Hi all I've got and input strings:
list_of_strings = ["apple", "orange ca", "pear sa", "banana sth"]
And I want to remove everything after multiple white spaces (more than 1), so end result is:
final_list_of_strings = ["apple", "orange ca", "pear", "banana"]
I've tried regex:
import re
regex_expression = r"(.*\s?)(\s{2,}.*)"
for name in list_of_strings:
regex_matching_groups = re.findall(regex_expression, name)
if regex_matching_groups:
name = regex_matching_groups[0][0]
but fails on multiple spaces ... Thank you for help!
You can use re.sub
in a list comprehension:
import re
list_of_strings = ["apple", "orange ca", "pear sa", "banana sth"]
list_of_strings = [re.sub(r'\s{2}.*', '', x, flags=re.S) for x in list_of_strings]
print(list_of_strings)
# -> ['apple', 'orange ca', 'pear', 'banana']
See the Python demo .
The \\s{2}.*
regex matches two whitespace chars and then the rest of the string (even if there are line break chars due to re.S
flag).
using a regular expression find the first word and optionally a second word with only one space between
list_of_strings = ["apple", "orange ca", "pear sa", "banana sth"]
my_list=[]
def find_phrase(list_of_strings):
for string in list_of_strings:
matches=re.findall(r"(\w+)( \w+)*", string)
if len(matches)>0:
my_list.append("".join([matches[0][0],matches[0][1]]))
return my_list
print(find_phrase(list_of_strings))
output:
['apple', 'orange ca', 'pear', 'banana']
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