使用Python替换列表中的值 [英] Replace values in list using Python

问题描述

我有一个列表，希望在其中condition()返回True的情况下用None替换值.

I have a list where I want to replace values with None where condition() returns True.

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

例如，如果条件检查bool(item％2)应该返回:

For example, if condition checks bool(item%2) should return:

[None, 1, None, 3, None, 5, None, 7, None, 9, None]

最有效的方法是什么?

推荐答案

使用列表理解功能构建新列表:

Build a new list with a list comprehension:

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，但实际上并没有节省时间:

You can modify the original list in-place if you want, but it doesn't actually save time:

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(Python 3.6.3)计时说明了非节时:

Here are (Python 3.6.3) timings demonstrating the non-timesave:

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的计时:

And Python 2.7.6 timings:

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

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