Identifying if one field is partial derived from another in R
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So given a pattern, say two letters, and the position of the pattern, is it a prefix, suffix or in the middle, I need to identify if a field is partial derived from another. So for example given the following dataset
data.V1 data.V2
1 GH GH1001
2 FD FD2002
3 TH 2345TH
4 ED ED56763
5 US 4345US
6 FG F6736tG
if LL is the pattern for column one where LL refers to two letters in this case. If the pattern for column 2 is LL#. This indicates that the position of the pattern is the first elements of each element in row 2. So in the dataset above rows 1,2&4 would obey the pattern.
I have tried if then statements but these did not work if the pattern was in the middle , #LL#. I have also tried the function regmathes but that did not work either.
r
asked Nov 23 '18 at 14:11
Sarah Sarah
65
add a comment |
So given a pattern, say two letters, and the position of the pattern, is it a prefix, suffix or in the middle, I need to identify if a field is partial derived from another. So for example given the following dataset
data.V1 data.V2
1 GH GH1001
2 FD FD2002
3 TH 2345TH
4 ED ED56763
5 US 4345US
6 FG F6736tG
if LL is the pattern for column one where LL refers to two letters in this case. If the pattern for column 2 is LL#. This indicates that the position of the pattern is the first elements of each element in row 2. So in the dataset above rows 1,2&4 would obey the pattern.
I have tried if then statements but these did not work if the pattern was in the middle , #LL#. I have also tried the function regmathes but that did not work either.
r
asked Nov 23 '18 at 14:11
Sarah Sarah
65
Useful link for trying out and better understanding regular expressions
– Nutle
Nov 23 '18 at 14:15
For example, in the link above, try[A-Z]{2},^[A-Z]{2}and[A-Z]{2}$for various entries.
– Nutle
Nov 23 '18 at 14:21
@Nutle thanks. Is it possible in regular expressions to specify where to start? For example if I wanted to check were the 3rd and 4th elements of an entry both letters.
– Sarah
Nov 23 '18 at 15:08
^for beginning,$for the end,.as a filler. So for the exactly third and 4th, try..[A-Z]{2}, note the two dots. Finally, questions with regex are very easily googleable, since they're a common standard everywhere
– Nutle
Nov 23 '18 at 15:13
add a comment |
So given a pattern, say two letters, and the position of the pattern, is it a prefix, suffix or in the middle, I need to identify if a field is partial derived from another. So for example given the following dataset
data.V1 data.V2
1 GH GH1001
2 FD FD2002
3 TH 2345TH
4 ED ED56763
5 US 4345US
6 FG F6736tG
if LL is the pattern for column one where LL refers to two letters in this case. If the pattern for column 2 is LL#. This indicates that the position of the pattern is the first elements of each element in row 2. So in the dataset above rows 1,2&4 would obey the pattern.
I have tried if then statements but these did not work if the pattern was in the middle , #LL#. I have also tried the function regmathes but that did not work either.
r
asked Nov 23 '18 at 14:11
Sarah Sarah
65
So given a pattern, say two letters, and the position of the pattern, is it a prefix, suffix or in the middle, I need to identify if a field is partial derived from another. So for example given the following dataset
data.V1 data.V2
1 GH GH1001
2 FD FD2002
3 TH 2345TH
4 ED ED56763
5 US 4345US
6 FG F6736tG
if LL is the pattern for column one where LL refers to two letters in this case. If the pattern for column 2 is LL#. This indicates that the position of the pattern is the first elements of each element in row 2. So in the dataset above rows 1,2&4 would obey the pattern.
I have tried if then statements but these did not work if the pattern was in the middle , #LL#. I have also tried the function regmathes but that did not work either.
r
r
asked Nov 23 '18 at 14:11
Sarah Sarah
65
asked Nov 23 '18 at 14:11
Sarah Sarah
65
asked Nov 23 '18 at 14:11
Sarah Sarah
65
asked Nov 23 '18 at 14:11
Sarah Sarah
65
asked Nov 23 '18 at 14:11
Sarah Sarah
65
65
Useful link for trying out and better understanding regular expressions
– Nutle
Nov 23 '18 at 14:15
For example, in the link above, try[A-Z]{2},^[A-Z]{2}and[A-Z]{2}$for various entries.
– Nutle
Nov 23 '18 at 14:21
@Nutle thanks. Is it possible in regular expressions to specify where to start? For example if I wanted to check were the 3rd and 4th elements of an entry both letters.
