Unpack a list with ternary operator

Question

I stumbled upon a weird behaviour when I want to unpack a list only if condition is true.

How can I use the unpacking (*) based on a condition?

Example:

def foo1(x1, x2):
    print(x1, x2)

def foo2(x):
    print(x)

l = [7,8]
foo1(*l if True else l) # this works
foo2(*l if False else l) # this does not work
foo1(l if not True else *l) # this does not work

Asked By: David

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Source

Answer 1

That’s not parsed as choosing between *l on the left and l on the right. l if True else l is evaluated, and then the result is unpacked, unconditionally.

There’s no way to have a function call expression conditionally either unpack or not unpack an argument. Something’s getting unpacked if you put a * there. What you can do is wrap l in another list and unpack that:

foo(*(l if condition else [l]))

Answered By: user2357112

Unpack a list with ternary operator

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