单行检查列表中的至少一项是否存在于另一个列表中? [英] one-liner to check if at least one item in list exists in another list?

问题描述

假设我有一个清单a=[1,2,3]我想知道其中是否至少有一个数字存在于另一个列表中，例如:b=[4,5,6,7,8,1]换句话说，我想知道列表 b 中是否存在 1,2 或 3 个.我现在可以做类似的事情

Lets say I have a list a=[1,2,3] And I want to know if at least one of the numbers in it exist in another list, like this one: b=[4,5,6,7,8,1] In other words, I want to know if 1,2 or 3 exist(s) in list b. I now I could do something like

def func(a, b):
    for i in a:
       if i in b:
          return True
    return False

但是有没有办法把它放在一行中让事情变得整洁?

But is there possibly a way to put that in one line to make things tidy?

推荐答案

有很多方法可以做到这一点.最直接的翻译是:

There are many ways to do this. The most direct translation is:

any_in = lambda a, b: any(i in b for i in a)

您还可以使用各种涉及集合的东西，例如:

You could also use various things involving sets, such as:

any_in = lambda a, b: bool(set(a).intersection(b))

(这取决于 a 的元素是否可以散列，但如果这是真的，从 a 和b 适用于这些方法中的任何一种).

(which depends on the elements of a being hashable, but if that's true, it'll probably be faster to make a set of the larger out of a and b for either of these approaches).

对于 Python 2.6 及更高版本，isdisjoint 比 intersection 更好，正如下面不同的人所指出的.很高兴了解这一点.:)

isdisjoint is better than intersection for Python 2.6 and above, as noted by various people below. Glad to learn about that. :)

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