字典

# 1.存储数据比较大
# 2.字典的查找速度比列表快
# 3.字典都是通过键来操作的，且键必须是唯一的。
# 4.# dic = {"键":"值","key":"value"}

字典的增删改查

字典增：
dic = {}
dic["键"]= 值  添加键值对

dic = {}
dic.setdefault("键",值) 有则更改，无则添加

字典删：
dic = {"111":[2,3,4],"222":[567]}
dic.pop("键") 通过键指定删除

dic = {"111":[2,3,4],"222":[567]}
dic.popitem() 随机删除键值对

dic = {"111":[2,3,4],"222":[567]}
del dic["111"] 指定删除键值对

dic = {"111":[2,3,4],"222":[567]}
dic.clear 删除整个字典

字典改：
dic = {"111":[2,3,4],"222":[567]}
dic[键] = 值  通过键修改值

字典查：
dic = {"111":[2,3,4],"222":[567]}
print(dic["111"]) 通过找键查询

大题示例：

要求得到的结果：{"河南":3,"河北":2,"北京":1,"内蒙":1}
id = [120121198903119561,
      120121198903110561,
      120121196903119561,
      130482198307144762,
      130482198307144662,
      110121197805144347,
      150121197502122799]
addrs = {
    120: "河南",
    130: "河北",
    110: "北京",
    150: "内蒙",
}
-----------------------------------------------------------------------------------------
答：
id = [120121198903119561,
      120121198903110561,
      120121196903119561,
      130482198307144762,
      130482198307144662,
      110121197805144347,
      150121197502122799]
addrs = {
    120: "河南",
    130: "河北",
    110: "北京",
    150: "内蒙",

dic = {}  #创建一个空的字典
for i in id:    循环id这个列表
    if dic.get(addrs.get(int(str(i)[0:3]))): 通过对i的切片拿到前三位做addrs的键，但是整数不支持切片需要str转换一下，切完后还要转回到整数做键，if判断dic这个字典中是否有这个值。
        dic[addrs.get(int(str(i)[0:3]))] += 1  如果这个字典中有就加一
    else:
        dic[addrs.get(int(str(i)[0:3]))] = 1   如果这个字典没有就创建1个，默认数为1.
print(dic)

字符串转字典

dic = {}
s = "{'key1':1,'key2':2,'key3':3}"
f = open("demo","a+",encoding="utf-8")
f.write(s)
f.seek(0)
f1 = f.read()
lst = f1[1:-1].replace("'","").split(",")
for i in lst:
    k,v = i.split(":")
    dic[k] = v
print(dic,type(dic))

原文地址：https://www.cnblogs.com/biyunfeng/p/11934575.html