检查条件是否满足列表中任何元素的Python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
};

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

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

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

any():

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

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):

Use any().

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

Python has a built in any() function for exactly this purpose.