How do I create a Python function with optional arguments?
Just use the
*args parameter, which allows you to pass as many arguments as you want after your
a,b,c. You would have to add some logic to map
c,d,e,f but its a "way" of overloading.
def myfunc(a,b, *args, **kwargs): for ar in args: print armyfunc(a,b,c,d,e,f)
And it will print values of
Similarly you could use the
kwargs argument and then you could name your parameters.
def myfunc(a,b, *args, **kwargs): c = kwargs.get('c', None) d = kwargs.get('d', None) #etcmyfunc(a,b, c='nick', d='dog', ...)
kwargs would have a dictionary of all the parameters that are key valued after
It is very easy just do this
def foo(a = None): print(a)
Instead of None you can type anything that should be in place if there was no argument for example if you will not write value of the parameter like this
foo() then it will print None because no argument is given and if you will GIVE it a argument like
foo("hello world") then it will print
hello world, OH YES i forgot to tell y'all that when these types of paramters i.e optional parameters, THESE ARGUMENTS NEEDS TO FOLLOW THE DEFAULT ARGUMENTS this means that, lets take the previous function and add another parameter
def foo(a = None, b): print(a)
Now if you'll execute your python file it is going to raise an exception saying that
Non-default arguments follows default arguments,
SyntaxError: non-default argument follows default argument
so you gotta put the optional or non-default argument after the arguments which are required
def foo (a, b=None): ... #This one is right(these threee dots are ellipsis which are somewhat like passdef foo(b=None, a): ... #and this isn't