Python字典增 python字典增删改查内存图解

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)