I was wondering if we could assert all elements in a list is not None, therefore while a = None will raise an error.
The sample list is [a, b, c]
I have tried assert [a, b, c] is not None, it will return True if any one of the elements is not None but not verifying all. Could you help figure it out? Thanks!!
Unless you have a weird element that claims it equals None:
assert None not in [a, b, c]
all and checking if each element is not None, but very close in terms of performance to all([a, b, c])
all([a, b, c]) is significantly less safe, though, as any false value (e.g., a zero) will make that return False.None check, which seems slow by comparison.Do you mean by all:
>>> assert all(i is not None for i in ['a', 'b', 'c'])
>>>
Or just simply:
assert None not in ['a', 'b', 'c']
P.S just noticed that @don'ttalkjustcode added the above.
Or with min:
>>> assert min(a, key=lambda x: x is not None, default=False)
>>>
default should be True, though I find min very weird for this. And for example it cries AssertionError for [0, 1, 2] even though there's no None. Btw the correct faster all(...) that I think you were trying earlier is as expected indeed faster, see @rv's benchmark.Midway in terms of performence between not in and all. Note that the sensible (go-to) version with all for this particular case will end up performing slow - but at least ahead of min
def assert_all_not_none(l):
for x in l:
if x is None:
return False
return True
Edit: here are some benchmarks for those intersted
from timeit import timeit
def all_not_none(l):
for b in l:
if b is None:
return False
return True
def with_min(l):
min(l, key=lambda x: x is not None, default=False)
def not_in(l):
return None not in l
def all1(l):
return all(i is not None for i in l)
def all2(l):
return all(False for i in l if i is None)
def all_truthy(l):
return all(l)
def any1(l):
return any(True for x in l if x is None)
l = ['a', 'b', 'c'] * 20_000
n = 1_000
# 0.63
print(timeit("all_not_none(l)", globals=globals(), number=n))
# 3.41
print(timeit("with_min(l)", globals=globals(), number=n))
# 1.66
print(timeit('all1(l)', globals=globals(), number=n))
# 0.63
print(timeit('all2(l)', globals=globals(), number=n))
# 0.63
print(timeit('any1(l)', globals=globals(), number=n))
# 0.26
print(timeit('all_truthy(l)', globals=globals(), number=n))
# 0.53
print(timeit('not_in(l)', globals=globals(), number=n))
Surprisingly the winner: all(list). Therfore, if you are certain list will not contain falsy values like empty string or zeros, nothing wrong with going with that.
all_not_none would normally be used inline (like if any(l):) without extra function call, while all_not_none would be used with a call (like if all_not_none(l)). But alright otherwise. Could also include the all(False ...) one I mentioned in a comment, should be faster for the same reason as this.all and it was really much faster. nice catch :)any should be not any.
assert [a, b, c] is not Nonewill pass even if all the elements areNone. The only thing thatis Noneis...None.ismeans the same object, not an equality check. It also is not possible to create more instances ofNone's type.None. Any list is not None, in fact, any object exceptNoneis not None.