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Removing rows if certain elements are found in character string
10 messages · jim holtman, arun, David Winsemius +3 more
Hello,
Try regular expressions instead.
In this data.frame, I've changed row nr.4 to have a row with 'D' as
first non-zero character.
dd <- read.table(text="
ch count
1 0000000000D0000000000000000000000000000000000000 0.007368
2 0000000000d0000000000000000000000000000000000000 0.002456
3 000000000T00000000000000000000000000000000000000 0.007368
4 000000000DT0000000000000000000000000000000000000 0.007368
5 000000000T00000000000000000000000000000000000000 0.002456
6 000000000Td0000000000000000000000000000000000000 0.002456
7 00000000T000000000000000000000000000000000000000 0.007368
8 00000000T0D0000000000000000000000000000000000000 0.007368
9 00000000T000000000000000000000000000000000000000 0.002456
10 00000000T0d0000000000000000000000000000000000000 0.002456
", header=TRUE)
dd
i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i2 <- grepl("^0*[Dd]", dd$ch)
dd[!i1, ]
dd[!i2, ]
dd[!(i1 | i2), ]
Hope this helps,
Rui Barradas
Em 02-07-2012 23:48, Claudia Penaloza escreveu:
I would like to remove rows from the following data frame (df) if there are
only two specific elements found in the df$ch character string (I want to
remove rows with only "0" & "D" or "0" & "d"). Alternatively, I would like
to remove rows if the first non-zero element is "D" or "d".
ch count
1 0000000000D0000000000000000000000000000000000000 0.007368;
2 0000000000d0000000000000000000000000000000000000 0.002456;
3 000000000T00000000000000000000000000000000000000 0.007368;
4 000000000TD0000000000000000000000000000000000000 0.007368;
5 000000000T00000000000000000000000000000000000000 0.002456;
6 000000000Td0000000000000000000000000000000000000 0.002456;
7 00000000T000000000000000000000000000000000000000 0.007368;
8 00000000T0D0000000000000000000000000000000000000 0.007368;
9 00000000T000000000000000000000000000000000000000 0.002456;
10 00000000T0d0000000000000000000000000000000000000 0.002456;
I tried the following but it doesn't work if there is more than one
character per string:
df <- df[!df$ch %in% c("0","D"),]
df <- df[!df$ch %in% c("0","d"),]
Any help greatly appreciated, Claudia [[alternative HTML version deleted]]
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You will have to change the 'i1' expression as follows:
i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i1 # matches strings with d & D in them
[1] TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# second string had 'd' & 'D' in it so it was TRUE above and FALSE below
i1new <- grepl("^([0D]*$|[0d]*$)", dd$ch)
i1new
[1] TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
I put a 'd' and 'D' in the second string and the original regular
expression is equivalent to
grepl("^[0dD]*$", dd$ch)
which will match strings containing d, D and 0. If you only want 'd'
or 'D' (and not both), then you will have to use the one in 'i1new'.
On Mon, Jul 2, 2012 at 7:24 PM, Rui Barradas <ruipbarradas at sapo.pt> wrote:
Hello,
Try regular expressions instead.
In this data.frame, I've changed row nr.4 to have a row with 'D' as first
non-zero character.
dd <- read.table(text="
ch count
1 0000000000D0000000000000000000000000000000000000 0.007368
2 0000000000d0000000000000000000000000000000000000 0.002456
3 000000000T00000000000000000000000000000000000000 0.007368
4 000000000DT0000000000000000000000000000000000000 0.007368
5 000000000T00000000000000000000000000000000000000 0.002456
6 000000000Td0000000000000000000000000000000000000 0.002456
7 00000000T000000000000000000000000000000000000000 0.007368
8 00000000T0D0000000000000000000000000000000000000 0.007368
9 00000000T000000000000000000000000000000000000000 0.002456
10 00000000T0d0000000000000000000000000000000000000 0.002456
", header=TRUE)
dd
i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i2 <- grepl("^0*[Dd]", dd$ch)
dd[!i1, ]
dd[!i2, ]
dd[!(i1 | i2), ]
Hope this helps,
Rui Barradas
Em 02-07-2012 23:48, Claudia Penaloza escreveu:
I would like to remove rows from the following data frame (df) if there
are
only two specific elements found in the df$ch character string (I want to
remove rows with only "0" & "D" or "0" & "d"). Alternatively, I would like
to remove rows if the first non-zero element is "D" or "d".
