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  • 菜鸟python---字典的增删改查操作

    python的数据结构之一
    字典 -- dict
    定义:
    dic = {"key":"value"} -- 键值对

    字典的作用:
    存储数据,大大量,将数据和数据起到关联作用

    dic = {"10":"苹果手机",
    "11":"苹果手机",
    15:"小米手机",
    15:"华为手机",
    (1,):"oppo手机",
    }
    print(dic)

    所有的操作都是通过键

    键:必须是不可变的数据类型(可哈希),且唯一 不可哈希就是可变数据类型
    值:任意

    字典是可变数据类型,无序的



    字典的增:
    1.暴力添加:
    dic["日阳"] = "曹洋" # 字典的添加,添加的是一个键值对
    dic["小妹"] = ["听歌","唱歌","吃","烤馕","大盘鸡","葡萄干"]
    print(dic)


    dic = {
    "日魔":["看动漫","健身","吃包子","吃大煎饼","吃大烧饼"],
    "炮手":"飞机",
    "豹哥":"贴膏药",
    "宝元":"宝剑",
    "alex":"吹牛逼"
    }
    2.有则不添加,无则添加
    dic.setdefault("元宝",["唱","跳","篮球","喝酒"])
    print(dic)

    setdefault 分为两步:
    1,先查看键是否在字典
    2,不存在的时候进行添加
    print(dic.setdefault("元宝"))

    字典的删除:
    dic = {
    "日魔":["看动漫","健身","吃包子","吃大煎饼","吃大烧饼"],
    "炮手":"飞机",
    "豹哥":"贴膏药",
    "宝元":"宝剑",
    "alex":"吹牛逼"
    }
    print(dic.pop("宝元")) #pop删除通过字典中的键进线删除 返回的也是被删除的值
    print(dic)

    dic.clear() # 清空
    print(dic)

    del dic # 删除的是整个容器
    print(dic)

    del dic["alex"] # 通过键进行删除
    print(dic)

    字典中是没有remove *****


    字典的改:

    dic = {
    "日魔":["看动漫","健身","吃包子","吃大煎饼","吃大烧饼"],
    "炮手":"飞机",
    "豹哥":"贴膏药",
    "宝元":"宝剑",
    "alex":"吹牛逼"
    }

    dic["alex"] = "dsb" # 有则就覆盖,没有添加
    print(dic)
    #
    dic1 = {"alex":"上过北大","wusir":"干过前端"}
    dic1.update(dic)
    print(dic1)

    字典的查:

    dic = {
    "日魔":["看动漫","健身","吃包子","吃大煎饼","吃大烧饼"],
    "炮手":"飞机",
    "豹哥":"贴膏药",
    "宝元":"宝剑",
    "alex":"吹牛逼"
    }

    print(dic.get("alex")) # 查询不到返回None
    print(dic.get("元宝","找不到啊")) # 查找不到的时候返回自己制定的内容
    print(dic.setdefault("alex")) # 查询不到返回None
    print(dic["alex"]) # 查询不到就报错了

    for i in dic: # 查看所有的键
    print(i)

    for i in dic: # 查看所有的值
    print(dic.get(i))

    print(dic.keys()) # 获取到的是一个高仿列表
    print(dic.values()) # 获取到的是一个高仿列表

    for i in dic.values(): # 高仿列表支持迭代
    print(i)

    print(dic.values()[3]) 高仿列表不支持索引

    lst = []
    for i in dic:
    lst.append(dic[i])
    print(lst)

    print(list(dic.values()))

    lst = []
    for i in dic:
    lst.append((i,dic[i]))
    print(lst)

    print(list(dic.items()))


    for i in dic:
    print(i,dic[i])

    for i in dic.items():
    print(i[0],i[1])


    解构:

    面试题:
    a = 10
    b = 20
    a,b = b,a
    print(a)
    print(b)

    a,b = 10,20
    print(a)
    print(b)

    a = 10,20
    print(a)
    print(10,20)

    a,b = (1,20)
    print(a)
    print(b)

    a,b = [11,20]
    print(a)
    print(b)

    a,b = "wc"
    print(a)
    print(b)

    dic = {"key1":2,"key2":4}
    a,b = dic
    print(a)
    print(b)

    解构的作用,

    lst = [1,2,3,4,5,6,7,8]
    lst1 = lst[:2] # 两个都是列表的时候才能够相加
    lst1.append(lst[4])
    for i in lst1:
    print(i)

    print(lst[:2] + list(str(lst[4])))

    lst = [1,2,3,4,5,6,7,8]
    a,b,c,d,e,*aa = lst # *是万能接受
    print(a,b,e)

    for i in dic.items():
    a,b = i
    print(a)
    print(b)

    for i in dic.items():
    a,b = i # 自己写的解构
    print(a)
    print(b)

    for a,b in dic.items():
    print(a)
    print(b)
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  • 原文地址:https://www.cnblogs.com/veryFuelong/p/13354969.html
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