FAQ: Function Arguments: *args and **kwargs - Variable number of arguments: *args

This community-built FAQ covers the “Variable number of arguments: *args” exercise from the lesson “Function Arguments: *args and **kwargs”.

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FAQs on the exercise Variable number of arguments: *args

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I didn’t know where to post this so I thought I’d try my luck here. I’m try to write a function to find the product of integer arguments…

def combinations(*args):
	num = 1
	for n in args:
		num *= n if n else 1
	return num

print(combinations(6, 7, 0))#, 42)

The code works great. However, below if my attempt at using list comprehension to solve the same problem but I am getting an error. Any assistance or pointers will be highly appreciated.

def combinations(*args):
	return [num *= n for n in args if n else 1]

print(combinations(6, 7, 0))#, 42)

A comprehension is a list object composed of expressions or expression evaluations. num *= n is not an expression, but a statement.

@mtf How then would i go about solving this? Assuming there is a way to do that

There is no practical way to do with a comprehension what your first example does… return a product.

Not sure one would give the name combinations to a function that returns a product of all its arguments. Again, the first example is a sensible approach, assuming zero is meant to be ignored.

@mtf Much appreciated

1 Like