Python regex remove everything except strings from list
I have string:
bdv. mot. g. vns. kilm.
And knowing list of strings like
important_strings_lst=['bdv.', 'dktv.', 'mot. g.', 'vyr. g.']
I want to get regex selection like:
bdv. mot. g.
I joined list and tried: idea from here
regex = re.compile(r'b(?!bdv.|dktv.|mot. g.|vyr. g.)w+', re.UNICODE)
regex.sub("", 'bdv. mot. g. vns. kilm.')
Got
'bdv. mot. . . .'
Changing places in regex with s also didn't work out. How to do it?
I could use something like [x for x in important_strings_lst if x in my_string] but I need good performance as this will be used with million rows of pandas dataframe with str.replace
python regex list replace
edited Nov 10 at 17:38
Sandeep Kadapa
5,667427
asked Nov 10 at 17:37
Lukas
355
add a comment |
I have string:
bdv. mot. g. vns. kilm.
And knowing list of strings like
important_strings_lst=['bdv.', 'dktv.', 'mot. g.', 'vyr. g.']
I want to get regex selection like:
bdv. mot. g.
I joined list and tried: idea from here
regex = re.compile(r'b(?!bdv.|dktv.|mot. g.|vyr. g.)w+', re.UNICODE)
regex.sub("", 'bdv. mot. g. vns. kilm.')
Got
'bdv. mot. . . .'
Changing places in regex with s also didn't work out. How to do it?
I could use something like [x for x in important_strings_lst if x in my_string] but I need good performance as this will be used with million rows of pandas dataframe with str.replace
python regex list replace
edited Nov 10 at 17:38
Sandeep Kadapa
5,667427
asked Nov 10 at 17:37
Lukas
355
add a comment |
I have string:
bdv. mot. g. vns. kilm.
And knowing list of strings like
important_strings_lst=['bdv.', 'dktv.', 'mot. g.', 'vyr. g.']
I want to get regex selection like:
bdv. mot. g.
I joined list and tried: idea from here
regex = re.compile(r'b(?!bdv.|dktv.|mot. g.|vyr. g.)w+', re.UNICODE)
regex.sub("", 'bdv. mot. g. vns. kilm.')
Got
'bdv. mot. . . .'
Changing places in regex with s also didn't work out. How to do it?
I could use something like [x for x in important_strings_lst if x in my_string] but I need good performance as this will be used with million rows of pandas dataframe with str.replace
python regex list replace
edited Nov 10 at 17:38
Sandeep Kadapa
5,667427
asked Nov 10 at 17:37
Lukas
355
I have string:
bdv. mot. g. vns. kilm.
And knowing list of strings like
important_strings_lst=['bdv.', 'dktv.', 'mot. g.', 'vyr. g.']
I want to get regex selection like:
bdv. mot. g.
I joined list and tried: idea from here
regex = re.compile(r'b(?!bdv.|dktv.|mot. g.|vyr. g.)w+', re.UNICODE)
regex.sub("", 'bdv. mot. g. vns. kilm.')
Got
'bdv. mot. . . .'
Changing places in regex with s also didn't work out. How to do it?
I could use something like [x for x in important_strings_lst if x in my_string] but I need good performance as this will be used with million rows of pandas dataframe with str.replace
python regex list replace
python regex list replace
edited Nov 10 at 17:38
Sandeep Kadapa
5,667427
asked Nov 10 at 17:37
Lukas
355
edited Nov 10 at 17:38
Sandeep Kadapa
5,667427
asked Nov 10 at 17:37
Lukas
355
edited Nov 10 at 17:38
Sandeep Kadapa
5,667427
edited Nov 10 at 17:38
Sandeep Kadapa
5,667427
edited Nov 10 at 17:38
Sandeep Kadapa
5,667427
5,667427
asked Nov 10 at 17:37
Lukas
355
asked Nov 10 at 17:37
Lukas
355
asked Nov 10 at 17:37
Lukas
355
355
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The . character has special meaning in regular expressions. You can use re.escape to make a string "safe" for use in a regular expression.
>>> import re
... important_strings=['bdv.', 'dktv.', 'mot. g.', 'vyr. g.']
