来自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']
Python求list中元素个数
list01 = ['a','b','c','a','c']
set01 = set(list01) print(set01) dict01 = {} for item in set01:
dict01.update({item:list01.count(item)})
print(dict01
'c', 'b', 'a'}
{'c': 2, 'b': 1, 'a': 2}