ch count
1 0000000000D0000000000000000000000000000000000000 0.007368;
2 0000000000d0000000000000000000000000000000000000 0.002456;
3 000000000T00000000000000000000000000000000000000 0.007368;
4 000000000TD0000000000000000000000000000000000000 0.007368;
5 000000000T00000000000000000000000000000000000000 0.002456;
6 000000000Td0000000000000000000000000000000000000 0.002456;
7 00000000T000000000000000000000000000000000000000 0.007368;
8 00000000T0D0000000000000000000000000000000000000 0.007368;
9 00000000T000000000000000000000000000000000000000 0.002456;
10 00000000T0d0000000000000000000000000000000000000 0.002456;
I tried the following but it doesn't work if there is more than one
character per string:
df <- df[!df$ch %in% c("0","D"),]
df <- df[!df$ch %in% c("0","d"),]
Any help greatly appreciated,
Claudia
[[alternative HTML version deleted]]
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______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it.
Hi,
I didn't think about the situation where D comes before T.? I changed my code a little to accommodate that.
dat2<-read.table(text="
1? 0000000000D0000000000000000000000000000000000000 0.007368;
2? 0000000000d0000000000000000000000000000000000000 0.002456;
3? 000000000T00000000000000000000000000000000000000 0.007368;
4? 000000000DT0000000000000000000000000000000000000 0.007368;
5? 000000000T00000000000000000000000000000000000000 0.002456;
6? 000000000Td0000000000000000000000000000000000000 0.002456;
7? 00000000T000000000000000000000000000000000000000 0.007368;
8? 00000000T0D0000000000000000000000000000000000000 0.007368;
9? 00000000T000000000000000000000000000000000000000 0.002456;
10 00000000T0d0000000000000000000000000000000000000 0.002456;
",sep="",header=FALSE)
colnames(dat2)<-c("num","Ch", "count")
dat2[grepl("0T|0Td|0TD",dat2$Ch),]
num?????????????????????????????????????????????? Ch???? count
3??? 3 000000000T00000000000000000000000000000000000000 0.007368;
5??? 5 000000000T00000000000000000000000000000000000000 0.002456;
6??? 6 000000000Td0000000000000000000000000000000000000 0.002456;
7??? 7 00000000T000000000000000000000000000000000000000 0.007368;
8??? 8 00000000T0D0000000000000000000000000000000000000 0.007368;
9??? 9 00000000T000000000000000000000000000000000000000 0.002456;
10? 10 00000000T0d0000000000000000000000000000000000000 0.002456;
A.K.
----- Original Message -----
From: Rui Barradas <ruipbarradas at sapo.pt>
To: Claudia Penaloza <claudiapenaloza at gmail.com>
Cc: r-help at r-project.org
Sent: Monday, July 2, 2012 7:24 PM
Subject: Re: [R] Removing rows if certain elements are found in character string
Hello,
Try regular expressions instead.
In this data.frame, I've changed row nr.4 to have a row with 'D' as
first non-zero character.
dd <- read.table(text="
ch? ? count
1? 0000000000D0000000000000000000000000000000000000 0.007368
2? 0000000000d0000000000000000000000000000000000000 0.002456
3? 000000000T00000000000000000000000000000000000000 0.007368
4? 000000000DT0000000000000000000000000000000000000 0.007368
5? 000000000T00000000000000000000000000000000000000 0.002456
6? 000000000Td0000000000000000000000000000000000000 0.002456
7? 00000000T000000000000000000000000000000000000000 0.007368
8? 00000000T0D0000000000000000000000000000000000000 0.007368
9? 00000000T000000000000000000000000000000000000000 0.002456
10 00000000T0d0000000000000000000000000000000000000 0.002456
", header=TRUE)
dd
i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i2 <- grepl("^0*[Dd]", dd$ch)
dd[!i1, ]
dd[!i2, ]
dd[!(i1 | i2), ]
Hope this helps,
Rui Barradas
Em 02-07-2012 23:48, Claudia Penaloza escreveu:
I would like to remove rows from the following data frame (df) if there are only two specific elements found in the df$ch character string (I want to remove rows with only "0" & "D" or "0" & "d"). Alternatively, I would like to remove rows if the first non-zero element is "D" or "d". ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ch? ? count 1? 0000000000D0000000000000000000000000000000000000 0.007368; 2? 0000000000d0000000000000000000000000000000000000 0.002456; 3? 000000000T00000000000000000000000000000000000000 0.007368; 4? 000000000TD0000000000000000000000000000000000000 0.007368; 5? 000000000T00000000000000000000000000000000000000 0.002456; 6? 000000000Td0000000000000000000000000000000000000 0.002456; 7? 00000000T000000000000000000000000000000000000000 0.007368; 8? 00000000T0D0000000000000000000000000000000000000 0.007368; 9? 00000000T000000000000000000000000000000000000000 0.002456; 10 00000000T0d0000000000000000000000000000000000000 0.002456; I tried the following but it doesn't work if there is more than one character per string:
df <- df[!df$ch %in% c("0","D"),]
df <- df[!df$ch %in% c("0","d"),]
Any help greatly appreciated, Claudia ??? [[alternative HTML version deleted]]
______________________________________________ R-help at 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 at 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.
