版权声明:本文为博主原创文章,转载请附上博文链接! https://blog.csdn.net/coco_link/article/details/88351315
速度快,但排序变化
# set函数
L = ['b','c','d','b','c','a','a']
set(L)
Out[21]: {'a', 'b', 'c', 'd'}
list(set(L))
Out[22]: ['a', 'c', 'b', 'd']
{}.fromkeys(l1).keys()
Out[23]: ['a', 'c', 'b', 'd']
保持原有排序
# list 的 sort 方法
l1 = ['b','c','d','b','c','a','a']
l2 = list(set(l1))
l2.sort(key=l1.index)
print l2
Out[28]: ['b', 'c', 'd', 'a']
l1 = ['b','c','d','b','c','a','a']
sorted(set(l1),key=l1.index)
Out[31]: ['b', 'c', 'd', 'a']
遍历
l1 = ['b','c','d','b','c','a','a']
l2 = []
for i in l1:
if not i in l2:
l2.append(i)
print l2
l1 = ['b','c','d','b','c','a','a']
l2 = []
[l2.append(i) for i in l1 if not i in l2]
print l2