I've a list ['test_x', 'text', 'x']. Is there any way to use regex in python to find if the string inside the list contains either '_x' or 'x'?
'x' should be a single string and not be part of a word without _.
The output should result in ['test_x', 'x'].
Thanks.
Using one line comprehension:
l = ['test_x', 'text', 'x']
result = [i for i in l if '_x' in i or 'x' == i]
'x' == i is needed since I don't want words which has x in between. Only _x or if str itself is x. Answer works though. Thanks.You can use regexp this way:
import re
print(list(filter(lambda x: re.findall(r'_x|^x$',x),l)))
The regexp searches for exact patterns ('_x' or 'x') within each element of the list. applies the func to each element of the iterable.
You can make your expression more genric this way:
print(list(filter(lambda x: re.findall(r'[^A-Za-z]x|^\W*x\W*$',x),l)))
Here am telling python to search for expressions which DON't start with A to Z or a to z but end in x OR search for expressions that start and end with 0 or more non-word characters but have x in between. You can refer this quick cheatsheet on regular expressions https://www.debuggex.com/cheatsheet/regex/python
[re.findall('x|_x', s) for s in your_list]
