grep capture python function definition

Question

I have tried:

$ grep -Po "def[[:space:]]+\K(.*)\(" code.py

The output should be a list of function names.

I think it should be correct, but it's only capturing one extra bracket. I would like to get rid off it in the expression (not postprocessing). Please, explain the syntax, you will use.

add a sample input and output... I think grep -Po 'def[[:space:]]+\K[^(]+' code.py is what you need — Sundeep
– Sundeep, Commented Apr 30, 2017 at 5:14
You realize this is missing any arguments to the method, right? So do you just want the function name or the entire definition name(args) ? — Burhan Khalid
– Burhan Khalid, Commented Apr 30, 2017 at 5:14
It seems like the ast module might be a better bet for what you're trying to do. — pvg
– pvg, Commented Apr 30, 2017 at 5:15
@pvg I know it exists, but for me quick grep is sufficient. I practise grep. — xralf
– xralf, Commented Apr 30, 2017 at 5:18
@xralf I get that but... There's a thing built into python that reliably parses python. Alternatively, you can look at what [whatnot]tags or your favourite python syntax highlighter has for python regexes. — pvg
– pvg, Commented Apr 30, 2017 at 5:20

jordanm · Accepted Answer · 2017-04-30 06:10:57Z

1

You can omit the "(" from the resulting string by putting it in a lookahead (?=<expression>):

grep -Po "def[[:space:]]+\K(.*)(?=\()" code.py

answered Apr 30, 2017 at 6:10

jordanm

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