时间比较：

Question

我有以下代码（元组列表）：

x = [(None, None, None), (None, None, None), (None, None, None)]

我怎么知道这基本上评估为False？

换句话说，我怎么能这样做：

if not x: # x should evaluate to False
    # do something

Answer 1

使用嵌套的any()来电：

if not any(any(inner) for inner in x):

仅当传递给它的中所有元素都是any()时，

False才会返回False。只有当所有元素都为假时，not any()才是True：

>>> x = [(None, None, None), (None, None, None), (None, None, None)]
>>> not any(any(inner) for inner in x)
True
>>> x = [(None, None, None), (None, None, None), (None, None, 1)]
>>> not any(any(inner) for inner in x)
False

Answer 2

您可以将any与itertools.chain一起使用，它比嵌套any更快，因为不需要python for-loop：

很少有其他选择：

not any (chain.from_iterable(x))
not any (chain(*x))
not any(map(any,x))
not any(imap(any,x))   #itertools.imap

时间比较：

>>> from itertools import chain,imap
>>> x = [(None, None, None), (None, None, None), (None, None, None)]
>>> %timeit not any (chain.from_iterable(x))       #winner
100000 loops, best of 3: 1.08 us per loop
>>> %timeit not any (chain(*x))
1000000 loops, best of 3: 1.29 us per loop
>>> %timeit not any(any(inner) for inner in x)
100000 loops, best of 3: 1.76 us per loop
>>> %timeit not any(map(any,x))
100000 loops, best of 3: 1.5 us per loop
>>> %timeit not any(imap(any,x))
100000 loops, best of 3: 1.37 us per loop


>>> x = [(None, None, None), (None, None, None), (None, None, None)]*1000
>>> %timeit not any (chain.from_iterable(x))      #winner
1000 loops, best of 3: 462 us per loop
>>> %timeit not any (chain(*x))
1000 loops, best of 3: 537 us per loop
>>> %timeit not any(any(inner) for inner in x)
1000 loops, best of 3: 1.26 ms per loop
>>> %timeit not any(map(any,x))
1000 loops, best of 3: 672 us per loop
>>> %timeit not any(imap(any,x))
1000 loops, best of 3: 765 us per loop

>>> x = [(None, None, None), (None, None, None), (None, None, None)]*10**5
>>> %timeit not any (chain.from_iterable(x))       #winner
10 loops, best of 3: 50.1 ms per loop        
>>> %timeit not any (chain(*x))                  
10 loops, best of 3: 70.3 ms per loop
>>> %timeit not any(any(inner) for inner in x)
10 loops, best of 3: 127 ms per loop
>>> %timeit not any(map(any,x))
10 loops, best of 3: 76.2 ms per loop
>>> %timeit not any(imap(any,x))
1 loops, best of 3: 64.9 ms per loop

检查嵌套iterables中的所有元素是否都计算为False

2 个答案:

时间比较：