Question

Assume you have a function like that:

def example(var1=None,var2=None,var3=None,*multi_values):
    print (var1,var2,var3,*multi_values)

Can you get around calling all the optional parameters and just adding stuff to the last one?

Example:

>>> multi=range(3)
>>> example(???)
(None,"hi",None,(1,2,3))

I don’t want to do this:

>>> multi=range(3)
>>> example(None,None,None,*multi) #bad -> this doesn't use the default values

Answer 1

One option is to change you prototype to take an iterable as multi_values argument, instead of accepting an arbitrary number of parameters:

def example(var1=None, var2=None, var3=None, multi_values=None):
    if multi_values is None:
        multi_values = []
    # whatever

Then you can call the function as

example(multi_values=multi)

Another option is to accept var1 through var3 only as keywords arguments:

def example(*multi_values, **kwargs):
    var1 = kwargs.pop("var1", None)
    var2 = kwargs.pop("var2", None)
    var3 = kwargs.pop("var3", None)
    if kwargs:
        raise TypeError("Unknown keyword arguments")
    # whatever

In Python 3.x, this is supported by adding var1 through var3 after the *multi_values argument:

def example(*multi_values, var1=None, var2=None, var3=None):
    # whatever

This does basically the same as the previous example, with the additional advantage of better introspection.