Python calling out positional argument

Question

I have a python function (python 2.5) with a definition like:

def Foo(a, b='Silly', c='Walks', d='Spam', e='Eggs', f='Ni', *addl):

Where addl can be any number of strings (filenames) to do something with.

I'm fine with all of the defaults, but I have filenames to put in addl.

I am inclined to do something like:

Foo('[email protected]', addl=('File1.txt', 'File2.txt'))

But that gets the following error:

TypeError: Foo() got an unexpected keyword argument 'addl'

Is there a syntax where I can succinctly call Foo with just the first required parameter and my (variable number of) additional strings? Or am I stuck redundantly specifying all of the defaults before I can break into the addl argument range?

For the sake of argument, the function definition is not able to be modified or refactored.

Just unpack tuple, it should work: Foo('[email protected]', *('File1.txt', 'File2.txt')) — AndreyT, Apr 04 '16 at 14:40
@AndreyT: That doesn't work. It assigns `'File1.txt'` to `a`, and `'File2.txt'` to `b`. — zondo, Apr 04 '16 at 14:42
Yep, you right. So, in case of 2.5 possible solution is usage kwargs instead of b,c,d,e,f parameters. — AndreyT, Apr 04 '16 at 14:50
@AndreyT It is not specific to `Python 2.5`, right? Can we say that it hold true for `Python>2.5`? I tried this on `Python 2.7` and assigns `File2.txt` to `b` as well. — malhar, Apr 04 '16 at 14:53
@AndreyT: The problem with that would be the unmitigated affect on all the other code using `Foo` — MellowNello, Apr 04 '16 at 14:55
@malhar: It's bit specific to 2.x. In 3.x we can use folow definition: def f(a, *args, b=1, c=2, etc=3) — AndreyT, Apr 04 '16 at 15:00

AndreyT · Accepted Answer · 2016-04-04T15:15:48.297

2

Ok, if function can't be modified, then why not create wrapper?

def foo_with_defaults(a, *addl):
    Foo(a, *(Foo.func_defaults + addl))

foo_with_defaults('[email protected]', *('File1.txt', 'File2.txt'))

edited Apr 04 '16 at 15:15

answered Apr 04 '16 at 14:56

AndreyT

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"For the sake of argument, the function definition is not able to be modified or refactored." – Kevin Apr 04 '16 at 14:56
`Foo.func_defaults` was the key piece I was missing. Thanks! – MellowNello Apr 04 '16 at 15:27

apr · Answer 2 · 2016-04-04T15:02:29.180

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To my knowledge it is not possible.

First you need to pass required (a) and default (b, c, d, e, f) arguments, and then every additional argument is passed as *addl:

Foo('[email protected]', 'Silly', 'Walks', 'Spam', 'Eggs', 'Ni', 'File1.txt', 'File2.txt')

edited Apr 04 '16 at 15:02

answered Apr 04 '16 at 14:51

apr

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So your answer to "is there a syntax where I can succinctly call Foo with just the first required parameter and my (variable number of) additional strings?" is "no"? Are you certain this is the one and only approach? – Kevin Apr 04 '16 at 14:52
Sorry for not answering the question precisely. To my knowledge it is not possible. I will edit my answer. – apr Apr 04 '16 at 14:58

score 0 · Answer 3 · answered Apr 04 '16 at 15:05

You are trying to use addl as a keyword argument, but you defined it as a positional argument. You should add one more * if you want to call the function like Foo(addl=('File1.txt', 'File2.txt')):

def Foo(a, b='Silly', c='Walks', d='Spam', e='Eggs', f='Ni', **addl):

Python calling out positional argument

3 Answers3