What is the most pythonic way to check if multiple variables are not None?

Question

If I have a construct like this:

def foo():
    a=None
    b=None
    c=None

    #...loop over a config file or command line options...

    if a is not None an         


        

Answer 1

I know this is an old question, but I wanted to add an answer which I believe is better.

If all elements which have to be checked are hashable, you could use a set instead of a list or tuple.

>>> None not in {1, 84, 'String', (6, 'Tuple'), 3}

This is much faster than the methods in the other answers.

>>> import timeit
>>> timeit.timeit("all(v is not None for v in [1, 84, 'String', (6, 'Tuple'), 3])")
1.7880705000000034
>>> timeit.timeit("None not in [1, 84, 'String', (6, 'Tuple'), 3]")
0.35424169999998867
>>> timeit.timeit("None not in (1, 84, 'String', (6, 'Tuple'), 3)")
0.3454340999999772
>>> timeit.timeit("None not in {1, 84, 'String', (6, 'Tuple'), 3}")
0.09577370000002361

Another advantage of this method is that it gives you the correct answer even if someone defines the __eq__ method of a class to always return True. (Of course, if they define the __hash__ method to return hash(None), this method won't work. But nobody should do that, because it would defeat the purpose of defining a hash.)

class my_int(int):
    def __init__(self, parent):
        super().__init__()

    def __eq__(self, other):
        return True

    def __hash__(self):
        return hash(super())

print(all(v is not None for v in [1, my_int(6), 2])) # True (correct)
print(None not in [1, my_int(6), 2])                 # False (wrong)
print(None not in (1, my_int(6), 2))                 # False (wrong)
print(None not in {1, my_int(6), 2})                 # True (correct)