m0_56588134的博客

#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] 4 5 #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][

#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=-23 aa=A() print(aa.reverse(a))

#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

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=0 count=1 while True: b=a.find(“i”,b) if b==-1: break if count>1: break