How can I programmatically access the default argument values of a method in Python? For example, in the following
def test(arg1='Foo'):
pass
how can I access the string 'Foo' inside test?
4 Answers
They are stored in test.func_defaults (python 2) and in test.__defaults__ (python 3).
As @Friedrich reminds me, Python 3 has "keyword only" arguments, and for those the defaults are stored in function.__kwdefaults__
1 Comment
Friedrich -- Слава Україні
See also
__kwdefaults__.Consider:
def test(arg1='Foo'):
pass
In [48]: test.func_defaults
Out[48]: ('Foo',)
.func_defaults gives you the default values, as a sequence, in order that the arguments appear in your code.
Apparently, func_defaults may have been removed in python 3.
1 Comment
Rob Wouters
I think
func_defaults only works on Python 2.x. __defaults__ seems to work on Python 2.7 and 3.2.Ricardo Cárdenes is on the right track. Actually getting to the function test inside test is going to be a lot more tricky. The inspect module will get you further, but it is going to be ugly: Python code to get current function into a variable?
As it turns out, you can refer to test inside the function:
def test(arg1='foo'):
print test.__defaults__[0]
Will print out foo. But refering to test will only work, as long as test is actually defined:
>>> test()
foo
>>> other = test
>>> other()
foo
>>> del test
>>> other()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 2, in test
NameError: global name 'test' is not defined
So, if you intend on passing this function around, you might really have to go the inspect route :(
2 Comments
Rob Wouters
I was under that impression as well, turns out that
test is in test's local scope as pointed out in Ricardo's comment on my answer.
Evgeni Sergeev
If we do
def test2(): print locals(), '\n\n', globals(), we see that test2 is in globals, and there is nothing in locals.This isn't very elegant (at all), but it does what you want:
def test(arg1='Foo'):
print(test.__defaults__)
test(arg1='Bar')
Works with Python 3.x too.
2 Comments
Ricardo Cárdenes
Why
globals()? test is local in its own scope, no need for that.Rob Wouters
@RicardoCárdenes, you're right. I did not know that, thank you. Fixing it now.
arg1?arg1when callingtest, thenarg1will default to'Foo'arg1can be overwritten, right?