How to distinguish unassigned variable from zero in Python?

Question

Some external code runs my function of the following code:

def __init__(self,weights=None,threshold=None):

    print "weights: ", weights
    print "threshold: ", threshold

    if weights:
        print "weights assigned"
        self.weights = weights
    if threshold:
        print "threshold assigned"
        self.threshold = threshold

And this code outputs:

weights:  [1, 2]
threshold:  0
weights assigned

I.e. print operator behaves like threshold is zero, while if operator behaves like it was not defined.

What is the correct interpretation? What is happening? What is the state of threshold parameter and how to recognize it?

I'm not sure what you're asking. `0` is false in a boolean context in Python. — Daniel Roseman, Jun 19 '16 at 15:05
So, they are passing zero for `threshold`. And what is assigned to `self.threshold` in result? — Dims, Jun 19 '16 at 15:09
There is not such thing as an undefined variable in Python. I am guessing you mean `threshold=None` means undefined, but it means `threshold` is default assigned `None` if no argument is given. Both `None` and `0` are considered `False`. — totoro, Jun 19 '16 at 15:10
@DanielRoseman and what is "nothing"? The same as `None` or not? — Dims, Jun 19 '16 at 17:38
No. It is not defined at all. Because the code never enters the block where it is defined. — Daniel Roseman, Jun 19 '16 at 17:59
@DanielRoseman Is it possible to make that nothing for omitted parameters? — Dims, Jun 19 '16 at 20:32
@DanielRoseman. no I mean how to define optional parameter for a function, which appears in the state of "nothing" if not passed? — Dims, Jun 19 '16 at 20:54

tzaman · Answer 1 · 2016-06-19T15:12:40.777

6

Use if weights is not None instead of if weights.

More detail: when you say if weights you're asking Python to evaluate weights in a boolean context, and many things can be "false-equivalent" (or "falsy") including 0, empty strings, empty containers, etc. If you want to only check for a None value, you have to do that explicitly.

edited Jun 19 '16 at 15:12

answered Jun 19 '16 at 15:06

tzaman

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Martin Hallén · Answer 2 · 2016-06-19T15:14:00.753

0

You can explicitly test for a None value.

def __init__(self,weights=None,threshold=None):
    print "weights: ", weights
    print "threshold: ", threshold

    if weights is not None:
        print "weights assigned"
        self.weights = weights
    if threshold is not None:
        print "threshold assigned"
        self.threshold = threshold

edited Jun 19 '16 at 15:14

answered Jun 19 '16 at 15:07

Martin Hallén

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`if not weights == None:` isn't it the same as `if weights != None`? Also, you can test for the `type()` too, though that's not very pythonic. – Moon Cheesez Jun 19 '16 at 15:11
1

You should generally use identity comparison against `None`, not equality. (i.e. `is None` or `is not None` instead of `==` and `!=`. – tzaman Jun 19 '16 at 15:12
You are correct. I fixed the answer. Thx! – Martin Hallén Jun 19 '16 at 15:14

How to distinguish unassigned variable from zero in Python?

2 Answers2