On Jul 2, 2012, at 6:48 PM, Claudia Penaloza wrote:
I would like to remove rows from the following data frame (df) if
there are
only two specific elements found in the df$ch character string (I
want to
remove rows with only "0" & "D" or "0" & "d"). Alternatively, I
would like
to remove rows if the first non-zero element is "D" or "d".
ch count
1 0000000000D0000000000000000000000000000000000000 0.007368;
2 0000000000d0000000000000000000000000000000000000 0.002456;
3 000000000T00000000000000000000000000000000000000 0.007368;
4 000000000TD0000000000000000000000000000000000000 0.007368;
5 000000000T00000000000000000000000000000000000000 0.002456;
6 000000000Td0000000000000000000000000000000000000 0.002456;
7 00000000T000000000000000000000000000000000000000 0.007368;
8 00000000T0D0000000000000000000000000000000000000 0.007368;
9 00000000T000000000000000000000000000000000000000 0.002456;
10 00000000T0d0000000000000000000000000000000000000 0.002456;
I tried the following but it doesn't work if there is more than one
character per string:
df <- df[!df$ch %in% c("0","D"),]
df <- df[!df$ch %in% c("0","d"),]
You seem to be missing test cases for the second set of conditions but
this works for the first set (and might for the second):
> dat[ grepl("[^0dD]", dat$ch) & ! grepl("^0+d|^0^D", dat$ch) , ]
ch count
3 000000000T00000000000000000000000000000000000000 0.007368
4 000000000TD0000000000000000000000000000000000000 0.007368
5 000000000T00000000000000000000000000000000000000 0.002456
6 000000000Td0000000000000000000000000000000000000 0.002456
7 00000000T000000000000000000000000000000000000000 0.007368
8 00000000T0D0000000000000000000000000000000000000 0.007368
9 00000000T000000000000000000000000000000000000000 0.002456
10 00000000T0d0000000000000000000000000000000000000 0.002456
David Winsemius, MD West Hartford, CT
Hello, Inline. Em 03-07-2012 01:15, jim holtman escreveu:
You will have to change the 'i1' expression as follows:
i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i1 # matches strings with d & D in them
[1] TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# second string had 'd' & 'D' in it so it was TRUE above and FALSE below
i1new <- grepl("^([0D]*$|[0d]*$)", dd$ch)
i1new
[1] TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
Right, apparently, I forgot that grep is greedy, and the test cases were not complete.
I put a 'd' and 'D' in the second string and the original regular
expression is equivalent to
grepl("^[0dD]*$", dd$ch)
This is only for the first request, and does not solve cases where there
are characters other than '0', 'd' or 'D', but 'd' or 'D' are the first
non-zero. This is the case of my 4th row, changed from the OP's data
example.
My regexpr for 'i2' is equivalent to this one, that I believe is more
readable:
i2b <- grepl("^0{0,}[Dd]", dd$ch)
First a zero, that might occur zero or more times, then a 'd' or 'D',
then and til the end, irrelevant.
which will match strings containing d, D and 0. If you only want 'd' or 'D' (and not both), then you will have to use the one in 'i1new'.
To the OP: bottom line, use Jim's 'i1new' and my 'i2' or 'i2b'. Rui Barradas
On Mon, Jul 2, 2012 at 7:24 PM, Rui Barradas <ruipbarradas at sapo.pt> wrote:
Hello,
Try regular expressions instead.
