Python word match

Question

我有一個url列表，我試圖用特定的關鍵詞來說明word1和word2，以及一個停用詞表[stop1，stop2，stop3]。有沒有一種方法可以在不使用許多條件的情況下過濾鏈接？當我使用每個停用詞的條件時，我得到了正確的輸出，這看起來不是一個可行的選擇。以下是蠻力法：

for link in url: 
    if word1 or word2 in link: 
     if stop1 not in link: 
      if stop2 not in link: 
       if stop3 not in link: 
        links.append(link) 

Answer 1

如果您可以舉一個例子，那麼它會有所幫助。如果我們把網址範例像

def urlSearch(): 
    word = [] 
    end_words = ['gmail', 'finance'] 
    Key_word = ['google'] 
    urlList= ['google.com//d/gmail', 'google.com/finance', 'google.com/sports', 'google.com/search'] 
    for i in urlList: 
     main_part = i.split('/',i.count('/')) 
     if main_part[len(main_part) - 1] in end_words: 
      word = [] 
      for k in main_part[:-1]: 
       for j in k.split('.'): 
        word.append(j) 
      print (word) 
     for p in Key_word: 
      if p in word: 
       print ("Url is: " + i) 

urlSearch() 

Answer 2

這裏有一對夫婦，如果我是你的情況，我會考慮的選項。

您可以使用內置的any和all功能列表解析來過濾掉從列表中選擇不需要的網址：

urls = ['http://somewebsite.tld/word', 
     'http://somewebsite.tld/word1', 
     'http://somewebsite.tld/word1/stop3', 
     'http://somewebsite.tld/word2', 
     'http://somewebsite.tld/word2/stop2', 
     'http://somewebsite.tld/word3', 
     'http://somewebsite.tld/stop3/word1', 
     'http://somewebsite.tld/stop4/word1'] 

includes = ['word1', 'word2'] 
excludes = ['stop1', 'stop2', 'stop3'] 

filtered_url_list = [url for url in urls if any(include in url for include in includes) if all(exclude not in url for exclude in excludes)] 

或者你可以做一個函數，它接受一個URL作爲參數，並返回那些True你想保留的URL和False你不這樣做，那麼傳遞函數的URL的未篩選列表一起內置的filter功能：

def urlfilter(url): 
    includes = ['word1', 'word2'] 
    excludes = ['stop1', 'stop2', 'stop3'] 
    for include in includes: 
     if include in url: 
      for exclude in excludes: 
       if exclude in url: 
        return False 
      else: 
       return True 

urls = ['http://somewebsite.tld/word', 
     'http://somewebsite.tld/word1', 
     'http://somewebsite.tld/word1/stop3', 
     'http://somewebsite.tld/word2', 
     'http://somewebsite.tld/word2/stop2', 
     'http://somewebsite.tld/word3', 
     'http://somewebsite.tld/stop3/word1', 
     'http://somewebsite.tld/stop4/word1'] 

filtered_url_list = filter(urlfilter, urls) 

Answer 3

我會用集和列表理解：

must_in = set([word1, word2]) 
musnt_in = set([stop1, stop2, stop3]) 
links = [x for x in url if must_in & set(x) and not (musnt_in & set(x))] 
print links 

上述代碼可以用任何數目的字被用於停止，不限於兩個單詞（WORD1，單詞2）和三個站（停止1，停止2，停止3） 。