查找只出現一次的單詞

Question

我只檢索文件中的唯一單詞，這裏是我迄今爲止的內容，但是有沒有更好的方法可以在大O表示法中實現這一點？眼下這爲n的平方。如果你想找到的所有獨特的文字和考慮foo一樣foo.

def retHapax(): 
    file = open("myfile.txt") 
    myMap = {} 
    uniqueMap = {} 
    for i in file: 
     myList = i.split(' ') 
     for j in myList: 
      j = j.rstrip() 
      if j in myMap: 
       del uniqueMap[j] 
      else: 
       myMap[j] = 1 
       uniqueMap[j] = 1 
    file.close() 
    print uniqueMap 

Answer 1

嘗試使用此方法獲得的唯一字的file.using Counter

from collections import Counter 
with open("myfile.txt") as input_file: 
    word_counts = Counter(word for line in input_file for word in line.split()) 
>>> [word for (word, count) in word_counts.iteritems() if count==1] 
-> list of unique words (words that appear exactly once) 

Answer 2

，你需要去掉標點符號。

from collections import Counter 
from string import punctuation 

with open("myfile.txt") as f: 
    word_counts = Counter(word.strip(punctuation) for line in f for word in line.split()) 

print([word for word, count in word_counts.iteritems() if count == 1]) 

如果你想忽略大小寫，你還需要使用line.lower()。如果你想準確地得到獨特的單詞，那麼除了在空白處分割行之外，還有更多的涉及。

Answer 3

你可以稍微修改你的邏輯和（使用套例如，而不是類型的字典），它從獨特的前進第二次出現：

words = set() 
unique_words = set() 
for w in (word.strip() for line in f for word in line.split(' ')): 
    if w in words: 
     continue 
    if w in unique_words: 
     unique_words.remove(w) 
     words.add(w) 
    else: 
     unique_words.add(w) 
print(unique_words) 

Answer 4

我會去與collections.Counter的做法，但如果你只想使用set S，那麼你可以通過這樣做：

with open('myfile.txt') as input_file: 
    all_words = set() 
    dupes = set() 
    for word in (word for line in input_file for word in line.split()): 
     if word in all_words: 
      dupes.add(word) 
     all_words.add(word) 

    unique = all_words - dupes 

鑑於輸入：

one two three 
two three four 
four five six 

具有的輸出：

{'five', 'one', 'six'}