用字符串按任意顺序匹配数组元素

Question

我对python还很陌生，正在尝试找出tweet是否有任何查找元素。你知道吗

如果我能找到“猫”这个词，它应该和“猫”匹配，可爱的小猫也可以任意搭配。但从我的理解来看，我找不到解决办法。感谢您的指导。你知道吗

import re
lookup_table = ['cats', 'cute kittens', 'dog litter park']
tweets = ['that is a cute cat',
          'kittens are cute',
          'that is a cute kitten',
          'that is a dog litter park',
          'no wonder that dog park is bad']
for tweet in tweets:
    lookup_found = None
    print re.findall(r"(?=(" + '|'.join(lookup_table) + r"))", tweet.lower())

输出

['cat']
[]
[]
['dog litter park']
[]

预期产量：

that is a cute cat > cats
kittens are cute > cute kittens
this is a cute kitten > cute kittens
that is a dog litter park > dog litter park
no wonder that dog park is bad > dog litter park

Answer 1

对于仅为一个单词文本的查找单词，可以使用

for word in tweet

对于像“可爱的小猫”这样的查找词，您可以在其中查看任何顺序。只需将单词拆分并在tweet字符串中查找即可。你知道吗

这是我尝试过的，它不是有效的，但它的工作。试着运行它。你知道吗

lookup_table = ['cat', 'cute kitten', 'dog litter park']
tweets = ['that is a cute cat',
          'kittens are cute',
          'that is a cute kitten',
          'that is a dog litter park',
          'no wonder that dog park is bad']

for word in lookup_table:
    for tweet in tweets:
        if " " in word:
            temp = word.split(sep=" ")
        else:
            temp = [word]
        for x in temp:
            if x in tweet:
                print(tweet)
                break