– Sarah
Nov 23 '18 at 15:08
^for beginning,$for the end,.as a filler. So for the exactly third and 4th, try..[A-Z]{2}, note the two dots. Finally, questions with regex are very easily googleable, since they're a common standard everywhere
– Nutle
Nov 23 '18 at 15:13
add a comment |
Useful link for trying out and better understanding regular expressions
– Nutle
Nov 23 '18 at 14:15
For example, in the link above, try[A-Z]{2},^[A-Z]{2}and[A-Z]{2}$for various entries.
– Nutle
Nov 23 '18 at 14:21
@Nutle thanks. Is it possible in regular expressions to specify where to start? For example if I wanted to check were the 3rd and 4th elements of an entry both letters.
– Sarah
Nov 23 '18 at 15:08
^for beginning,$for the end,.as a filler. So for the exactly third and 4th, try..[A-Z]{2}, note the two dots. Finally, questions with regex are very easily googleable, since they're a common standard everywhere
– Nutle
Nov 23 '18 at 15:13
Useful link for trying out and better understanding regular expressions
– Nutle
Nov 23 '18 at 14:15
Useful link for trying out and better understanding regular expressions
– Nutle
Nov 23 '18 at 14:15
For example, in the link above, try
[A-Z]{2}, ^[A-Z]{2} and [A-Z]{2}$ for various entries.– Nutle
Nov 23 '18 at 14:21
For example, in the link above, try
[A-Z]{2}, ^[A-Z]{2} and [A-Z]{2}$ for various entries.– Nutle
Nov 23 '18 at 14:21
@Nutle thanks. Is it possible in regular expressions to specify where to start? For example if I wanted to check were the 3rd and 4th elements of an entry both letters.
– Sarah
Nov 23 '18 at 15:08
@Nutle thanks. Is it possible in regular expressions to specify where to start? For example if I wanted to check were the 3rd and 4th elements of an entry both letters.
– Sarah
Nov 23 '18 at 15:08
^ for beginning, $ for the end, . as a filler. So for the exactly third and 4th, try ..[A-Z]{2}, note the two dots. Finally, questions with regex are very easily googleable, since they're a common standard everywhere– Nutle
Nov 23 '18 at 15:13
^ for beginning, $ for the end, . as a filler. So for the exactly third and 4th, try ..[A-Z]{2}, note the two dots. Finally, questions with regex are very easily googleable, since they're a common standard everywhere– Nutle
Nov 23 '18 at 15:13
add a comment |
1 Answer
1
active
oldest
votes
apply(df,1,function(x) grepl(paste0("^",x["data.V1"]),x["data.V2"]))
For every row (that's the 1 in apply), this will check whether the contents of data.V1 appears right at the beginning of data.V2 (^ means the beginning of a string for regular expressions).
Result:
1 2 3 4 5 6
TRUE TRUE FALSE TRUE FALSE FALSE
Replace the grepl first argument with the following for:
End of string: paste0(x["data.V1"],"$")
Middle of string: paste0(".+",x["data.V1"],".+")
After n characters (n defined elsewhere): paste0(".{",n,"}",x["data.V1"])
(for the last one, the form "{n}" means the last character is repeated n times. Since it is preceded by ".", it means any n characters.)
answered Nov 23 '18 at 14:16
iodiod
4,3562724
Thanks.That worked great if the pattern was at the beginning. Would it be possible to do something similar if the pattern was in the middle or end of a entry? For example row 5 in the dataset above.
– Sarah
Nov 23 '18 at 14:39
see the edited answer above: just replace the first argument of the grepl with either option.
– iod
Nov 23 '18 at 16:12
@Sarah don't forget to upvote and accept!
– iod
Nov 23 '18 at 16:29
@ iod. Just wondering is it possible to generalize the statement. For example I tried apply(data,1,function(x) grepl(paste0('rep(.,n)+',x["V1"],'...+'),x["V3"])). So that I can specify the position outside the function.
– Sarah
Nov 23 '18 at 17:09
Not sure what you're trying to achieve with that rep() call. As noted in the comments to your question - it's a good idea to familiarize yourself with regex syntax.
– iod
Nov 23 '18 at 17:19
|
show 3 more comments
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1 Answer
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apply(df,1,function(x) grepl(paste0("^",x["data.V1"]),x["data.V2"]))
For every row (that's the 1 in apply), this will check whether the contents of data.V1 appears right at the beginning of data.V2 (^ means the beginning of a string for regular expressions).
