python中list去重

比较容易记忆的是用内置的set
l1 = ['b','c','d','b','c','a','a']
l2 = list(set(l1))
print l2

还有一种据说速度更快的，没测试过两者的速度差别
l1 = ['b','c','d','b','c','a','a']
l2 = {}.fromkeys(l1).keys()
print l2

这两种都有个缺点，祛除重复元素后排序变了：
['a', 'c', 'b', 'd']



如果想要保持他们原来的排序：
用list类的sort方法
l1 = ['b','c','d','b','c','a','a']
l2 = list(set(l1))
l2.sort(key=l1.index)
print l2
也可以这样写
l1 = ['b','c','d','b','c','a','a']
l2 = sorted(set(l1),key=l1.index)
print l2


也可以用遍历
l1 = ['b','c','d','b','c','a','a']
l2 = []
for i in l1:
if not i in l2:
l2.append(i)
print l2
上面的代码也可以这样写
l1 = ['b','c','d','b','c','a','a']
l2 = []
[l2.append(i) for i in l1 if not i in l2]
print l2

这样就可以保证排序不变了：

['b', 'c', 'd', 'a']

l1 = ['b','c','d','b','c','a','a']
l2 = ['b']
for i in l1:
    print 'i=====',i
    s = 0
    for j in l2:
        print 'j=',j
        print 'l2=',l2
        if(j==i):##当只是判断i和j的部分相同时，也可判断
            break
        else:
            s+=1
    if(s == len(l2)):
        l2.append(i)