你可以做一个丑陋的列表理解黑客。

[l[i] for i in range(len(l)) if l.index(l[i]) == i]

你可以做一个丑陋的列表理解黑客。

[l[i] for i in range(len(l)) if l.index(l[i]) == i]

1. 这些解决方案很好…… 为了在保留秩序的同时删除重复项，本页其他地方提出了优秀的解决方案:

seen = set()
[x for x in seq if not (x in seen or seen.add(x))]

以及变化，例如:

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

确实很受欢迎，因为它们简单、极简，并部署了正确的哈希以获得最佳效率。关于这些方法的主要抱怨似乎是，将方法see .add(x)“返回”的不变量None用作逻辑表达式中的常量(因此是多余的/不必要的)值(只是为了它的副作用)是笨拙和/或令人困惑的。

2. …but they waste one hash lookup per iteration. Surprisingly, given the amount of discussion and debate on this topic, there is actually a significant improvement to the code that seems to have been overlooked. As shown, each "test-and-set" iteration requires two hash lookups: the first to test membership x not in seen and then again to actually add the value seen.add(x). Since the first operation guarantees that the second will always be successful, there is a wasteful duplication of effort here. And because the overall technique here is so efficient, the excess hash lookups will likely end up being the most expensive proportion of what little work remains.

3.相反，让布景完成它的工作吧! 注意，上面的例子只调用set。加上预见，这样做总是会导致集合成员的增加。集合本身永远没有机会拒绝副本;我们的代码片段实际上已经篡夺了这个角色。使用显式的两步测试和设置代码剥夺了set自身排除这些重复的核心能力。

4. 单哈希查找代码: 下面的版本将每次迭代的哈希查找次数减少了一半，从两次减少到只有一次。

seen = set()
[x for x in seq if len(seen) < len(seen.add(x) or seen)]

我觉得如果你想维持秩序，

你可以试试这个:

list1 = ['b','c','d','b','c','a','a']    
list2 = list(set(list1))    
list2.sort(key=list1.index)    
print list2

或者类似地，你可以这样做:

list1 = ['b','c','d','b','c','a','a']  
list2 = sorted(set(list1),key=list1.index)  
print list2 

你还可以这样做:

list1 = ['b','c','d','b','c','a','a']    
list2 = []    
for i in list1:    
    if not i in list2:  
        list2.append(i)`    
print list2

它也可以写成这样:

list1 = ['b','c','d','b','c','a','a']    
list2 = []    
[list2.append(i) for i in list1 if not i in list2]    
print list2 

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])

5倍更快减少变种，但更复杂

>>> l = [5, 6, 6, 1, 1, 2, 2, 3, 4]
>>> reduce(lambda r, v: v in r[1] and r or (r[0].append(v) or r[1].add(v)) or r, l, ([], set()))[0]
[5, 6, 1, 2, 3, 4]

解释:

default = (list(), set())
# use list to keep order
# use set to make lookup faster

def reducer(result, item):
    if item not in result[1]:
        result[0].append(item)
        result[1].add(item)
    return result

>>> reduce(reducer, l, default)[0]
[5, 6, 1, 2, 3, 4]