python对嵌套list取值 python列表嵌套元组

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flyingsmiling 2023-06-10 00:11:03

文章标签 python对嵌套list取值取值元组 文章分类 Python 后端开发

names = ["Zhangsan"," Lisi"," Wangwu"," Zhoumazi"]

import copy

names = ["1Zhangsan"," #Lisi"," xWangwu",["Wangshanfeng","Paopao"], "Zhoumazi"]

print(names[0👎2])

print(names[👎2])

print(names[:])for i in names:

print(i)

name2 = copy.deepcopy(names)#深深地复制

name2 = names.copy()#复制不全

print(names)

print(name2)

names[2] = "王五"

names[3][0] ="王珊凤"

print(names)

print(name2)

names.append("Wusangui") #追加一个人到这个列表里

names.insert(1,"Wusangui") #想插入一个人在某某的位置

names.insert(4,"Zhangsan") #想插入一个人在某某的后面位置

names[2] = "Zhaolaotou" #想把某某换成另一个人print(names[0],names[2])

print(names[1:4]) #切片

print(names[0:3]) #切片 取前三位

print(names[:3]) #切片 取前面三个

print(names[-1]) #切片 不知道多长的情况下取值

print(names[-2:]) #切片 取最后面的两个值names.remove("Lisi")#删除Lisi

del names[1]#删除Lisi

names.pop() #删除最后一个值

names.pop(1)#删除Lisi

print(names)

print(names.index("Zhaolaotou") ) #通过王五的名字找到它所在的位置

print(names[names.index("Zhaolaotou")])

print(names.count("Zhangsan"))

names.clear()

names.reverse()#列表反转

names.sort() #排序

names2 = [1,2,3,4]

names.extend(names2)#合并

print(names,names2)