Python 检查任何列表中的所有项是否在字符串中的最佳方法

我有一个列表，如

mylist=[[“1”、“2”]、[“abc”、“def”]]

和一个字符串，如

mystr=“1 2”

我想检查

mylist

中是否有任何列表的所有字符串都在字符串中

我通过以下几点做到了这一点

if True in [all(keyword in mystr for keyword in keywords) for keywords in mylist]:
    print("yes")

---

有没有更快的方法？

是的，使用

any

if any(all(keyword in mystr for keyword in keywords) for keywords in mylist):
    print("yes")

这会更快，因为它一看到真实值就停止迭代（短路）。它还有一个副作用，就是更容易阅读

谢谢你提到这一点

基准：

>>> min(repeat(lambda: any(all(map(matcher, keywords)) for keywords in mylist), repeat=20))
1.1285329000002093
>>> min(repeat(lambda: any(all(map(mystr.__contains__, keywords)) for keywords in mylist), repeat=20))
1.2246240000004036
>>> min(repeat(lambda: any(all(keyword in mystr for keyword in keywords) for keywords in mylist), repeat=20))
1.3369910999999775
>>> min(repeat(lambda: True in [all(keyword in mystr for keyword in keywords) for keywords in mylist], repeat=20))
1.726889200000187

>>> min(repeat(lambda: any(all(map(matcher, keywords)) for keywords in mylist), repeat=20))
1.1285329000002093
>>> min(repeat(lambda: any(all(map(mystr.__contains__, keywords)) for keywords in mylist), repeat=20))
1.2246240000004036
>>> min(repeat(lambda: any(all(keyword in mystr for keyword in keywords) for keywords in mylist), repeat=20))
1.3369910999999775
>>> min(repeat(lambda: True in [all(keyword in mystr for keyword in keywords) for keywords in mylist], repeat=20))
1.726889200000187