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，祛除重複元素後排序變了： [‘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 上面的程式碼也可以這樣寫 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’]