リスト内の単語を数えるには？

Question

from bs4 import BeautifulSoup 
import urllib2 
# Imported libraries for future use. 
response = urllib2.urlopen('http://www.nytimes.com').read() 
soup = BeautifulSoup(response,"lxml") 

host = [] 
#created empty list to append future words extracted from data set. 
for story_heading in soup.find_all(class_="story-heading"): 
    story_title = story_heading.text.replace("\n", " ").strip() 
    new_story_title = story_title.encode('utf-8') 


    parts = new_story_title.split()[0] 

    i=['a','A','an','An','the','The','from','From','to','To','when','When','what','What','on','On','for','For'] 
    if parts not in i: 
     host.append(parts) 
    else: 
     pass 
#now i have to calculate the number of repeated words in the file and calcute the number of repeatation.  
print host 

作成したリストの繰り返し単語の数を計算する方法を教えてください。 実際には上記のコードについてもかなり混乱しています。 誰かが私が間違ったことを説明できるのであれば、それは感謝するでしょう。

Answer 1

あなたは行うことができますとcount

d = {i: host.count(i) for i in set(host)} 
print(d) 

Answer 2

使用のコレクションにCounter方法モジュール：

from bs4 import BeautifulSoup 
from collections import Counter 
import urllib2 
# Imported libraries for future use. 
response = urllib2.urlopen('http://www.nytimes.com').read() 
soup = BeautifulSoup(response,"lxml") 

host = [] 
#created empty list to append future words extracted from data set. 
for story_heading in soup.find_all(class_="story-heading"): 
    story_title = story_heading.text.replace("\n", " ").strip() 
    new_story_title = story_title.encode('utf-8') 


    parts = new_story_title.split()[0] 

    i=['a','A','an','An','the','The','from','From','to','To','when','When','what','What','on','On','for','For'] 
    if parts not in i: 
     host.append(parts) 
    else: 
     pass 
#now i have to calculate the number of repeated words in the file and calcute the number of repeatation.  
print Counter(host) 

出力：

>>> ================================ RESTART ================================ 
>>> 
Counter({'North': 2, 'Trump': 1, 'U.S.': 1, 'Kasich-Cruz': 1, '8': 1, 'Court': 1, 'Where': 1, 'Your': 1, 'Forget': 1}) 
>>> 

Answer 3

あなたは以下のコードを見ることができますlisを使わないt理解。これは理解しやすいはずです。

host = ['Hello','foo','bar','World','foo','Hello'] 
dict1 = {} 
host_unique = list(set(host)) 
for i in host_unique: 
    dict[i] = host.count(i) 

Answer 4

要素のセットを反復辞書内包表記の使用：

• 、大文字と小文字を区別したバージョンを（ "何" = "何を"！）：

  occurrences = { item: host.count(item) for item in set(host) } 
  

• 大文字と小文字を区別しないバージョン（ "What" == "what"）：

  occurrences = { item: host.count(item) for item in set(item.lower() for item in host) } 
  

  この場合、辞書キーも小文字の要素になります。

Answer 5

用途：

lst = ['hi', 'Hio', 'Hi', 'hello', 'there' ] 
s = set() 
map(lambda x: s.add(x.lower()), lst) 
print(len(s)) 

OR

lst = ['hi', 'Hio', 'Hi', 'hello', 'there' ] 
s = set() 
for item in lst: 
    s.add(item.lower()) 
print(len(s))