Get the characters after a certain pattern in R - regex
1
I have a data frame with one column:
df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
df$cat
How can I extract the characters that appear after c(", sometimes there is only one backslash and sometimes there's 2. Similarly with the characters, sometimes the characters are 2 and sometimes they are 3. e.g. BP2, BP etc.
So far I have tried:
substr(x = df$cat, start = 4, stop = 6)
But this results in:
""BP" "BP2" "BPT" "CN""
And I only want the output to show
"BPT" "BP2" "BPT" "CN"
regex string rstudio text-extraction
asked Nov 13 '18 at 13:47
SyedSyed
8310
|
show 3 more comments
1
I have a data frame with one column:
df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
df$cat
How can I extract the characters that appear after c(", sometimes there is only one backslash and sometimes there's 2. Similarly with the characters, sometimes the characters are 2 and sometimes they are 3. e.g. BP2, BP etc.
So far I have tried:
substr(x = df$cat, start = 4, stop = 6)
But this results in:
""BP" "BP2" "BPT" "CN""
And I only want the output to show
"BPT" "BP2" "BPT" "CN"
regex string rstudio text-extraction
asked Nov 13 '18 at 13:47
SyedSyed
8310
2
Trylapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))
– Wiktor Stribiżew
Nov 13 '18 at 13:55
Orgsub('^c\(|\)$|\\(")', '\1', df$cat)
– Wiktor Stribiżew
Nov 13 '18 at 14:00
Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one
– Syed
Nov 13 '18 at 14:08
1
Maybelapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Orunlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP
– Wiktor Stribiżew
Nov 13 '18 at 14:13
Just let know if the firstBPin your expected output is a typo or not, I understand you need"BPT" "BP2" "BPT" "CN"and not"BP" "BP2" "BPT" "CN"
– Wiktor Stribiżew
Nov 13 '18 at 14:21
|
show 3 more comments
1
1
1
I have a data frame with one column:
df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
df$cat
How can I extract the characters that appear after c(", sometimes there is only one backslash and sometimes there's 2. Similarly with the characters, sometimes the characters are 2 and sometimes they are 3. e.g. BP2, BP etc.
So far I have tried:
substr(x = df$cat, start = 4, stop = 6)
But this results in:
""BP" "BP2" "BPT" "CN""
And I only want the output to show
"BPT" "BP2" "BPT" "CN"
regex string rstudio text-extraction
asked Nov 13 '18 at 13:47
SyedSyed
8310
I have a data frame with one column:
df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
df$cat
How can I extract the characters that appear after c(", sometimes there is only one backslash and sometimes there's 2. Similarly with the characters, sometimes the characters are 2 and sometimes they are 3. e.g. BP2, BP etc.
So far I have tried:
substr(x = df$cat, start = 4, stop = 6)
But this results in:
""BP" "BP2" "BPT" "CN""
And I only want the output to show
"BPT" "BP2" "BPT" "CN"
regex string rstudio text-extraction
regex string rstudio text-extraction
asked Nov 13 '18 at 13:47
SyedSyed
8310
asked Nov 13 '18 at 13:47
SyedSyed
8310
edited Nov 13 '18 at 15:35
Syed
asked Nov 13 '18 at 13:47
SyedSyed
8310
asked Nov 13 '18 at 13:47
SyedSyed
8310
asked Nov 13 '18 at 13:47
SyedSyed
8310
8310
2
Trylapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))
– Wiktor Stribiżew
Nov 13 '18 at 13:55
Orgsub('^c\(|\)$|\\(")', '\1', df$cat)
– Wiktor Stribiżew
Nov 13 '18 at 14:00
Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one
– Syed
Nov 13 '18 at 14:08
1
Maybelapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Orunlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP
– Wiktor Stribiżew
Nov 13 '18 at 14:13
Just let know if the firstBPin your expected output is a typo or not, I understand you need"BPT" "BP2" "BPT" "CN"and not"BP" "BP2" "BPT" "CN"
– Wiktor Stribiżew
Nov 13 '18 at 14:21
|
show 3 more comments
2
Trylapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))
– Wiktor Stribiżew
Nov 13 '18 at 13:55
Orgsub('^c\(|\)$|\\(")', '\1', df$cat)
– Wiktor Stribiżew
Nov 13 '18 at 14:00
Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one
– Syed
Nov 13 '18 at 14:08
1
Maybelapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Orunlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP
– Wiktor Stribiżew
Nov 13 '18 at 14:13
Just let know if the firstBPin your expected output is a typo or not, I understand you need"BPT" "BP2" "BPT" "CN"and not"BP" "BP2" "BPT" "CN"
– Wiktor Stribiżew
Nov 13 '18 at 14:21
2
2
Try
lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))– Wiktor Stribiżew
Nov 13 '18 at 13:55
Try
lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))– Wiktor Stribiżew
Nov 13 '18 at 13:55
Or
gsub('^c\(|\)$|\\(")', '\1', df$cat)– Wiktor Stribiżew
Nov 13 '18 at 14:00
Or
gsub('^c\(|\)$|\\(")', '\1', df$cat)– Wiktor Stribiżew
Nov 13 '18 at 14:00
Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one
– Syed
Nov 13 '18 at 14:08
Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one
– Syed
Nov 13 '18 at 14:08
1
1
Maybe
lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Or unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP– Wiktor Stribiżew
Nov 13 '18 at 14:13
Maybe
lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Or unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP– Wiktor Stribiżew
Nov 13 '18 at 14:13
Just let know if the first
BP in your expected output is a typo or not, I understand you need "BPT" "BP2" "BPT" "CN" and not "BP" "BP2" "BPT" "CN"– Wiktor Stribiżew
Nov 13 '18 at 14:21
Just let know if the first
BP in your expected output is a typo or not, I understand you need "BPT" "BP2" "BPT" "CN" and not "BP" "BP2" "BPT" "CN"– Wiktor Stribiżew
Nov 13 '18 at 14:21
|
show 3 more comments
1 Answer
1
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oldest
votes
1
You may use
df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))
## => [1] "BPT" "BP2" "BPT" "CN"
See the R demo online.
