从列表中删除 None 值而不删除 0 值 [英] remove None value from a list without removing the 0 value

问题描述

这是我开始的来源.

我的列表

L = [0, 23, 234, 89, None, 0, 35, 9]

当我运行这个时:

L = filter(None, L)

我得到这个结果

[23, 234, 89, 35, 9]

但这不是我需要的，我真正需要的是:

[0, 23, 234, 89, 0, 35, 9]

因为我正在计算数据的百分位数，而 0 有很大的不同.

如何在不删除 0 值的情况下从列表中删除 None 值?

解决方案

>>>L = [0, 23, 234, 89, 无, 0, 35, 9]>>>[x 代表 L 中的 x，如果 x 不是 None][0, 23, 234, 89, 0, 35, 9]

只是为了好玩，这里介绍了如何在不使用 lambda 的情况下调整 filter 来执行此操作，(我不推荐此代码 - 它仅用于科学目的)

<预><代码>>>>from 操作符导入 is_not>>>从 functools 导入部分>>>L = [0, 23, 234, 89, 无, 0, 35, 9]>>>过滤器(部分(is_not，无)，L)[0, 23, 234, 89, 0, 35, 9]

This was my source I started with.

My List

L = [0, 23, 234, 89, None, 0, 35, 9]

When I run this :

L = filter(None, L)

I get this results

[23, 234, 89, 35, 9]

But this is not what I need, what I really need is :

[0, 23, 234, 89, 0, 35, 9]

Because I'm calculating percentile of the data and the 0 make a lot of difference.

How to remove the None value from a list without removing 0 value?

解决方案

>>> L = [0, 23, 234, 89, None, 0, 35, 9]
>>> [x for x in L if x is not None]
[0, 23, 234, 89, 0, 35, 9]

Just for fun, here's how you can adapt filter to do this without using a lambda, (I wouldn't recommend this code - it's just for scientific purposes)

>>> from operator import is_not
>>> from functools import partial
>>> L = [0, 23, 234, 89, None, 0, 35, 9]
>>> filter(partial(is_not, None), L)
[0, 23, 234, 89, 0, 35, 9]

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