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RSS订阅原创 python基础
#获取[[[]]]li=[1,[2,3,[4,5]]]print(li[0])print(li[1])print(li[1][2])print(li[1][2][0])print(li[1][2][1])回车1[2, 3, [4, 5]][4, 5]45#获取[[]，[]，[]]的值li=[[1,2,3,10],[4,5],[7,8]]print(li[0])print(li[0][0])print(li[1][1])print(li[2][0])print(li[0][
2021-04-26 17:16:47
117
原创 Python基础
#将数字颠倒class A():def reverse(self,a):b=str(a)if a>=0:return b[::-1]else:b=str(-a)return -int(b[::-1])a=-23aa=A()print(aa.reverse(a))
2021-04-23 03:53:01
200
原创 python基础
#li={“h”:1,“g”:3,“k”:2}#li=list(li.items())#li.sort(key=lambda x:x[1])#print(li)def a(x):return x[1]li={“h”:1,“g”:3,“k”:2}li=list(li.items())li.sort(key=a)print(li)回车[(‘h’, 1), (‘k’, 2), (‘g’, 3)]#用key返回出x[索引]，以索引1排序def a(num):for i in range(2
2021-04-17 01:57:36
175
原创 python基础知识
a=“this is”b=list(a)print(b)回车[‘t’, ‘h’, ‘i’, ‘s’, ’ ', ‘i’, ‘s’]a=“this is”b=a.split()print(b)回车[‘this’, ‘is’]#以上是b=list(a)和b=a.split的区别#如何用find找到第二个i的索引a=“this is i i”b=0count=1while True:b=a.find(“i”,b)if b==-1:breakif count>1:break
2021-04-17 01:51:03
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