检查列表列表中是否存在项目的最佳方法是什么？

Question

我有这样的示例列表:

example_list = [['aaa'], ['fff', 'gg'], ['ff'], ['', 'gg']]

现在,我检查它是否有这样的空字符串:

has_empty = False;
for list1 in example_list:
    for val1 in list1:
        if val1 == '':
            has_empty = True

print(has_empty)

这可以正常,因为它打印True,但寻找更多的pythonik方法？

Answer 1

你可以使用itertools.chain.from_iterable:

>>> from itertools import chain
>>> example_list = [['aaa'], ['fff', 'gg'], ['ff'], ['', 'gg']]
>>> '' in chain.from_iterable(example_list)
True

如果内部列表更大(超过100个项目),那么使用any生成器将比上面的示例更快,因为使用Python for循环的速度惩罚可以通过快速in操作来补偿:

>>> any('' in x for x in example_list)
True

时间比较:

>>> example_list = [['aaa']*1000, ['fff', 'gg']*1000, ['gg']*1000]*10000 + [['']*1000]
>>> %timeit '' in chain.from_iterable(example_list)
1 loops, best of 3: 706 ms per loop
>>> %timeit any('' in x for x in example_list)
1 loops, best of 3: 417 ms per loop

# With smaller inner lists for-loop makes `any()` version little slow

>>> example_list = [['aaa'], ['fff', 'gg'], ['gg', 'kk']]*10000 + [['']]
>>> %timeit '' in chain.from_iterable(example_list)
100 loops, best of 3: 2 ms per loop
>>> %timeit any('' in x for x in example_list)
100 loops, best of 3: 2.65 ms per loop