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)