检查条件是否满足列表中任何元素的Python方法

问题描述

我在Python中有一个列表，我想检查是否有任何负数. Specman具有用于列表的has()方法，

I have a list in Python, and I want to check if any elements are negative. Specman has the has() method for lists which does:

x: list of uint;
if (x.has(it < 0)) {
    // do something
};

it是一个Specman关键字，依次映射到列表的每个元素.

Where it is a Specman keyword mapped to each element of the list in turn.

我觉得这很优雅.我浏览了Python文档找不到类似的东西.我能想到的最好的是:

I find this rather elegant. I looked through the Python documentation and couldn't find anything similar. The best I could come up with was:

if (True in [t < 0 for t in x]):
    # do something

我觉得这很不雅致.在Python中有更好的方法吗?

I find this rather inelegant. Is there a better way to do this in Python?

推荐答案

any():

if any(t < 0 for t in x):
    # do something

此外，如果要使用"True in ..."，请将其设为生成器表达式，这样就不会占用O(n)内存:

Also, if you're going to use "True in ...", make it a generator expression so it doesn't take O(n) memory:

if True in (t < 0 for t in x):

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