查找只出现一次的单词

Question

我只检索文件中的唯一单词，这里是我迄今为止的内容，但是有没有更好的方法可以在大O表示法中实现这一点？眼下这为n的平方。如果你想找到的所有独特的文字和考虑foo一样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

尝试使用此方法获得的唯一字的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

，你需要去掉标点符号。

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]) 

如果你想忽略大小写，你还需要使用line.lower()。如果你想准确地得到独特的单词，那么除了在空白处分割行之外，还有更多的涉及。

Answer 3

你可以稍微修改你的逻辑和（使用套例如，而不是类型的字典），它从独特的前进第二次出现：

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

我会去与collections.Counter的做法，但如果你只想使用set S，那么你可以通过这样做：

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 

鉴于输入：

one two three 
two three four 
four five six 

具有的输出：

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