Fix Python – Is there a way to pass optional parameters to a function?

The Python 2 documentation, 7.6. Function definitions gives you a couple of ways to detect whether a caller supplied an optional parameter.

First, you can use special formal parameter syntax *. If the function definition has a formal parameter preceded by a single *, then Python populates that parameter with any positional parameters that aren’t matched by preceding formal parameters (as a tuple). If the function definition has a formal parameter preceded by **, then Python populates that parameter with any keyword parameters that aren’t matched by preceding formal parameters (as a dict). The function’s implementation can check the contents of these parameters for any “optional parameters” of the sort you want.

For instance, here’s a function opt_fun which takes two positional parameters x1 and x2, and looks for another keyword parameter named “optional”.

>>> def opt_fun(x1, x2, *positional_parameters, **keyword_parameters):
...     if ('optional' in keyword_parameters):
...         print 'optional parameter found, it is ', keyword_parameters['optional']
...     else:
...         print 'no optional parameter, sorry'
... 
>>> opt_fun(1, 2)
no optional parameter, sorry
>>> opt_fun(1,2, optional="yes")
optional parameter found, it is  yes
>>> opt_fun(1,2, another="yes")
no optional parameter, sorry

Second, you can supply a default parameter value of some value like None which a caller would never use. If the parameter has this default value, you know the caller did not specify the parameter. If the parameter has a non-default value, you know it came from the caller.