如果你经常使用pandas，并且美学优先于性能，那么考虑内置函数pandas. series .drop_duplicate:

    import pandas as pd
    import numpy as np

    uniquifier = lambda alist: pd.Series(alist).drop_duplicates().tolist()

    # from the chosen answer 
    def f7(seq):
        seen = set()
        seen_add = seen.add
        return [ x for x in seq if not (x in seen or seen_add(x))]

    alist = np.random.randint(low=0, high=1000, size=10000).tolist()

    print uniquifier(alist) == f7(alist)  # True

时间:

    In [104]: %timeit f7(alist)
    1000 loops, best of 3: 1.3 ms per loop
    In [110]: %timeit uniquifier(alist)
    100 loops, best of 3: 4.39 ms per loop

对于不可哈希类型(例如列表的列表)，基于MizardX的:

def f7_noHash(seq)
    seen = set()
    return [ x for x in seq if str( x ) not in seen and not seen.add( str( x ) )]

一个简单的递归解决方案:

def uniquefy_list(a):
    return uniquefy_list(a[1:]) if a[0] in a[1:] else [a[0]]+uniquefy_list(a[1:]) if len(a)>1 else [a[0]]

就地方法

这个方法是二次的，因为我们对列表中的每个元素都有一个线性查找(由于del，我们必须加上重新排列列表的代价)。

也就是说，如果我们从列表的末尾开始，并向原点前进，删除出现在其左侧子列表中的每一项，就有可能在原地操作

这个想法在代码中很简单

for i in range(len(l)-1,0,-1): 
    if l[i] in l[:i]: del l[i] 

实现的简单测试

In [91]: from random import randint, seed                                                                                            
In [92]: seed('20080808') ; l = [randint(1,6) for _ in range(12)] # Beijing Olympics                                                                 
In [93]: for i in range(len(l)-1,0,-1): 
    ...:     print(l) 
    ...:     print(i, l[i], l[:i], end='') 
    ...:     if l[i] in l[:i]: 
    ...:          print( ': remove', l[i]) 
    ...:          del l[i] 
    ...:     else: 
    ...:          print() 
    ...: print(l)
[6, 5, 1, 4, 6, 1, 6, 2, 2, 4, 5, 2]
11 2 [6, 5, 1, 4, 6, 1, 6, 2, 2, 4, 5]: remove 2
[6, 5, 1, 4, 6, 1, 6, 2, 2, 4, 5]
10 5 [6, 5, 1, 4, 6, 1, 6, 2, 2, 4]: remove 5
[6, 5, 1, 4, 6, 1, 6, 2, 2, 4]
9 4 [6, 5, 1, 4, 6, 1, 6, 2, 2]: remove 4
[6, 5, 1, 4, 6, 1, 6, 2, 2]
8 2 [6, 5, 1, 4, 6, 1, 6, 2]: remove 2
[6, 5, 1, 4, 6, 1, 6, 2]
7 2 [6, 5, 1, 4, 6, 1, 6]
[6, 5, 1, 4, 6, 1, 6, 2]
6 6 [6, 5, 1, 4, 6, 1]: remove 6
[6, 5, 1, 4, 6, 1, 2]
5 1 [6, 5, 1, 4, 6]: remove 1
[6, 5, 1, 4, 6, 2]
4 6 [6, 5, 1, 4]: remove 6
[6, 5, 1, 4, 2]
3 4 [6, 5, 1]
[6, 5, 1, 4, 2]
2 1 [6, 5]
[6, 5, 1, 4, 2]
1 5 [6]
[6, 5, 1, 4, 2]

In [94]:                                                                                                                             

x = [1, 2, 1, 3, 1, 4]

# brute force method
arr = []
for i in x:
  if not i in arr:
    arr.insert(x[i],i)

# recursive method
tmp = []
def remove_duplicates(j=0):
    if j < len(x):
      if not x[j] in tmp:
        tmp.append(x[j])
      i = j+1  
      remove_duplicates(i)

      

remove_duplicates()

使用_sorted_ a numpy数组的相对有效方法:

b = np.array([1,3,3, 8, 12, 12,12])    
numpy.hstack([b[0], [x[0] for x in zip(b[1:], b[:-1]) if x[0]!=x[1]]])

输出:

array([ 1,  3,  8, 12])