Python

names = ["AA","BB","CC","DD","DD","FF"]

print(names[1]) #取值
print(names[0:4]) #切片
print(names[-2:])#以坐标的方式取值,左闭右开,不写默认正/负无穷
names.append("EE")#在最后追加字符串

names.insert(2,"FF")#在计算机下标为2 的位置前面加字符串

names.remove("FF")#移除第一个字符串
del  names[0]#移除字符串
names.pop()#默认删除最后一个字符串,可以加计算机下标
print(names.index("DD"))获取第一个字符串的计算机位置
print(names)
print(names.count("DD"))#获取该字符串的数量
#names.clear()#清除数组
names.reverse()#倒置数组
print(names)
数组操作
#Author:Wuyc

goods = (["苹果笔记本电脑",4200,4200111],
         ["华为手机",2199],
         ["字领露肩蓝色蕾丝",388.00],
         ["Bettychow女装",328.00],
         ["HP 笔记本",3900],
         ["F七匹狼夹克外套",459.00 ],)
'''
for name,money in enumerate(goods):
    print(name,money);#把下标取出来
'''

BuyGoods = [];
i = 0
for item in goods:
    i = i + 1
    print(str(i) + ":" , item[0],item[1])
salary = int(input("工资:"));
print("输入 T 退出 !")
print("输入 P 打印商品!")
while True:
    input1 = input("商品编号:");
    if (input1 == "T"):
        print("--------Goods List-------")
        for item in BuyGoods:
            print(item)
            # break;
            exit()
    elif (input1 == "P"):
        for item in BuyGoods:
            print(item)
    if input1.isdigit():
        good_num = int(input1)
        if good_num > len(goods):
            print("没有这个商品")
        else:
            good_money = goods[good_num-1][1]
            if salary >= good_money:
                BuyGoods.append(goods[good_num-1])
                salary = salary - good_money
                print("买了个 {goodname},花了 {goodmoney}是,余额为:33[31;1m{YE}33[0m 元".format(goodname=goods[good_num-1][0],goodmoney=goods[good_num-1][1],YE=salary))
            else:
                print("33[41;1m余额不足,重新选择!33[0m")
购物商城
原文地址:https://www.cnblogs.com/Jacob-Wu/p/9490778.html