打印包含另一个列表中所有字符串的子列表，而不必是直接匹配

Question

search_terms = ['word','cow','horse']

library = [['desk','chair','lamp'],['cow','horse','word 223','barn']]

我希望能够在库中打印包含search_terms中所有术语的所有列表。

所以使用上面的search_terms列表会在库中打印第二个子列表，即使'word 223'只包含'word'，但不是直接匹配。

我不会总是拥有相同数量的字符串...

感谢任何愿意帮助我的人！

感谢falsetru帮助我解决第一个问题！

Answer 1

要获得点击量，请使用列表理解：

search_terms = ['word', 'cow', 'horse']

library = [['desk', 'chair', 'lamp'],
           ['cow', 'horse', 'word 223', 'barn']]

hits = [l for l in library if 
        all(any(t in s for s in l) 
            for t in search_terms)]

这可以如下工作

1. for l;
中的每个子列表library {li> for all条t search_terms; {li> if any s中l的{​​{1}}包含它;
2. 将l保留在新列表hits中。

Answer 2

>>> search_terms = ['word','cow','horse']
>>> library = [['desk','chair','lamp'],['cow','horse','word 223','barn']]
>>> from itertools import chain
>>> list(chain(*library))
['desk', 'chair', 'lamp', 'cow', 'horse', 'word 223', 'barn']
>>> [word for word in search_terms if word in list(chain(*library))]
['cow', 'horse']
>>> [l for l in library if any(word for word in search_terms if word in l)]
[['cow', 'horse', 'word 223', 'barn']]