... regex = re.compile('|'.join(re.escape(s) for s in important_strings))
... regex.findall('bdv. mot. g. vns. kilm.')
['bdv.', 'mot. g.']
Pandas has its own findall which should work like re.findall
answered Nov 10 at 17:51
Håken Lid
10.5k62441
1
@perreal, your comment above is not clear, can your pls make it clear.
– pygo
Nov 10 at 18:02
Pandas series indeed hasstr.findallmethod. Andre.escaperemoves dots. What is left is list instead of string. But may I get out with this.
– Lukas
Nov 10 at 18:17
1
.str.findall('|'.join(re.escape(s) for s in important_strings)).str.join(' ')
– Lukas
Nov 10 at 18:23
You can benchmark to test if findall is faster than your original negative lookahead. I try to avoid using lookaround assertions in my regular expressions because they are often hard to read/understand and in some cases they can be very slow, if the regex engine is forced to do a lot of backtracking.
– Håken Lid
Nov 10 at 18:38
add a comment |
Maybe split string
bdv. mot. g. vns. kilm.
using your list and remove from oryginal string what left after spliting.
answered Nov 10 at 18:08
user10403681
11
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The . character has special meaning in regular expressions. You can use re.escape to make a string "safe" for use in a regular expression.
>>> import re
... important_strings=['bdv.', 'dktv.', 'mot. g.', 'vyr. g.']
... regex = re.compile('|'.join(re.escape(s) for s in important_strings))
... regex.findall('bdv. mot. g. vns. kilm.')
['bdv.', 'mot. g.']
Pandas has its own findall which should work like re.findall
answered Nov 10 at 17:51
Håken Lid
10.5k62441
1
@perreal, your comment above is not clear, can your pls make it clear.
– pygo
Nov 10 at 18:02
Pandas series indeed hasstr.findallmethod. Andre.escaperemoves dots. What is left is list instead of string. But may I get out with this.
– Lukas
Nov 10 at 18:17
1
.str.findall('|'.join(re.escape(s) for s in important_strings)).str.join(' ')
– Lukas
Nov 10 at 18:23
You can benchmark to test if findall is faster than your original negative lookahead. I try to avoid using lookaround assertions in my regular expressions because they are often hard to read/understand and in some cases they can be very slow, if the regex engine is forced to do a lot of backtracking.
– Håken Lid
Nov 10 at 18:38
add a comment |
The . character has special meaning in regular expressions. You can use re.escape to make a string "safe" for use in a regular expression.
>>> import re
... important_strings=['bdv.', 'dktv.', 'mot. g.', 'vyr. g.']
... regex = re.compile('|'.join(re.escape(s) for s in important_strings))
... regex.findall('bdv. mot. g. vns. kilm.')
['bdv.', 'mot. g.']
Pandas has its own findall which should work like re.findall
answered Nov 10 at 17:51
Håken Lid
10.5k62441
1
@perreal, your comment above is not clear, can your pls make it clear.
– pygo
Nov 10 at 18:02
Pandas series indeed hasstr.findallmethod. Andre.escaperemoves dots. What is left is list instead of string. But may I get out with this.
– Lukas
Nov 10 at 18:17
1
.str.findall('|'.join(re.escape(s) for s in important_strings)).str.join(' ')
– Lukas
Nov 10 at 18:23
You can benchmark to test if findall is faster than your original negative lookahead. I try to avoid using lookaround assertions in my regular expressions because they are often hard to read/understand and in some cases they can be very slow, if the regex engine is forced to do a lot of backtracking.
– Håken Lid
Nov 10 at 18:38
add a comment |
The . character has special meaning in regular expressions. You can use re.escape to make a string "safe" for use in a regular expression.
>>> import re
... important_strings=['bdv.', 'dktv.', 'mot. g.', 'vyr. g.']
... regex = re.compile('|'.join(re.escape(s) for s in important_strings))
... regex.findall('bdv. mot. g. vns. kilm.')
['bdv.', 'mot. g.']
Pandas has its own findall which should work like re.findall
answered Nov 10 at 17:51
Håken Lid
10.5k62441
The . character has special meaning in regular expressions. You can use re.escape to make a string "safe" for use in a regular expression.
>>> import re
... important_strings=['bdv.', 'dktv.', 'mot. g.', 'vyr. g.']