Notes
gsub('\', '', df$cat, fixed=TRUE)removes all backslashes. You may usegsub('\"', '"', df$cat, fixed=TRUE)if you only plan to remove backslashes before".eval(parse(text=x))[[1]]parses the vector and returns the first itemlapplyhelps traverse the whole data you have. See Using sapply and lapply.
answered Nov 13 '18 at 15:22
Wiktor StribiżewWiktor Stribiżew
316k16134215
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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1
You may use
df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))
## => [1] "BPT" "BP2" "BPT" "CN"
See the R demo online.
Notes
gsub('\', '', df$cat, fixed=TRUE)removes all backslashes. You may usegsub('\"', '"', df$cat, fixed=TRUE)if you only plan to remove backslashes before".eval(parse(text=x))[[1]]parses the vector and returns the first itemlapplyhelps traverse the whole data you have. See Using sapply and lapply.
answered Nov 13 '18 at 15:22
Wiktor StribiżewWiktor Stribiżew
316k16134215
add a comment |
1
You may use
df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))
## => [1] "BPT" "BP2" "BPT" "CN"
See the R demo online.
Notes
gsub('\', '', df$cat, fixed=TRUE)removes all backslashes. You may usegsub('\"', '"', df$cat, fixed=TRUE)if you only plan to remove backslashes before".eval(parse(text=x))[[1]]parses the vector and returns the first itemlapplyhelps traverse the whole data you have. See Using sapply and lapply.
answered Nov 13 '18 at 15:22
Wiktor StribiżewWiktor Stribiżew
316k16134215
add a comment |
1
1
1
You may use
df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))
## => [1] "BPT" "BP2" "BPT" "CN"
See the R demo online.
Notes
gsub('\', '', df$cat, fixed=TRUE)removes all backslashes. You may usegsub('\"', '"', df$cat, fixed=TRUE)if you only plan to remove backslashes before".eval(parse(text=x))[[1]]parses the vector and returns the first itemlapplyhelps traverse the whole data you have. See Using sapply and lapply.
answered Nov 13 '18 at 15:22
Wiktor StribiżewWiktor Stribiżew
316k16134215
You may use
df <- data.frame(cat = c("c(\"BPT\", "BP")", "c("BP2", "BP")", "c("BPT", "BP")", "c("CN", "NC")"))
df$cat <- as.character(df$cat)
unlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))
## => [1] "BPT" "BP2" "BPT" "CN"
See the R demo online.
Notes
gsub('\', '', df$cat, fixed=TRUE)removes all backslashes. You may usegsub('\"', '"', df$cat, fixed=TRUE)if you only plan to remove backslashes before".eval(parse(text=x))[[1]]parses the vector and returns the first itemlapplyhelps traverse the whole data you have. See Using sapply and lapply.
answered Nov 13 '18 at 15:22
Wiktor StribiżewWiktor Stribiżew
316k16134215
answered Nov 13 '18 at 15:22
Wiktor StribiżewWiktor Stribiżew
316k16134215
answered Nov 13 '18 at 15:22
Wiktor StribiżewWiktor Stribiżew
316k16134215
answered Nov 13 '18 at 15:22
Wiktor StribiżewWiktor Stribiżew
316k16134215
316k16134215
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2
Try
lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x)))– Wiktor Stribiżew
Nov 13 '18 at 13:55
Or
gsub('^c\(|\)$|\\(")', '\1', df$cat)– Wiktor Stribiżew
Nov 13 '18 at 14:00
Thanks @WiktorStribiżew. I tried both, the first one outputs both the texts - BPT and BP. The 2nd piece is also outputting both and I only need one - the first one
– Syed
Nov 13 '18 at 14:08
1
Maybe
lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]])? Orunlist(lapply(gsub('\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))? See ideone.com/xImbjP– Wiktor Stribiżew
Nov 13 '18 at 14:13
Just let know if the first
BPin your expected output is a typo or not, I understand you need"BPT" "BP2" "BPT" "CN"and not"BP" "BP2" "BPT" "CN"– Wiktor Stribiżew
Nov 13 '18 at 14:21