How to match any string from a list of strings in regular expressions in python?
Question:
Lets say I have a list of strings,
string_lst = ['fun', 'dum', 'sun', 'gum']
I want to make a regular expression, where at a point in it, I can match any of the strings i have in that list, within a group, such as this:
import re
template = re.compile(r".*(elem for elem in string_lst).*")
template.match("I love to have fun.")
What would be the correct way to do this? Or would one have to make multiple regular expressions and match them all separately to the string?
Answers:
Join the list on the pipe character |, which represents different options in regex.
string_lst = ['fun', 'dum', 'sun', 'gum']
x="I love to have fun."
print re.findall(r"(?=("+'|'.join(string_lst)+r"))", x)
Output: ['fun']
You cannot use match as it will match from start.
Using search you will get only the first match. So use findall instead.
Also use lookahead if you have overlapping matches not starting at the same point.
Except for the regular expression, you can use list comprehension, hope it’s not off the topic.
import re
def match(input_string, string_list):
words = re.findall(r'w+', input_string)
return [word for word in words if word in string_list]
>>> string_lst = ['fun', 'dum', 'sun', 'gum']
>>> match("I love to have fun.", string_lst)
['fun']
You should make sure to escape the strings correctly before combining into a regex
>>> import re
>>> string_lst = ['fun', 'dum', 'sun', 'gum']
>>> x = "I love to have fun."
>>> regex = re.compile("(?=(" + "|".join(map(re.escape, string_lst)) + "))")
>>> re.findall(regex, x)
['fun']
regex module has named lists (sets actually):
#!/usr/bin/env python
import regex as re # $ pip install regex
p = re.compile(r"L<words>", words=['fun', 'dum', 'sun', 'gum'])
if p.search("I love to have fun."):
print('matched')
Here words is just a name, you can use anything you like instead.
.search() methods is used instead of .* before/after the named list.
To emulate named lists using stdlib’s re module:
#!/usr/bin/env python
import re
words = ['fun', 'dum', 'sun', 'gum']
longest_first = sorted(words, key=len, reverse=True)
p = re.compile(r'(?:{})'.format('|'.join(map(re.escape, longest_first))))
if p.search("I love to have fun."):
print('matched')
re.escape() is used to escape regex meta-characters such as .*? inside individual words (to match the words literally).
sorted() emulates regex behavior and it puts the longest words first among the alternatives, compare:
>>> import re
>>> re.findall("(funny|fun)", "it is funny")
['funny']
>>> re.findall("(fun|funny)", "it is funny")
['fun']
>>> import regex
>>> regex.findall(r"L<words>", "it is funny", words=['fun', 'funny'])
['funny']
>>> regex.findall(r"L<words>", "it is funny", words=['funny', 'fun'])
['funny']
In line with @vks reply – I feel this actually does the complete task…
finds = re.findall(r"(?=(b" + '\b|\b'.join(string_lst) + r"b))", x)
Adding word boundary completes the task!
Lets say I have a list of strings,
string_lst = ['fun', 'dum', 'sun', 'gum']
I want to make a regular expression, where at a point in it, I can match any of the strings i have in that list, within a group, such as this:
import re
template = re.compile(r".*(elem for elem in string_lst).*")
template.match("I love to have fun.")
What would be the correct way to do this? Or would one have to make multiple regular expressions and match them all separately to the string?
Join the list on the pipe character |, which represents different options in regex.
string_lst = ['fun', 'dum', 'sun', 'gum']
x="I love to have fun."
print re.findall(r"(?=("+'|'.join(string_lst)+r"))", x)
Output: ['fun']
You cannot use match as it will match from start.
Using search you will get only the first match. So use findall instead.
Also use lookahead if you have overlapping matches not starting at the same point.
Except for the regular expression, you can use list comprehension, hope it’s not off the topic.
import re
def match(input_string, string_list):
words = re.findall(r'w+', input_string)
return [word for word in words if word in string_list]
>>> string_lst = ['fun', 'dum', 'sun', 'gum']
>>> match("I love to have fun.", string_lst)
['fun']
You should make sure to escape the strings correctly before combining into a regex
>>> import re
>>> string_lst = ['fun', 'dum', 'sun', 'gum']
>>> x = "I love to have fun."
>>> regex = re.compile("(?=(" + "|".join(map(re.escape, string_lst)) + "))")
>>> re.findall(regex, x)
['fun']
regex module has named lists (sets actually):
#!/usr/bin/env python
import regex as re # $ pip install regex
p = re.compile(r"L<words>", words=['fun', 'dum', 'sun', 'gum'])
if p.search("I love to have fun."):
print('matched')
Here words is just a name, you can use anything you like instead.
.search() methods is used instead of .* before/after the named list.
To emulate named lists using stdlib’s re module:
#!/usr/bin/env python
import re
words = ['fun', 'dum', 'sun', 'gum']
longest_first = sorted(words, key=len, reverse=True)
p = re.compile(r'(?:{})'.format('|'.join(map(re.escape, longest_first))))
if p.search("I love to have fun."):
print('matched')
re.escape() is used to escape regex meta-characters such as .*? inside individual words (to match the words literally).
sorted() emulates regex behavior and it puts the longest words first among the alternatives, compare:
>>> import re
>>> re.findall("(funny|fun)", "it is funny")
['funny']
>>> re.findall("(fun|funny)", "it is funny")
['fun']
>>> import regex
>>> regex.findall(r"L<words>", "it is funny", words=['fun', 'funny'])
['funny']
>>> regex.findall(r"L<words>", "it is funny", words=['funny', 'fun'])
['funny']
In line with @vks reply – I feel this actually does the complete task…
finds = re.findall(r"(?=(b" + '\b|\b'.join(string_lst) + r"b))", x)
Adding word boundary completes the task!