I have a a filename below and I want to extract year and _TEXT part.
fle_2019-11-17A17-21-09.01(_TEXT).txt
I am able to do this using two regex and then join the results.
(?<=\_)(\d{4})(?=\-)
This gives me year
(?<=\()(.*)(?=\)) This gives me _TEXT
Is there a way to get this from a single expression?
One option is to use 2 capturing groups. Depending on what you would allow to match before the first underscore, you could for example use a character class to match word characters without an underscore [^\W_]+
^[^\W_]+_(\d{4})-[\w.-]+\(([^)]+)\)\.\w+$
In parts
^ Start of string[^\W_]+ Match 1+ word chars except __ Match the _(\d{4}) Capture group 1, match 1+ digits-[\w.-]+ Match - and 1+ word chars, . or - (extend the character class with what you would allow to match\( Match (
([^)]+) Capture group 2, match 1+ times any char except )\) Match )\.\w+ Match a . and 1+ word chars$ End of stringFor example
import re
regex = r"^[^\W_]+_(\d{4})-[\w.-]+\(([^)]+)\)\.\w+$"
test_str = "fle_2019-11-17A17-21-09.01(_TEXT).txt"
print(re.findall(regex, test_str))
Output
[('2019', '_TEXT')]
In the interest of simplicity, we could just try using re.findall with an alternation which captures either a 4 digit year or a file name:
file = "fle_2019-11-17A17-21-09.01(_TEXT).txt"
parts = re.findall(r'\d{4}(?=-\d{2})|(?<=\().*?(?=\))', file)
print(parts)
This prints:
['2019', '_TEXT']
I like this approach because the output already yields separate logical values for the year and file name.