Result:
1 2 3 4 5 6
TRUE TRUE FALSE TRUE FALSE FALSE
Replace the grepl first argument with the following for:
End of string: paste0(x["data.V1"],"$")
Middle of string: paste0(".+",x["data.V1"],".+")
After n characters (n defined elsewhere): paste0(".{",n,"}",x["data.V1"])
(for the last one, the form "{n}" means the last character is repeated n times. Since it is preceded by ".", it means any n characters.)
answered Nov 23 '18 at 14:16
iodiod
4,3562724
Thanks.That worked great if the pattern was at the beginning. Would it be possible to do something similar if the pattern was in the middle or end of a entry? For example row 5 in the dataset above.
– Sarah
Nov 23 '18 at 14:39
see the edited answer above: just replace the first argument of the grepl with either option.
– iod
Nov 23 '18 at 16:12
@Sarah don't forget to upvote and accept!
– iod
Nov 23 '18 at 16:29
@ iod. Just wondering is it possible to generalize the statement. For example I tried apply(data,1,function(x) grepl(paste0('rep(.,n)+',x["V1"],'...+'),x["V3"])). So that I can specify the position outside the function.
– Sarah
Nov 23 '18 at 17:09
Not sure what you're trying to achieve with that rep() call. As noted in the comments to your question - it's a good idea to familiarize yourself with regex syntax.
– iod
Nov 23 '18 at 17:19
|
show 3 more comments
apply(df,1,function(x) grepl(paste0("^",x["data.V1"]),x["data.V2"]))
For every row (that's the 1 in apply), this will check whether the contents of data.V1 appears right at the beginning of data.V2 (^ means the beginning of a string for regular expressions).
Result:
1 2 3 4 5 6
TRUE TRUE FALSE TRUE FALSE FALSE
Replace the grepl first argument with the following for:
End of string: paste0(x["data.V1"],"$")
Middle of string: paste0(".+",x["data.V1"],".+")
After n characters (n defined elsewhere): paste0(".{",n,"}",x["data.V1"])
(for the last one, the form "{n}" means the last character is repeated n times. Since it is preceded by ".", it means any n characters.)
answered Nov 23 '18 at 14:16
iodiod
4,3562724
Thanks.That worked great if the pattern was at the beginning. Would it be possible to do something similar if the pattern was in the middle or end of a entry? For example row 5 in the dataset above.
– Sarah
Nov 23 '18 at 14:39
see the edited answer above: just replace the first argument of the grepl with either option.
– iod
Nov 23 '18 at 16:12
@Sarah don't forget to upvote and accept!
– iod
Nov 23 '18 at 16:29
@ iod. Just wondering is it possible to generalize the statement. For example I tried apply(data,1,function(x) grepl(paste0('rep(.,n)+',x["V1"],'...+'),x["V3"])). So that I can specify the position outside the function.
– Sarah
Nov 23 '18 at 17:09
Not sure what you're trying to achieve with that rep() call. As noted in the comments to your question - it's a good idea to familiarize yourself with regex syntax.
– iod
Nov 23 '18 at 17:19
|
show 3 more comments
apply(df,1,function(x) grepl(paste0("^",x["data.V1"]),x["data.V2"]))
For every row (that's the 1 in apply), this will check whether the contents of data.V1 appears right at the beginning of data.V2 (^ means the beginning of a string for regular expressions).
Result:
1 2 3 4 5 6
TRUE TRUE FALSE TRUE FALSE FALSE
Replace the grepl first argument with the following for:
End of string: paste0(x["data.V1"],"$")
Middle of string: paste0(".+",x["data.V1"],".+")
After n characters (n defined elsewhere): paste0(".{",n,"}",x["data.V1"])
(for the last one, the form "{n}" means the last character is repeated n times. Since it is preceded by ".", it means any n characters.)
answered Nov 23 '18 at 14:16
iodiod
4,3562724
apply(df,1,function(x) grepl(paste0("^",x["data.V1"]),x["data.V2"]))
For every row (that's the 1 in apply), this will check whether the contents of data.V1 appears right at the beginning of data.V2 (^ means the beginning of a string for regular expressions).
Result:
1 2 3 4 5 6
TRUE TRUE FALSE TRUE FALSE FALSE
Replace the grepl first argument with the following for:
End of string: paste0(x["data.V1"],"$")
Middle of string: paste0(".+",x["data.V1"],".+")
After n characters (n defined elsewhere): paste0(".{",n,"}",x["data.V1"])
(for the last one, the form "{n}" means the last character is repeated n times. Since it is preceded by ".", it means any n characters.)
answered Nov 23 '18 at 14:16
iodiod
4,3562724
edited Nov 23 '18 at 17:41
answered Nov 23 '18 at 14:16
iodiod
4,3562724
answered Nov 23 '18 at 14:16
iodiod
4,3562724
answered Nov 23 '18 at 14:16
iodiod
4,3562724
4,3562724
Thanks.That worked great if the pattern was at the beginning. Would it be possible to do something similar if the pattern was in the middle or end of a entry? For example row 5 in the dataset above.
