How to pass in a variable to a function under a condition

[Six1Seven1]

Hi,

Sorry for all the posts. I have run into a problem. This stems back to my novice programming experience... My question is, how do I pass in an argument to a function only sometimes, only when a condition is true. Here is an example...

#call the function----------------------------

if checkbox == True:

    Function1(x, y, z)

else:

    Function1(x,y)

#define the function-----------------------

def Function1(x,y,z)       <----- Problem! variable z is only present when checkbox is TRUE.

Thus, i get an error. How do i define the function to only pass in the variable 'Z' when the checkbox is true? Sorry, I am still inexperienced in programming.

Thanks!

[ekinsb · Accepted solution]

Python supports optional arguments in a function.  For your function

def Function1(x, y, z):

You can make the last argument optional by defining a default value for it.  If you don't provide a value when calling the function the default value will be used.  The first example below will set z to zero.

def Function1(x, y, z=0):

If you need to know if the value was assigned by the caller or is the default you could use something else because in the example above, zero would be a valid value to use.  The example below uses None and then you could check for that in the function to see if a value was provided or not.

def Function1(x, y, z=None):

In both cases the function would be called using:

Function1(23, 3.6)

or

Function1(21, 34.6, 1.2)

[Six1Seven1]

Brian, that worked great.

Thank you very much