Python keyword arguments referencing each other

Question

When I try the following I get an error

def test_func(key1=2.7, key2=key1*3.5):
    print(key1, key2)

NameError: name 'key1' is not defined

My solution would be something like

def test_func(key1=2.7, key2=None):
    if not key2:
        key2 = key1*3.5

    print(key1, key2)

but this looks kind of ugly to me. Does anybody have a better solution?

edit:

so my final solution is

def test_func(key1=2.7, key2=None):
    if key2 is not None:
        key2 = key1*3.5

    print(key1, key2)

thanks for all answers

maybe you should even change that to an explicit `is key2 is not None`, otherwise users will be unable to pass `0` as the second argument. — Niklas B., Feb 24 '13 at 11:05
the numbers were just an example, the function i use looks quite different, but you are right. i did not think on that. — Peter, Feb 24 '13 at 11:12

score 5 · Accepted Answer · edited May 23 '17 at 11:53

5

Nope, there is no better solution.

Function argument definitions can be expressions, but they are evaluated only once (which sometimes surprises people, see "Least Astonishment" and the Mutable Default Argument).

edited May 23 '17 at 11:53

Community

answered Feb 24 '13 at 11:04

Martijn Pieters

score 0 · Answer 2 · answered Feb 24 '13 at 11:24

0

My solution would be:

def t(key1=1, **kwargs):
    key2 = kwargs.get('key2', key1*2.7)

getting key2 from kwargs and get data from it with the default key1*2.7

answered Feb 24 '13 at 11:24

Julian Hille

That has the disadvantage that the user will not get an error if he passes an unexpected parameter, like `key3`. It will just silently ignore it. – Niklas B. Feb 24 '13 at 11:28

2 Answers2