– Sarah
Nov 23 '18 at 14:39
see the edited answer above: just replace the first argument of the grepl with either option.
– iod
Nov 23 '18 at 16:12
@Sarah don't forget to upvote and accept!
– iod
Nov 23 '18 at 16:29
@ iod. Just wondering is it possible to generalize the statement. For example I tried apply(data,1,function(x) grepl(paste0('rep(.,n)+',x["V1"],'...+'),x["V3"])). So that I can specify the position outside the function.
– Sarah
Nov 23 '18 at 17:09
Not sure what you're trying to achieve with that rep() call. As noted in the comments to your question - it's a good idea to familiarize yourself with regex syntax.
– iod
Nov 23 '18 at 17:19
|
show 3 more comments
Thanks.That worked great if the pattern was at the beginning. Would it be possible to do something similar if the pattern was in the middle or end of a entry? For example row 5 in the dataset above.
– Sarah
Nov 23 '18 at 14:39
see the edited answer above: just replace the first argument of the grepl with either option.
– iod
Nov 23 '18 at 16:12
@Sarah don't forget to upvote and accept!
– iod
Nov 23 '18 at 16:29
@ iod. Just wondering is it possible to generalize the statement. For example I tried apply(data,1,function(x) grepl(paste0('rep(.,n)+',x["V1"],'...+'),x["V3"])). So that I can specify the position outside the function.
– Sarah
Nov 23 '18 at 17:09
Not sure what you're trying to achieve with that rep() call. As noted in the comments to your question - it's a good idea to familiarize yourself with regex syntax.
– iod
Nov 23 '18 at 17:19
Thanks.That worked great if the pattern was at the beginning. Would it be possible to do something similar if the pattern was in the middle or end of a entry? For example row 5 in the dataset above.
– Sarah
Nov 23 '18 at 14:39
Thanks.That worked great if the pattern was at the beginning. Would it be possible to do something similar if the pattern was in the middle or end of a entry? For example row 5 in the dataset above.
– Sarah
Nov 23 '18 at 14:39
see the edited answer above: just replace the first argument of the grepl with either option.
– iod
Nov 23 '18 at 16:12
see the edited answer above: just replace the first argument of the grepl with either option.
– iod
Nov 23 '18 at 16:12
@Sarah don't forget to upvote and accept!
– iod
Nov 23 '18 at 16:29
@Sarah don't forget to upvote and accept!
– iod
Nov 23 '18 at 16:29
@ iod. Just wondering is it possible to generalize the statement. For example I tried apply(data,1,function(x) grepl(paste0('rep(.,n)+',x["V1"],'...+'),x["V3"])). So that I can specify the position outside the function.
– Sarah
Nov 23 '18 at 17:09
@ iod. Just wondering is it possible to generalize the statement. For example I tried apply(data,1,function(x) grepl(paste0('rep(.,n)+',x["V1"],'...+'),x["V3"])). So that I can specify the position outside the function.
– Sarah
Nov 23 '18 at 17:09
Not sure what you're trying to achieve with that rep() call. As noted in the comments to your question - it's a good idea to familiarize yourself with regex syntax.
– iod
Nov 23 '18 at 17:19
Not sure what you're trying to achieve with that rep() call. As noted in the comments to your question - it's a good idea to familiarize yourself with regex syntax.
– iod
Nov 23 '18 at 17:19
|
show 3 more comments
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Useful link for trying out and better understanding regular expressions
– Nutle
Nov 23 '18 at 14:15
For example, in the link above, try
[A-Z]{2},^[A-Z]{2}and[A-Z]{2}$for various entries.– Nutle
Nov 23 '18 at 14:21
@Nutle thanks. Is it possible in regular expressions to specify where to start? For example if I wanted to check were the 3rd and 4th elements of an entry both letters.
– Sarah
Nov 23 '18 at 15:08
^for beginning,$for the end,.as a filler. So for the exactly third and 4th, try..[A-Z]{2}, note the two dots. Finally, questions with regex are very easily googleable, since they're a common standard everywhere– Nutle
Nov 23 '18 at 15:13