... regex = re.compile('|'.join(re.escape(s) for s in important_strings))
... regex.findall('bdv. mot. g. vns. kilm.')
['bdv.', 'mot. g.']
Pandas has its own findall which should work like re.findall
answered Nov 10 at 17:51
Håken Lid
10.5k62441
edited Nov 10 at 18:17
answered Nov 10 at 17:51
Håken Lid
10.5k62441
answered Nov 10 at 17:51
Håken Lid
10.5k62441
answered Nov 10 at 17:51
Håken Lid
10.5k62441
10.5k62441
1
@perreal, your comment above is not clear, can your pls make it clear.
– pygo
Nov 10 at 18:02
Pandas series indeed hasstr.findallmethod. Andre.escaperemoves dots. What is left is list instead of string. But may I get out with this.
– Lukas
Nov 10 at 18:17
1
.str.findall('|'.join(re.escape(s) for s in important_strings)).str.join(' ')
– Lukas
Nov 10 at 18:23
You can benchmark to test if findall is faster than your original negative lookahead. I try to avoid using lookaround assertions in my regular expressions because they are often hard to read/understand and in some cases they can be very slow, if the regex engine is forced to do a lot of backtracking.
– Håken Lid
Nov 10 at 18:38
add a comment |
1
@perreal, your comment above is not clear, can your pls make it clear.
– pygo
Nov 10 at 18:02
Pandas series indeed hasstr.findallmethod. Andre.escaperemoves dots. What is left is list instead of string. But may I get out with this.
– Lukas
Nov 10 at 18:17
1
.str.findall('|'.join(re.escape(s) for s in important_strings)).str.join(' ')
– Lukas
Nov 10 at 18:23
You can benchmark to test if findall is faster than your original negative lookahead. I try to avoid using lookaround assertions in my regular expressions because they are often hard to read/understand and in some cases they can be very slow, if the regex engine is forced to do a lot of backtracking.
– Håken Lid
Nov 10 at 18:38
1
1
@perreal, your comment above is not clear, can your pls make it clear.
– pygo
Nov 10 at 18:02
@perreal, your comment above is not clear, can your pls make it clear.
– pygo
Nov 10 at 18:02
Pandas series indeed has
str.findall method. And re.escape removes dots. What is left is list instead of string. But may I get out with this.– Lukas
Nov 10 at 18:17
Pandas series indeed has
str.findall method. And re.escape removes dots. What is left is list instead of string. But may I get out with this.– Lukas
Nov 10 at 18:17
1
1
.str.findall('|'.join(re.escape(s) for s in important_strings)).str.join(' ')– Lukas
Nov 10 at 18:23
.str.findall('|'.join(re.escape(s) for s in important_strings)).str.join(' ')– Lukas
Nov 10 at 18:23
You can benchmark to test if findall is faster than your original negative lookahead. I try to avoid using lookaround assertions in my regular expressions because they are often hard to read/understand and in some cases they can be very slow, if the regex engine is forced to do a lot of backtracking.
– Håken Lid
Nov 10 at 18:38
You can benchmark to test if findall is faster than your original negative lookahead. I try to avoid using lookaround assertions in my regular expressions because they are often hard to read/understand and in some cases they can be very slow, if the regex engine is forced to do a lot of backtracking.
– Håken Lid
Nov 10 at 18:38
add a comment |
Maybe split string
bdv. mot. g. vns. kilm.
using your list and remove from oryginal string what left after spliting.
answered Nov 10 at 18:08
user10403681
11
add a comment |
Maybe split string
bdv. mot. g. vns. kilm.
using your list and remove from oryginal string what left after spliting.
answered Nov 10 at 18:08
user10403681
11
add a comment |
Maybe split string
bdv. mot. g. vns. kilm.
using your list and remove from oryginal string what left after spliting.
answered Nov 10 at 18:08
user10403681
11
Maybe split string
bdv. mot. g. vns. kilm.
using your list and remove from oryginal string what left after spliting.
answered Nov 10 at 18:08
user10403681
11
answered Nov 10 at 18:08
user10403681
11
answered Nov 10 at 18:08
user10403681
11
answered Nov 10 at 18:08
user10403681
11
11
add a comment |
add a comment |
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