allen, Joel; Green, Hyatt; McManus, Michael
Subject: grep(pattern = each element of a vector) ?
Hi,
I have a large dataframe that contains species names. I have a second
dataframe that contains species names and some additional info, called 'Class',
about each species. I would like match the species name is the first data
frame with the 'Class' information contained in the second. Since the species
names are often formatted differently between the data sets, merge doesn't work
well. grep does the trick, but the function needs to be called separately for
each observation in the first data frame. I put grep into a loop, but this is
too slow. Is there a way to run grep repeatedly without resorting to a loop?
Possibly something in the apply family?
df1 - data.frame(Taxa = c('blue', 'red', NA))
df2 - data.frame(Taxa = c( 'blue', 'red', NA), Class = c('Z', 'HI', 'A'))
index - NULL
for (i in 1:length(df1$Taxa)) {
index[i] - grep(df1$Taxa[1], df2$Taxa)
}
index
sessionInfo()
R version 3.0.1 (2013-05-16)
Platform: i386-w64-mingw32/i386 (32-bit)
==
Jake J. Beaulieu, PhD
US Environmental Protection Agency
National Risk Management Research Lab
26 W. Martin Luther King Drive
Cincinnati, OH 45268
USA
513-569-7842 (desk)
513-487-2511 (fax)
beaulieu.j...@epa.govmailto:beaulieu.j...@epa.gov
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] grep pattern
try this using strsplit:
x - round(runif(10)*10, digits=0)
y - as.Date(x, origin=1970-01-01)
str(y)
Class 'Date' num [1:10] 26551 37212 57285 90821 20168 ...
y1 - as.character(y)
str(y1)
chr [1:10] 2042-09-11 2071-11-19 2126-11-04 2218-08-30
2025-03-21 2215-12-22 ...
x - strsplit(y1, '-')
x[1:3]
[[1]]
[1] 2042 09 11
[[2]]
[1] 2071 11 19
[[3]]
[1] 2126 11 04
x.1 - sapply(x, '[', 3)
str(x.1)
chr [1:10] 11 19 04 30 21 22 24 03 31 02
On Tue, May 24, 2011 at 10:19 AM, Kang Min ngokang...@gmail.com wrote:
I have another question -
I'd like to extract dates from a vector of -mm-dd, so I just want
the dd.
x - round(runif(10)*10, digits=0)
y - as.Date(x, origin=1970-01-01)
I tried this based on the code that Jim provided, but it just printed
the whole date. I think I just need to tweak it a little, but haven't
been able to figure it out.
y[grep([[:digit:]]{2}$, y)]
Thanks.
Kang Min
On May 23, 7:22 am, jim holtman jholt...@gmail.com wrote:
If you want to only match names of length 6, you will have to use
thispattern:
x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ, ZAAZ, ZAZ,
+ ZZAZ, ZRITEZ)
# match exactly values of length 6
len6 - ^Z[[:alpha:]]{4}Z$
grep(len6, x)
[1] 2 5 9
On Sun, May 22, 2011 at 5:10 PM, Kang Min ngokang...@gmail.com wrote:
Thanks!
On May 21, 7:09 am, David Winsemius dwinsem...@comcast.net wrote:
On May 20, 2011, at 11:57 AM, Kang Min wrote:
Hi all,
I'm trying to subset apatternin a vector. Each argument has 6
letters, and I need those that start with Z and end with Z.
e.g.
x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ)
I've looked up other discussions but still can't seem to find the
answer.
You may need to study the regex page a bit longer
the ^ is the beginning of a string
.+ will math can arbitrarily long string of anything
and $ indicates the end of a string
x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ)
grep(^Z.+Z$, x)
[1] 2 5
grep(^Z.+Z$, x, value=TRUE)
[1] ZFHJKZ ZKFLPZ
Thanks.
Kangmin
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
guidehttp://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
r-h...@r-project.org mailing
listhttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
guidehttp://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
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Data Munger Guru
What is the problem that you are trying to solve?
__
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PLEASE do read the posting guidehttp://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.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
__
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] grep pattern
I have another question -
I'd like to extract dates from a vector of -mm-dd, so I just want
the dd.
x - round(runif(10)*10, digits=0)
y - as.Date(x, origin=1970-01-01)
I tried this based on the code that Jim provided, but it just printed
the whole date. I think I just need to tweak it a little, but haven't
been able to figure it out.
y[grep([[:digit:]]{2}$, y)]
Thanks.
