小编典典

检查列表中是否存在项目的最佳方法?

python

我有这样的示例列表:

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

现在,我检查它是否具有空字符串,如下所示:

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

print(has_empty)

这可以正常工作,因为它可以打印True,但是是否需要更多的pythonik方法?


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2020-12-20

共1个答案

小编典典

您可以使用itertools.chain.from_iterable

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

万一如果内部列表更大(超过100个项目),则any与生成器一起使用的速度将比上面的示例更快,因为这样,使用Python
for循环的速度代价将由快速in操作来补偿:

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

时序比较:

>>> 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
2020-12-20