Question

我是python的新手，并试图找到一条推文是否有任何查找元素。

例如。如果我能找到它应该与猫匹配的单词，那么可爱的小猫可以任意顺序匹配。但据我所知，我无法找到解决方案。任何指导表示赞赏。

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']
[]

预期产出：

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

对于只有一个单词文本的查找单词，您可以使用

for word in tweet

对于像“可爱的小猫”这样的查找单词，您希望任何订单。只需将该单词拆分并在推文字符串中查找即可。

这是我尝试过的，它效率不高但是它的工作原理。尝试运行它。

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

Answer 2

我将如何做到这一点。我认为lookup_table不需要太严格，我们可以避免复数;

import re
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 data in lookup_table:
    words=data.split(" ")
    for word in words:
        result=re.findall(r'[\w\s]*' + word + '[\w\s]*',','.join(tweets))
        if len(result)>0:
            print(result)

Answer 3

问题1：

单数/复数： 只是为了让事情滚动，我会使用inflect，一个python包来摆脱奇异＆amp;复数等...

问题2：

拆分和加入： 我写了一个小脚本来演示如何使用它，没有经过强大的测试，但应该让你感动

import inflect 
p = inflect.engine()
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:
    matched = []
    for lt in lookup_table:
            match_result = [lt for mt in lt.split() for word in tweet.split() if p.compare(word, mt)]
            if any(match_result):
                matched.append(" ".join(match_result))
    print tweet, '>>' , matched

使用字符串以任意顺序匹配数组元素

3 个答案: