Mixing
split character at deliminator conditionally negative lookahead assertion
I want to split a string at . or : unless the next character is )
Following this question: R strsplit: Split based on character except when a specific character follows why isn't
strsplit("Glenelg (Vic.)",'\.|:(?!\))', perl = TRUE)
returning
[[1]]
[1] "Glenelg (Vic)"
instead it splits at the ., like so:
[1] "Glenelg (Vic" ")"
r regex pcre
asked Nov 19 '18 at 23:15
R.M.R.M.
1,0451924
add a comment |
I want to split a string at . or : unless the next character is )
Following this question: R strsplit: Split based on character except when a specific character follows why isn't
strsplit("Glenelg (Vic.)",'\.|:(?!\))', perl = TRUE)
returning
[[1]]
[1] "Glenelg (Vic)"
instead it splits at the ., like so:
[1] "Glenelg (Vic" ")"
r regex pcre
asked Nov 19 '18 at 23:15
R.M.R.M.
1,0451924
add a comment |
I want to split a string at . or : unless the next character is )
Following this question: R strsplit: Split based on character except when a specific character follows why isn't
strsplit("Glenelg (Vic.)",'\.|:(?!\))', perl = TRUE)
returning
[[1]]
[1] "Glenelg (Vic)"
instead it splits at the ., like so:
[1] "Glenelg (Vic" ")"
r regex pcre
asked Nov 19 '18 at 23:15
R.M.R.M.
1,0451924
I want to split a string at . or : unless the next character is )
Following this question: R strsplit: Split based on character except when a specific character follows why isn't
strsplit("Glenelg (Vic.)",'\.|:(?!\))', perl = TRUE)
returning
[[1]]
[1] "Glenelg (Vic)"
instead it splits at the ., like so:
[1] "Glenelg (Vic" ")"
r regex pcre
r regex pcre
asked Nov 19 '18 at 23:15
R.M.R.M.
1,0451924
asked Nov 19 '18 at 23:15
R.M.R.M.
1,0451924
asked Nov 19 '18 at 23:15
R.M.R.M.
1,0451924
asked Nov 19 '18 at 23:15
R.M.R.M.
1,0451924
asked Nov 19 '18 at 23:15
R.M.R.M.
1,0451924
1,0451924
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
It is not grouped correctly. .|:(?!)) matches a . anywhere in a string or a : not followed with ). If you group . and : patterns, '(?:\.|:)(?!\))', it will work.
However, you may use a better regex version based on a character class:
strsplit("Glenelg (Vic.)",'[.:](?!\))', perl = TRUE)
[[1]]
[1] "Glenelg (Vic.)"
Here, [.:](?!)) matches either . or : that are both not immediately followed with ).
See the regex demo.
answered Nov 19 '18 at 23:17
Wiktor StribiżewWiktor Stribiżew
310k16131206
add a comment |
You can also use stringr:
stringr::str_split("Glenelg (Vic.)","[\.:](?!\))")
[[1]]
[1] "Glenelg (Vic.)"
answered Nov 20 '18 at 0:00
JoséJosé
501715
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is not grouped correctly. .|:(?!)) matches a . anywhere in a string or a : not followed with ). If you group . and : patterns, '(?:\.|:)(?!\))', it will work.
However, you may use a better regex version based on a character class:
strsplit("Glenelg (Vic.)",'[.:](?!\))', perl = TRUE)
[[1]]
[1] "Glenelg (Vic.)"
Here, [.:](?!)) matches either . or : that are both not immediately followed with ).
See the regex demo.
answered Nov 19 '18 at 23:17
Wiktor StribiżewWiktor Stribiżew
310k16131206
add a comment |
It is not grouped correctly. .|:(?!)) matches a . anywhere in a string or a : not followed with ). If you group . and : patterns, '(?:\.|:)(?!\))', it will work.
However, you may use a better regex version based on a character class:
strsplit("Glenelg (Vic.)",'[.:](?!\))', perl = TRUE)
[[1]]
[1] "Glenelg (Vic.)"
Here, [.:](?!)) matches either . or : that are both not immediately followed with ).
See the regex demo.
answered Nov 19 '18 at 23:17
Wiktor StribiżewWiktor Stribiżew
310k16131206
add a comment |
It is not grouped correctly. .|:(?!)) matches a . anywhere in a string or a : not followed with ). If you group . and : patterns, '(?:\.|:)(?!\))', it will work.
However, you may use a better regex version based on a character class:
strsplit("Glenelg (Vic.)",'[.:](?!\))', perl = TRUE)
[[1]]
[1] "Glenelg (Vic.)"
Here, [.:](?!)) matches either . or : that are both not immediately followed with ).
See the regex demo.
answered Nov 19 '18 at 23:17
Wiktor StribiżewWiktor Stribiżew
310k16131206
It is not grouped correctly. .|:(?!)) matches a . anywhere in a string or a : not followed with ). If you group . and : patterns, '(?:\.|:)(?!\))', it will work.
However, you may use a better regex version based on a character class:
strsplit("Glenelg (Vic.)",'[.:](?!\))', perl = TRUE)
[[1]]
[1] "Glenelg (Vic.)"
Here, [.:](?!)) matches either . or : that are both not immediately followed with ).
See the regex demo.
answered Nov 19 '18 at 23:17
Wiktor StribiżewWiktor Stribiżew
310k16131206
answered Nov 19 '18 at 23:17
Wiktor StribiżewWiktor Stribiżew
310k16131206
answered Nov 19 '18 at 23:17
Wiktor StribiżewWiktor Stribiżew
310k16131206
answered Nov 19 '18 at 23:17
Wiktor StribiżewWiktor Stribiżew
310k16131206
310k16131206
add a comment |
add a comment |
You can also use stringr:
stringr::str_split("Glenelg (Vic.)","[\.:](?!\))")
[[1]]
[1] "Glenelg (Vic.)"
answered Nov 20 '18 at 0:00
JoséJosé
501715
add a comment |
You can also use stringr:
stringr::str_split("Glenelg (Vic.)","[\.:](?!\))")
[[1]]
[1] "Glenelg (Vic.)"
answered Nov 20 '18 at 0:00
JoséJosé
501715
add a comment |
You can also use stringr:
stringr::str_split("Glenelg (Vic.)","[\.:](?!\))")
[[1]]
[1] "Glenelg (Vic.)"
answered Nov 20 '18 at 0:00
JoséJosé
501715
You can also use stringr:
stringr::str_split("Glenelg (Vic.)","[\.:](?!\))")
[[1]]
[1] "Glenelg (Vic.)"
answered Nov 20 '18 at 0:00
JoséJosé
501715
answered Nov 20 '18 at 0:00
JoséJosé
501715
answered Nov 20 '18 at 0:00
JoséJosé
501715
answered Nov 20 '18 at 0:00
JoséJosé
501715
501715
add a comment |
add a comment |
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