我有一个列表,其中我希望将值替换为None,其中condition()返回True。
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
例如,如果条件检查bool(项%2)应返回:
[None, 1, None, 3, None, 5, None, 7, None, 9, None]
执行此操作的最有效方法是什么?
发布于 2009-10-08 19:58:51
使用列表理解构建一个新列表:
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,但这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
以下是演示非timesave的(Python 3.6.3)计时:
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 µs ± 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计时:
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 µs 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 µs per loop
发布于 2009-10-08 19:58:58
ls = [x if (condition) else None for x in ls]
发布于 2009-10-09 10:54:17
这里有另一种方法:
>>> L = range (11)
>>> map(lambda x: x if x%2 else None, L)
[None, 1, None, 3, None, 5, None, 7, None, 9, None]
https://stackoverflow.com/questions/1540049
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