In this data.frame, I've changed row nr.4 to have a row with 'D' as first
non-zero character.
dd <- read.table(text="
ch count
1 0000000000D0000000000000000000000000000000000000 0.007368
2 0000000000d0000000000000000000000000000000000000 0.002456
3 000000000T00000000000000000000000000000000000000 0.007368
4 000000000DT0000000000000000000000000000000000000 0.007368
5 000000000T00000000000000000000000000000000000000 0.002456
6 000000000Td0000000000000000000000000000000000000 0.002456
7 00000000T000000000000000000000000000000000000000 0.007368
8 00000000T0D0000000000000000000000000000000000000 0.007368
9 00000000T000000000000000000000000000000000000000 0.002456
10 00000000T0d0000000000000000000000000000000000000 0.002456
", header=TRUE)
dd
i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i2 <- grepl("^0*[Dd]", dd$ch)
dd[!i1, ]
dd[!i2, ]
dd[!(i1 | i2), ]
Hope this helps,
Rui Barradas
Em 02-07-2012 23:48, Claudia Penaloza escreveu:
I would like to remove rows from the following data frame (df) if there
are
only two specific elements found in the df$ch character string (I want to
remove rows with only "0" & "D" or "0" & "d"). Alternatively, I would like
to remove rows if the first non-zero element is "D" or "d".
ch count
1 0000000000D0000000000000000000000000000000000000 0.007368;
2 0000000000d0000000000000000000000000000000000000 0.002456;
3 000000000T00000000000000000000000000000000000000 0.007368;
4 000000000TD0000000000000000000000000000000000000 0.007368;
5 000000000T00000000000000000000000000000000000000 0.002456;
6 000000000Td0000000000000000000000000000000000000 0.002456;
7 00000000T000000000000000000000000000000000000000 0.007368;
8 00000000T0D0000000000000000000000000000000000000 0.007368;
9 00000000T000000000000000000000000000000000000000 0.002456;
10 00000000T0d0000000000000000000000000000000000000 0.002456;
I tried the following but it doesn't work if there is more than one
character per string:
df <- df[!df$ch %in% c("0","D"),]
df <- df[!df$ch %in% c("0","d"),]
Any help greatly appreciated,
Claudia
[[alternative HTML version deleted]]
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
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Hello, I'm glad it helped. See answer inline. Em 03-07-2012 17:09, Claudia Penaloza escreveu:
Thank you Rui and Jim, both 'i1' and 'i1new' worked perfectly
because there are no instances of 'Dd' or 'dD' in the data set (that I
would/not want to include/exclude)... but I understand that 'i1new'
targets precisely what I want.
Why isn't a leader of zero's required for either 'i1' or 'i1new', as so?
i1newer <- grepl("^0{0,}[D]*$|^0{0,}[d]*$", dd$ch)
Because both 'i1' and 'i1new' test from beginning to end of string, allowing only '0' and either 'd' or 'D', but not both (i1new). So, there's no need to explicitly test for a string that begins with '0'. Rui Barradas
Thank you again,
Claudia
On Tue, Jul 3, 2012 at 2:06 AM, Rui Barradas <ruipbarradas at sapo.pt
<mailto:ruipbarradas at sapo.pt>> wrote:
Hello,
Inline.
Em 03-07-2012 01:15, jim holtman escreveu:
You will have to change the 'i1' expression as follows:
i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i1 # matches strings with d & D in them
[1] TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# second string had 'd' & 'D' in it so it was TRUE above and
FALSE below
i1new <- grepl("^([0D]*$|[0d]*$)", dd$ch)
i1new
[1] TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
Right, apparently, I forgot that grep is greedy, and the test cases
were not complete.
I put a 'd' and 'D' in the second string and the original regular
expression is equivalent to
grepl("^[0dD]*$", dd$ch)
This is only for the first request, and does not solve cases where
there are characters other than '0', 'd' or 'D', but 'd' or 'D' are
the first non-zero. This is the case of my 4th row, changed from the
OP's data example.
My regexpr for 'i2' is equivalent to this one, that I believe is
more readable:
i2b <- grepl("^0{0,}[Dd]", dd$ch)
First a zero, that might occur zero or more times, then a 'd' or
'D', then and til the end, irrelevant.
which will match strings containing d, D and 0. If you only
want 'd'
or 'D' (and not both), then you will have to use the one in 'i1new'.
To the OP: bottom line, use Jim's 'i1new' and my 'i2' or 'i2b'.
Rui Barradas
On Mon, Jul 2, 2012 at 7:24 PM, Rui Barradas
<ruipbarradas at sapo.pt <mailto:ruipbarradas at sapo.pt>> wrote:
Hello,
Try regular expressions instead.
