1. 字典是無序的，每次print（）都不一樣；
字典的鍵必須都是獨一無二的，字典不能作為字典的key；
可以通過索引方式查到指定元素；
不能切片，因為是無序的；
2. dict()舉例
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 循環 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)}
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