python如何替换列表中的量_使用Python 替换列表中的值

使用列表解析构建新列表：

new_items = [x if x % 2 else None for x in items]

如果需要,您可以就地修改原始列表,但实际上并不节省时间：

items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

for index, item in enumerate(items):

if not (item % 2):

items[index] = None

以下是(Python 3.6.3)演示非时间的时间：

In [1]: %%timeit

...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

...: for index, item in enumerate(items):

...: if not (item % 2):

...: items[index] = None

...:

1.06 ?s ± 33.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [2]: %%timeit

...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

...: new_items = [x if x % 2 else None for x in items]

...:

891 ns ± 13.6 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

和Python 2.7.6时序：

In [1]: %%timeit

...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

...: for index, item in enumerate(items):

...: if not (item % 2):

...: items[index] = None

...:

1000000 loops, best of 3: 1.27 ?s per loop

In [2]: %%timeit

...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

...: new_items = [x if x % 2 else None for x in items]

...:

1000000 loops, best of 3: 1.14 ?s per loop