Question

This is just a curiosity question for a “good” pythonic way to do this.

I have a function with an optional parameter ie.

def foo( a, b, c='en' ):
    print c

There is a dict with a bunch of info in it, and if a particular key is in the dict, I would like to pass it into foo to override c’s default, but if the key is not in the dict, I just want to use the default for c.

Obviously this will work…

if "SomeKey" in mydict:
    foo( val1, val2, mydict[ "SomeKey" ]
else:
    foo( val1, val2 )

And another option would be to do something like

params = [ val1, val2 ]
if "SomeKey" in mydict:
    params.append( mydict[ "SomeKey" ] )
foo( *params )

but there must be a slick, more pythonic way to do this? ie.

foo( val1, val2, mydict[ "SomeKey" ] if "SomeKey" in mydict else < use default > )

Thanks in advance!

Answer 1

Don’t use a “true” default value for this – instead, use a placeholder value:

def foo( a, b, c=None ):
    if c is None:
        c = 'en'
    print c

and then just call with…

foo( val1, val2, mydict.get("SomeKey") )

(.get() returns None if the key isn’t in the dict, by default)

---

If you can’t modify foo(), then you can do the more complex varargs path:

maybe_c = {'c': mydict["SomeKey"]} if "SomeKey" in mydict else {}
foo( val1, val2, **maybe_c )