In this data.frame, I've changed row nr.4 to have a row with
'D' as first
non-zero character.
dd <- read.table(text="
ch count
1 0000000000D0000000000000000000__000000000000000000 0.007368
2 0000000000d0000000000000000000__000000000000000000 0.002456
3 000000000T00000000000000000000__000000000000000000 0.007368
4 000000000DT0000000000000000000__000000000000000000 0.007368
5 000000000T00000000000000000000__000000000000000000 0.002456
6 000000000Td0000000000000000000__000000000000000000 0.002456
7 00000000T000000000000000000000__000000000000000000 0.007368
8 00000000T0D0000000000000000000__000000000000000000 0.007368
9 00000000T000000000000000000000__000000000000000000 0.002456
10 00000000T0d0000000000000000000__000000000000000000 0.002456
", header=TRUE)
dd
i1 <- grepl("^([0D]|[0d])*$", dd$ch)
i2 <- grepl("^0*[Dd]", dd$ch)
dd[!i1, ]
dd[!i2, ]
dd[!(i1 | i2), ]
Hope this helps,
Rui Barradas
Em 02-07-2012 23:48, Claudia Penaloza escreveu:
I would like to remove rows from the following data
frame (df) if there
are
only two specific elements found in the df$ch character
string (I want to
remove rows with only "0" & "D" or "0" & "d").
Alternatively, I would like
to remove rows if the first non-zero element is "D" or "d".
ch
count
1 0000000000D0000000000000000000__000000000000000000
0.007368;
2 0000000000d0000000000000000000__000000000000000000
0.002456;
3 000000000T00000000000000000000__000000000000000000
0.007368;
4 000000000TD0000000000000000000__000000000000000000
0.007368;
5 000000000T00000000000000000000__000000000000000000
0.002456;
6 000000000Td0000000000000000000__000000000000000000
0.002456;
7 00000000T000000000000000000000__000000000000000000
0.007368;
8 00000000T0D0000000000000000000__000000000000000000
0.007368;
9 00000000T000000000000000000000__000000000000000000
0.002456;
10 00000000T0d0000000000000000000__000000000000000000
0.002456;
I tried the following but it doesn't work if there is
more than one
character per string:
df <- df[!df$ch %in% c("0","D"),]
df <- df[!df$ch %in% c("0","d"),]
Any help greatly appreciated,
Claudia
[[alternative HTML version deleted]]
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<http://www.R-project.org/posting-guide.html>
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1 day later
Perhaps I've missed something, but if it's really true that the goal is to
remove rows if the first non-zero element is "D" or "d", then how about
this:
tmp <- gsub('0','',df$ch)
first <- substr(tmp,1,1)
subset(df, tolower(first) != 'd')
and of course it could be rolled up into a single expression, but I wrote
it in several steps to make it easy to follow. No need to wrap one's brain
around regular expressions (which is hard for me!)
-Don
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 7/2/12 3:48 PM, "Claudia Penaloza" <claudiapenaloza at gmail.com> wrote:
>I would like to remove rows from the following data frame (df) if there
>are
>only two specific elements found in the df$ch character string (I want to
>remove rows with only "0" & "D" or "0" & "d"). Alternatively, I would like
>to remove rows if the first non-zero element is "D" or "d".
>
>
> ch count
>1 0000000000D0000000000000000000000000000000000000 0.007368;
>2 0000000000d0000000000000000000000000000000000000 0.002456;
>3 000000000T00000000000000000000000000000000000000 0.007368;
>4 000000000TD0000000000000000000000000000000000000 0.007368;
>5 000000000T00000000000000000000000000000000000000 0.002456;
>6 000000000Td0000000000000000000000000000000000000 0.002456;
>7 00000000T000000000000000000000000000000000000000 0.007368;
>8 00000000T0D0000000000000000000000000000000000000 0.007368;
>9 00000000T000000000000000000000000000000000000000 0.002456;
>10 00000000T0d0000000000000000000000000000000000000 0.002456;
>
>
>I tried the following but it doesn't work if there is more than one
>character per string:
>
>>df <- df[!df$ch %in% c("0","D"),]
>>df <- df[!df$ch %in% c("0","d"),]
>
>Any help greatly appreciated,
>Claudia
>
> [[alternative HTML version deleted]]
>
>______________________________________________
>R-help at 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.