來自比較容易記憶的是用內置的set
l1 = ['b','c','d','b','c','a','a']
l2 = list(set(l1))
print l2
還有一種據說速度更快的，沒測試過兩者的速度差別
l1 = ['b','c','d','b','c','a','a']
l2 = {}.fromkeys(l1).keys()
print l2
這兩種都有個缺點，祛除重復元素后排序變了：
['a', 'c', 'b', 'd']
如果想要保持他們原來的排序：
用list類的sort方法
l1 = ['b','c','d','b','c','a','a']
l2 = list(set(l1))
l2.sort(key=l1.index)
print l2
也可以這樣寫
l1 = ['b','c','d','b','c','a','a']
l2 = sorted(set(l1),key=l1.index)
print l2
也可以用遍歷
l1 = ['b','c','d','b','c','a','a']
l2 = []
for i in l1:
if not i in l2:
l2.append(i)
print l2
上面的代碼也可以這樣寫
l1 = ['b','c','d','b','c','a','a']
l2 = []
[l2.append(i) for i in l1 if not i in l2]
print l2
這樣就可以保證排序不變了：
['b', 'c', 'd', 'a']