字典的进阶操作[00000000]嵌套循环

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# # 字典的进阶操作,循环和嵌套
# dic = {"key0": "v0",
#        "key1": "v1",
#        "key2": "v2",
#        "key3": "v3",
#        "key4": "v4",
#        "key5": "v5",
#        }
# # print(dic)
# # 1.可以用for循环,直接拿到key
# # for key in dic:
# #     print(key, dic[key])  # 拿出key,与value
# # # 2.希望把所有的key保存在列表中
# # print(list(dic.keys()))  # 拿到所有的key
# # #  list(dic.keys())  把所有的key放入列表中
# # # 3.希望把所有的value值放入一个列表中
# # print(list(dic.values()))  # 拿到所有的value值
# # 4.直接拿到字典的key和value
# # print(type(dic.items()))
# # t = {1: "a", 2: "b", 77: "b", 4: "d"}
# # print(type(t[77]))
# # for a in t:
# # #     print(t[77])
# # a, b = [1, 2]  # 元组或者列表都可以执行改操作,该操作被称为解包(解构)
# # # ValueError: too many values to unpack (expected 2)
# # print(a)
# # print(b)
# # for item in dic.items():
# for key, value in dic.items():  # 可以直接拿到字典的所有key和value值
#     #  程序可以简化为替换item变量
#     # print(item)  # 确定item中只有两项元素
#     # key, value = item
#     print(key, value)
    # print(type(value))
#  5.字典的嵌套(键值只有一个,固定的,值value可以嵌套
wangfeng = {
    "name": "汪峰",
    "age": 48,
    "wife": {
        "name": "章子怡",
        "age": 38,
        "assistant": {
            "name": "大熊",
            "age": 29
        }
    }
}
#  我需要汪峰的老婆的助手的年龄
age = wangfeng['wife']['assistant']['name']
print(age)
# 给汪峰的老婆的助手加一岁
wangfeng['wife']['assistant']['age'] = wangfeng['wife']['assistant']['age'] + 1
print(wangfeng)
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posted @   ascecc  阅读(37)  评论(0编辑  收藏  举报
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