info = {"k1" : "v1", "k2" : "v2"}
以键值对形式组成字典
字典里的value可以是任何值
列表不能作为字典的key
字典可以作为字典的key
元组可以作为字典的key
布尔值可以作为字典的key 但是重复的key只会显示一个，True容易和数字1重复，需注意！
字典是无序的
info = {
"k1": 18,
2 : True,
"k3": [
"li",
[],
(),
22,
33,
{
"kk1": "vv1", "kk2": "vv2", "kk3": (11, 22)
}
],
"k4": (11, 22, 33, 44)
}
v = info["k1"]
print(v)
info = {
"k1": 18,
2 : True,
"k3": [
"li",
[],
(),
22,
33,
{
"kk1": "vv1", "kk2": "vv2", "kk3": (11, 22)
}
],
"k4": (11, 22, 33, 44)
}
v = info["k3"][5]["kk3"][0]
print(v)
注意：通过索引key值去找value的话，如果该key不存在，会报错。
支持del删除
del info[‘k1‘]
支持for 循环
1.输出所有keys
info = {"k1": "v1", "k2": "v2"}
for item in info.keys():
print(item)
2.输出所有values值
info = {"k1": "v1", "k2": "v2"}
for item in info.values():
print(item)
3. 同时输出keys和values：
info = {"k1": "v1", "k2": "v2"}
for item in info.keys():
print(item, info[item])
或（这个更重要）items 方法
info = {"k1": "v1", "k2": "v2"}
for k, v in info.items():
print(k, v)
v = dict.fromkeys(["k1", 123, "999"], "爱你") print(v)
dic = {"k1": "v1"}
v = dic.get("k1")
v2 = dic.get("大白", "不存在")
print(v, v2)
dic = {"k1": "v1",
"k2": "v2"
}
v1 = dic.pop("k1")
v2 = dic.pop("k3", "不存在哦亲")
print(dic)
print(v1)
print(v2)
dic = {"k1": "v1",
"k2": "v2"
}
v = dic.popitem()
print(dic, v)
或
dic = {"k1": "v1",
"k2": "v2"
}
k, v = dic.popitem()
print(dic, k, v)
dic = {"k1": "v1",
"k2": "v2"
}
v = dic.setdefault("大白", "123")
v2 = dic.setdefault("k1", "11111111")
print(dic, v)
print(dic, v2)
dic = {"k1": "v1",
"k2": "v2"
}
dic.update({"k1": "11111", "k3": "liu"})
print(dic)
dic = {"k1": "v1",
"k2": "v2"
}
dic.update(k1=123, k3 = 345)
print(dic)
重要：keys() values() items() get update
补充：
in dic 默认找的是key
in dic.values() 能找到values
dic = {"k1": "v1",
"k2": "v2"
}
v = "k1" in dic
v2 = "v1" in dic.values()
print(v)
print(v2)
布尔值：在内存中就表现为0和1
转换：bool(...)
以下情况为假：None "" () [] {} 0 ====>False
原文：https://www.cnblogs.com/dabai123/p/10962772.html