Python - how to set default parameter equal to another parameter

Question

I have a function, which should have one optional parameter. I want the optional parameter to be the same as other parameter. Something like this:

def foo(arg1, arg2, src, dst=src):
  ...
  ...

Parameter dst is the optional one. The thing is when dst is not given when calling foo, it should be same as src.

Any ideas?

Thank you

Reut Sharabani · Accepted Answer · 2015-03-02 08:40:42Z

3

How about this?

def foo(arg1, arg2, src, dst=None):
    dst = dst if dst is not None else src

Test:

>>> def foo(arg1, arg2, src, dst=None):
...     dst = dst if dst is not None else src
...     print dst
... 
>>> foo(0, 0, "test")
test

Following @TanveerAlam 's comment, I don't want to make assumptions about your arguments (what if dst can be False?), I did use shorter versions in my original post and I'll leave them here for reference:

def foo(arg1, arg2, src, dst=None):
    dst = dst if dst else src

Or even:

def foo(arg1, arg2, src, dst=None):
    dst = dst or src

Note that these versions will prevent values other than None in dst (like False, [] ...)

edited Mar 2, 2015 at 8:40

answered Mar 2, 2015 at 8:31

Reut Sharabani

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2 Comments

Tanveer Alam Over a year ago

dst = dst if dst else src this would work too as bool value of None is False.

Kroltan Over a year ago

I would prefer dst = dst or src. A bit more clear and less repetitive, and relies on the same behaviour as dst if dst else src

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