时间:2021-07-01 10:21:17 帮助过:32人阅读
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']
希望本文所述对大家的Python程序设计有所帮助。