使用字符串以任意順序匹配數組元素

-2

我對python很陌生，試圖找到tweet是否有任何查找元素。使用字符串以任意順序匹配數組元素

例如，如果我能找到這個單詞貓，它應該匹配貓，也可以任意順序匹配可愛的小貓。但從我瞭解我無法找到解決方案。任何指導表示讚賞。

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())

輸出

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

預期輸出：

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

來源

2017-04-18 user6083088

?? ??使用單數形式。 –

你也應該告訴我們你實際需要的輸出。 –

@KarolyHorvath我不確定你的意思是 – user6083088

對於查找的話這是隻有一個字的文字，你可以使用

for word in tweet

而對於像查找單詞「可愛的小貓」，你在哪裏等任何訂單。只需將它分開並在推文字符串中查找即可。

這是我試過的，它效率不高，但它的工作。嘗試運行它。

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

來源

2017-04-18 10:16:38

這是我該怎麼做。我認爲lookup_table不必太嚴格，我們可以避免複數;

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)

來源

2017-04-18 10:27:36 gr8tech

問題1：

單/複數： 只是爲了讓事情滾動我會用活用，Python包擺脫單一&複數，例如...

問題2：

分裂和加入： 我寫了一個小腳本來演示率你如何使用它，沒有穩健測試，但應該讓你移動

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

來源

2017-04-18 11:19:06

使用字符串以任意順序匹配數組元素

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