如何獲得列表中的非重複元素

Question

我想要列表中的非重複元素。這裏是。

A=[1, 2, 3, 2, 5, 1] 

所需的輸出：[3,5]

set()給[1, 2, 3, 5]

請讓我知道要實現這個任務。

Answer 1

使用Counter和defaultdict從collections：

from collections import defaultdict 
from collections import Counter 

A = [1 ,2 ,3 ,2 ,5, 1] 
# create a mapping from each item to it's count 
counter = Counter(A) 

# now reverse the mapping by using a defaultdict for convenience 
countmap = defaultdict(list) 

for k, v in counter.iteritems(): 
    countmap[v].append(k) 

# take all values that occurred once 
print countmap[1] # [3, 5] 

如果你感興趣的映射是什麼，你可以打印出來：

的counter：

Counter({1: 2, 2: 2, 3: 1, 5: 1}) 

的countmap：

defaultdict(<type 'list'>, {1: [3, 5], 2: [1, 2]}) 

手動創建您可以使用此功能的計數器：

def make_counter(lst): 
    counter = dict() 
    for item in lst: 
     if item in counter: 
      counter[item] += 1 
     else: 
      counter[item] = 1 
    return counter 

make_counter(A) 

輸出：

{1: 2, 2: 2, 3: 1, 5: 1} 

Answer 2

可以使用count()方法從list對象獲得同等的計數，以1：

>>> A=[1, 2, 3, 2, 5, 1] 
>>> unique=[i for i in A if A.count(i)==1] 
>>> unique 
[3, 5] 

可替代地從collections模塊Counter()類可用於：

A = [1, 2, 3, 2, 5, 1] 

c = Counter(A) 

print [key for key, count in c.iteritems() if count==1] 

Answer 3

#!/usr/bin/python3 
from collections import Counter 
from functools import reduce 


# Initialize Variable 
A = [1, 2, 3, 2, 5, 1] 

---

# iterating style 
result1 = [key for key, count in Counter(A).items() if count == 1] 

---

# functional style 
result2 = reduce(
    lambda acc, pair: acc + [pair[0] if pair[1] == 1 else acc, 
    Counter(A).items(), [])