您可以通过使用集合简单地做到这一点。

步骤1:获取列表的不同元素 Step2获取列表的公共元素 3 .结合

In [1]: a = ["apples", "bananas", "cucumbers"]

In [2]: b = ["pears", "apples", "watermelons"]

In [3]: set(a).symmetric_difference(b).union(set(a).intersection(b))
Out[3]: {'apples', 'bananas', 'cucumbers', 'pears', 'watermelons'}

一个更好的方法是，

import pandas as pd

myList = [1, 2, 3, 1, 2, 5, 6, 7, 8]
cleanList = pd.Series(myList).drop_duplicates().tolist()
print(cleanList)

#> [1, 2, 3, 5, 6, 7, 8]

秩序得以维持。

这个人关心的是没有太多麻烦的订单(OrderdDict & others)。可能不是最python化的方式，也不是最短的方式，但却达到了目的:

def remove_duplicates(item_list):
    ''' Removes duplicate items from a list '''
    singles_list = []
    for element in item_list:
        if element not in singles_list:
            singles_list.append(element)
    return singles_list

Write a Python program to create a list of numbers by taking input from the user and then remove  the duplicates from the list. You can take input of non-zero numbers, with an appropriate  prompt, from the user until the user enters a zero to create the list assuming that the numbers  are non-zero.  
Sample Input: [10, 34, 18, 10, 12, 34, 18, 20, 25, 20]  
Output: [10, 34, 18, 12, 20, 25] 

 lst = []
print("ENTER ZERO NUMBER FOR EXIT !!!!!!!!!!!!")
print("ENTER LIST ELEMENTS  :: ")
while True:
    n = int(input())
    if n == 0 :
       print("!!!!!!!!!!! EXIT !!!!!!!!!!!!")
       break
    else :
        lst.append(n)
print("LIST ELEMENR ARE :: ",lst)
#dup = set()
uniq = []
for x in lst:
    if x not in uniq:
        uniq.append(x)
       # dup.add(x)
print("UNIQUE ELEMENTS IN LIST ARE :: ",uniq)

尝试使用集合:

import sets
t = sets.Set(['a', 'b', 'c', 'd'])
t1 = sets.Set(['a', 'b', 'c'])

print t | t1
print t - t1

def remove_duplicates(A):
   [A.pop(count) for count,elem in enumerate(A) if A.count(elem)!=1]
   return A

用于删除重复项的列表推导