如果你想删除重复项(就地编辑而不是返回新列表)而不使用内置set, dict，请检查此选项。键，唯一，计数器

>>> t = [1, 2, 3, 1, 2, 5, 6, 7, 8]
>>> for i in t:
...     if i in t[t.index(i)+1:]:
...         t.remove(i)
... 
>>> t
[3, 1, 2, 5, 6, 7, 8]

>>> t = [1, 2, 3, 1, 2, 5, 6, 7, 8]
>>> t
[1, 2, 3, 1, 2, 5, 6, 7, 8]
>>> s = []
>>> for i in t:
       if i not in s:
          s.append(i)
>>> s
[1, 2, 3, 5, 6, 7, 8]

创建一个新的列表，保持L中重复元素的第一个元素的顺序:

newlist = [ii for n,ii in enumerate(L) if ii not in L[:n]]

例如:如果L =[1,2,2,3,4,2,4,3,5]，则newlist将为[1,2,3,4,5]

这将在添加之前检查之前没有出现在列表中的每个新元素。 它也不需要进口。

不幸的是。这里的大多数答案要么不保持顺序，要么太长。这里有一个简单的、有序的答案。

s = [1,2,3,4,5,2,5,6,7,1,3,9,3,5]
x=[]

[x.append(i) for i in s if i not in x]
print(x)

这将得到x，删除重复项，但保留顺序。

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)

要删除重复的，将其设置为SET，然后再次将其设置为LIST，并打印/使用它。 一个集合保证有唯一的元素。例如:

a = [1,2,3,4,5,9,11,15]
b = [4,5,6,7,8]
c=a+b
print c
print list(set(c)) #one line for getting unique elements of c

输出将如下所示(在python 2.7中检查)

[1, 2, 3, 4, 5, 9, 11, 15, 4, 5, 6, 7, 8]  #simple list addition with duplicates
[1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 15] #duplicates removed!!