來自比較容易記憶的是用內置的setapp
l1 = ['b','c','d','b','c','a','a'] 測試
l2 = list(set(l1)) .net
print l2blog
還有一種聽說速度更快的,沒測試過二者的速度差異排序
l1 = ['b','c','d','b','c','a','a'] get
l2 = {}.fromkeys(l1).keys() 遍歷
print l2方法
這兩種都有個缺點,祛除重複元素後排序變了:sort
['a', 'c', 'b', 'd']index
若是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
也能夠這樣寫
l1 = ['b','c','d','b','c','a','a']
l2 = sorted(set(l1),key=l1.index)
print l2
也能夠用遍歷
l1 = ['b','c','d','b','c','a','a']
l2 = []
for i in l1:
if not i in l2:
l2.append(i)
print l2
上面的代碼也能夠這樣寫
l1 = ['b','c','d','b','c','a','a']
l2 = []
[l2.append(i) for i in l1 if not i in l2]
print l2
這樣就能夠保證排序不變了:
['b', 'c', 'd', 'a']