def thefunction(a=1,b=2,c=3):
pass
print allkeywordsof(thefunction) #allkeywordsof doesnt exist
which would give [a,b,c]
Is there a function like allkeywordsof?
I cant change anything inside, thefunction
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2duplicate of stackoverflow.com/questions/2677185/how-read-method-signatureXavier Combelle– Xavier Combelle2012-08-11 13:23:17 +00:00Commented Aug 11, 2012 at 13:23
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possible duplicate of Getting method parameter names in pythonMarcin– Marcin2012-08-11 17:58:47 +00:00Commented Aug 11, 2012 at 17:58
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3 Answers
I think you are looking for inspect.getargspec:
import inspect
def thefunction(a=1,b=2,c=3):
pass
argspec = inspect.getargspec(thefunction)
print(argspec.args)
yields
['a', 'b', 'c']
If your function contains both positional and keyword arguments, then finding the names of the keyword arguments is a bit more complicated, but not too hard:
def thefunction(pos1, pos2, a=1,b=2,c=3, *args, **kwargs):
pass
argspec = inspect.getargspec(thefunction)
print(argspec)
# ArgSpec(args=['pos1', 'pos2', 'a', 'b', 'c'], varargs='args', keywords='kwargs', defaults=(1, 2, 3))
print(argspec.args)
# ['pos1', 'pos2', 'a', 'b', 'c']
print(argspec.args[-len(argspec.defaults):])
# ['a', 'b', 'c']
5 Comments
user1513192
How can I make it give me the agrs only
jamylak
inspect.getargspec(thefunction).argsjamylak
argspec.args[-len(argspec.defaults):] this should be accepted because it answers the question.
Jon Clements
No disrespect to the OP or Ashwini, but I agree this should really be the accepted answer for future references...
nocibambi
Do you want something like this:
>>> def func(x,y,z,a=1,b=2,c=3):
pass
>>> func.func_code.co_varnames[-len(func.func_defaults):]
('a', 'b', 'c')
answered Aug 11, 2012 at 13:27
Ashwini Chaudhary
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8 Comments
Rostyslav Dzinko
co_varnames is a wrong decision, it also holds non-keyword arguments and topic-starter wants only keyword arguments
Ashwini Chaudhary
@RostyslavDzinko yeah you're right, but the problem is
inspect.getargspec() also returns non-keyword arguments.Rostyslav Dzinko
@user1513192, this answer doesn't fix the problem you described. If it fits your needs - redescribe your problem (My upper comment shows why)
Ashwini Chaudhary
@jamylak used this because I haven't learned
inspect yet and will surely learn it now. :)
naught101
In python 3
func_code has been renamed to __code__. |
You can do the following in order to get exactly what you are looking for.
>>>
>>> def funct(a=1,b=2,c=3):
... pass
...
>>> import inspect
>>> inspect.getargspec(funct)[0]
['a', 'b', 'c']
>>>
