Is there a way to get the parameter names a function takes?
def foo(bar, buz):
pass
magical_way(foo) == ["bar", "buz"]
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1Duplicate: stackoverflow.com/questions/582056/…, stackoverflow.com/questions/218616/…Sasha Chedygov– Sasha Chedygov2010-08-19 01:25:17 +00:00Commented Aug 19, 2010 at 1:25
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3Maybe one more is warranted, since I spent a good half an hour trying to Google this without hitting those dupes.Bemmu– Bemmu2010-08-19 02:03:37 +00:00Commented Aug 19, 2010 at 2:03
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3This is the first hit on Google now, thanks to the way OP asked the question. Making answers easier to find by framing the question better is definitely relevant to UX of those looking for answers.episodeyang– episodeyang2017-04-13 19:44:24 +00:00Commented Apr 13, 2017 at 19:44
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Btw. I have since learned that it's wrong to say "argument" here, as receiving variables are parameters, not arguments.Bemmu– Bemmu2019-05-31 05:47:48 +00:00Commented May 31, 2019 at 5:47
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2 Answers
Use the inspect module from Python's standard library (the cleanest, most solid way to perform introspection).
Specifically, inspect.getargspec(f) returns the names and default values of f's arguments -- if you only want the names and don't care about special forms *a, **k,
import inspect
def magical_way(f):
return inspect.getargspec(f)[0]
completely meets your expressed requirements.
8 Comments
John La Rooy
you can also use
return inspect.getargspec(f).args
Luke Murray
maybe you can edit this since getargspec is now deprecated and inspect.signature() is favored instead
juan Isaza
inspect.getargspec is deprecated as of Python 3.6. Instead one should use inspect.signature. Any one knows how to use it?, thanks
lovesh
@juanIsaza
inspect.signature(f).parameters
Mauricio Arboleda-Zapata
@juanIsaza alternative you can use inspect.getfullargspec(f).args or inspect.getfullargspec(f)[0] (python 3.7)
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>>> import inspect
>>> def foo(bar, buz):
... pass
...
>>> inspect.getargspec(foo)
ArgSpec(args=['bar', 'buz'], varargs=None, keywords=None, defaults=None)
>>> def magical_way(func):
... return inspect.getargspec(func).args
...
>>> magical_way(foo)
['bar', 'buz']
2 Comments
Wolph
Note that
.args only works in Python 2.6 and up. For older verions you have to use [0] instead.
Chinmay Kanchi
+1 for providing the
magical_way() function.