Kang Min
On May 23, 7:22 am, jim holtman jholt...@gmail.com wrote:
If you want to only match names of length 6, you will have to use thispattern:
x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ, ZAAZ, ZAZ,
+ ZZAZ, ZRITEZ)
# match exactly values of length 6
len6 - ^Z[[:alpha:]]{4}Z$
grep(len6, x)
[1] 2 5 9
On Sun, May 22, 2011 at 5:10 PM, Kang Min ngokang...@gmail.com wrote:
Thanks!
On May 21, 7:09 am, David Winsemius dwinsem...@comcast.net wrote:
On May 20, 2011, at 11:57 AM, Kang Min wrote:
Hi all,
I'm trying to subset apatternin a vector. Each argument has 6
letters, and I need those that start with Z and end with Z.
e.g.
x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ)
I've looked up other discussions but still can't seem to find the
answer.
You may need to study the regex page a bit longer
the ^ is the beginning of a string
.+ will math can arbitrarily long string of anything
and $ indicates the end of a string
x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ)
grep(^Z.+Z$, x)
[1] 2 5
grep(^Z.+Z$, x, value=TRUE)
[1] ZFHJKZ ZKFLPZ
Thanks.
Kangmin
__
r-h...@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
guidehttp://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
r-h...@r-project.org mailing
listhttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
r-h...@r-project.org mailing list
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PLEASE do read the posting guidehttp://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?
__
r-h...@r-project.org mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guidehttp://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] grep pattern
Thanks! On May 21, 7:09 am, David Winsemius dwinsem...@comcast.net wrote: On May 20, 2011, at 11:57 AM, Kang Min wrote: Hi all, I'm trying to subset a pattern in a vector. Each argument has 6 letters, and I need those that start with Z and end with Z. e.g. x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ) I've looked up other discussions but still can't seem to find the answer. You may need to study the regex page a bit longer the ^ is the beginning of a string .+ will math can arbitrarily long string of anything and $ indicates the end of a string x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ) grep(^Z.+Z$, x) [1] 2 5 grep(^Z.+Z$, x, value=TRUE) [1] ZFHJKZ ZKFLPZ Thanks. Kangmin __ r-h...@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ r-h...@r-project.org mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guidehttp://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] grep pattern
If you want to only match names of length 6, you will have to use this pattern:
x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ, ZAAZ, ZAZ,
+ ZZAZ, ZRITEZ)
# match exactly values of length 6
len6 - ^Z[[:alpha:]]{4}Z$
grep(len6, x)
[1] 2 5 9
On Sun, May 22, 2011 at 5:10 PM, Kang Min ngokang...@gmail.com wrote:
Thanks!
On May 21, 7:09 am, David Winsemius dwinsem...@comcast.net wrote:
On May 20, 2011, at 11:57 AM, Kang Min wrote:
Hi all,
I'm trying to subset a pattern in a vector. Each argument has 6
letters, and I need those that start with Z and end with Z.
e.g.
x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ)
I've looked up other discussions but still can't seem to find the
answer.
You may need to study the regex page a bit longer
the ^ is the beginning of a string
.+ will math can arbitrarily long string of anything
and $ indicates the end of a string
x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ)
grep(^Z.+Z$, x)
[1] 2 5
grep(^Z.+Z$, x, value=TRUE)
[1] ZFHJKZ ZKFLPZ
Thanks.
Kangmin
__
r-h...@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
r-h...@r-project.org mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
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--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] grep pattern
Hi all, I'm trying to subset a pattern in a vector. Each argument has 6 letters, and I need those that start with Z and end with Z. e.g. x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ) I've looked up other discussions but still can't seem to find the answer. Thanks. Kangmin __ 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] grep pattern
On May 20, 2011, at 11:57 AM, Kang Min wrote: Hi all, I'm trying to subset a pattern in a vector. Each argument has 6 letters, and I need those that start with Z and end with Z. e.g. x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ) I've looked up other discussions but still can't seem to find the answer. You may need to study the regex page a bit longer the ^ is the beginning of a string .+ will math can arbitrarily long string of anything and $ indicates the end of a string x - c(ZFHSJK, ZFHJKZ,ZIOPWE,ZLKJSD,ZKFLPZ) grep(^Z.+Z$, x) [1] 2 5 grep(^Z.+Z$, x, value=TRUE) [1] ZFHJKZ ZKFLPZ Thanks. Kangmin __ 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. David Winsemius, MD West Hartford, CT __ 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.
