最节省时间/空间的方法来检查整数列表中的所有元素是否为0

Question

This is an implementation question for Python 2.7

Say I have a list of integers called nums , and I need to check if all values in nums are equal to zero. nums contains many elements (ie more than 10000), with many repeating values.

Using all() :

if all(n == 0 for n in set(nums)):  # I assume this conversion from list to set helps?
    # do something

Using set subtraction:

if set(nums) - {0} == set([]):
    # do something

Edit: better way to do the above approach, courtesy of user U9-Forward

if set(nums) == {0}:
    # do something

How do the time and space complexities compare for each of these approaches? Is there a more efficient way to check this?

Note: for this case, I am trying to avoid using numpy/pandas.

Answer 1

Any set conversion of nums won't help as it will iterate the entire list:

if all(n == 0 for n in nums):
    # ...

is just fine as it stops at the first non-zero element, disregarding the remainder.

Asymptotically, all these approaches are linear with random data. Implementational details (no repeated function calls on the generator) makes not any(nums) even faster, but that relies on the absence of any other falsy elements but 0 , eg '' or None .

Answer 2

not any(nums) is probably the fastest because it will stop when/if it finds any non-zero element.

Performance comparison:

a = range(10000)
b = [0] * 10000
%timeit not any(a) # 72 ns, fastest for non-zero lists
%timeit not any(b) # 33 ns, fastest for zero lists
%timeit all(n == 0 for n in a) # 365 ns
%timeit all(n == 0 for n in b) # 350 µs
%timeit set(a)=={0} # 228 µs
%timeit set(b)=={0} # 58 µs

Answer 3

如果可以使用numpy，则(np.array(nums) == 0).all()应该这样做。

Answer 4

Additionally to @schwobaseggl's answer, second example could be even better:

if set(nums)=={0}:
    # do something