Python: pass "not" as a lambda function

Question

I need to pass a function as a parameter that works as the boolean "not". I tried something like this but it didn't work because not isn't a function.

theFunction(callback=not) # Doesn't work :(

I need to do the following, but I wonder if there exists any predefined function that does this simple job, so that I don't have to redefine it like this:

theFunction(callback=lambda b: not b, anotherCallback=lambda b: not b)

Note: I can't change the fact that I have to pass a function like this because it's an API call.

[**`operator.not_`**](https://docs.python.org/3/library/operator.html#operator.not_) — Peter Wood, Feb 04 '18 at 18:13
Instead of your lambda example you could've still been mostly idiomatic with `def not_(b): return not b` and then `theFunction(callback=not_, anotherCallBack=not_)`. — Andras Deak -- Слава Україні, Feb 04 '18 at 20:54

score 26 · Accepted Answer · answered Feb 04 '18 at 18:13

26

Yes, there is the operator module: https://docs.python.org/3.6/library/operator.html

import operator
theFunction(callback=operator.not_)

answered Feb 04 '18 at 18:13

quamrana

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score 14 · Answer 2 · answered Feb 04 '18 at 18:15

not is not a function, but a keyword. So that means you can not pass a reference. There are good reasons, since it allows Python to "short circuit" certain expressions.

You can however use the not_ (with underscore) of the operator package:

from operator import not_

theFunction(callback=not_, anotherCallback=not_)

Python: pass "not" as a lambda function

2 Answers2