字典dict()

1. 字典是无序的，每次print（）都不一样；

字典的键必须都是独一无二的，字典不能作为字典的key；

可以通过索引方式查到指定元素；

不能切片，因为是无序的；

2. dict()举例

dict1 = {'K1':18,2:True,'K3':[1,[],(),2,3,{'k1':'v1','kk3':(11,22)}],'K4':(11,22,33,44)}
v = dict1['K3'][5]['kk3'][0]
print(v)
————————————————————
11

　　

del dict1['K3'][5]['k1']
print(dict1)
————————————————
{'K1': 18, 2: True, 'K3': [1, [], (), 2, 3, {'kk3': (11, 22)}], 'K4': (11, 22, 33, 44)}

3.for 循环 keys()

dict1 = {'K1':18,2:True,'K3':[1,[],(),2,3,{'k1':'v1','kk3':(11,22)}],'K4':(11,22,33,44)}
for item in dict1:#=dict1.keys()
    print(item)
——————————————
K1
2
K3
K4

4.　values()

for item in dict1.values():
    print(item)
______________
18
True
[1, [], (), 2, 3, {'k1': 'v1', 'kk3': (11, 22)}]
(11, 22, 33, 44)

5. items()

for k,v in dict1.items():
    print(k,v)
_____________
K1 18
2 True
K3 [1, [], (), 2, 3, {'k1': 'v1', 'kk3': (11, 22)}]
K4 (11, 22, 33, 44)

6. update()

dict1 = {'K1':18,2:True,'K3':[1,[],(),2,3,{'k1':'v1','kk3':(11,22)}],'K4':(11,22,33,44)}
dict1.update(K1=111,K3=123)
print(dict1)
_______________________
{'K1': 111, 2: True, 'K3': 123, 'K4': (11, 22, 33, 44)}