collections.ChainMap類合併字典或映射

## 使用update()方法或者ChainMap類合併字典或映射

 1 # 使用update()方法合併
 2 
 3 a = {'x': 1, 'z': 3}
 4 b = {'y': 2, 'z': 4}
 5 merged = dict(b) # 創建一個新字典
 6 print(merged)
 7 # {'y': 2, 'z': 4}
 8 merged.update(a) # 更新字典數據(合併)
 9 print(merged)
10 # {'y': 2, 'z': 3, 'x': 1}
11 
12 a['x'] = 10     # 對原有字典的改變不會影響合併後的字典
13 print(merged)
14 # {'y': 2, 'z': 3, 'x': 1}
15 
16 
17 
18 # 使用collections.ChainMap()類
19 from collections import ChainMap
20 
21 
22 a = {'x': 1, 'z': 3}
23 b = {'y': 2, 'z': 4}
24 c = ChainMap(a, b)
25 print(c)
26 # ChainMap({'x': 1, 'z': 3}, {'y': 2, 'z': 4})
27 
28 
29 # 數據的讀取總是從第一個字典開始查找，找不到再從第二個中查找
30 print(c['x'])
31 # 1
32 print(c['y'])
33 # 2
34 print(c['z'])
35 # 3
36 
37 
38 # 數據的增加或者刪除操作總是針對第一個字典
39 c['w'] = 4
40 print(c)
41 # ChainMap({'x': 1, 'z': 3, 'w': 4}, {'y': 2, 'z': 4})
42 
43 del c['w']
44 print(c)
45 # ChainMap({'x': 1, 'z': 3}, {'y': 2, 'z': 4})
46 
47 del c['y']      # 無法刪除第二個字典中的鍵值對
48 # KeyError: 'y'
49 # KeyError: "Key not found in the first mapping: 'y'"
50 
51 
52 # 對原有字典的操作會影響合併的字典
53 a = {'x': 1, 'z': 3}
54 b = {'y': 2, 'z': 4}
55 c = ChainMap(a, b)
56 print(c)
57 # ChainMap({'x': 1, 'z': 3}, {'y': 2, 'z': 4})
58 print(c['x'])
59 # 1
60 print(c['y'])
61 # 2
62 print(c['z'])
63 # 3
64 
65 a['x'] = 5  # 修改原有字典的值
66 print(c['x'])  # 合併的字典受到影響
67 # 5
68 
69 
70 # ChainMap()對象支持大多數字典的操作
71 print(len(c))
72 # 3
73 print(list(c.keys()))
74 # ['x', 'y', 'z']
75 print(list(c.values()))
76 # [5, 2, 3]
77 
78 
79 # 添加或刪除新的字典
80 values = ChainMap()
81 values['x'] = 1
82 values = values.new_child()  # 添加一個空白字典
83 print(values)
84 # ChainMap({}, {'x': 1})
85 values['x'] = 2  # 給空白字典添加內容
86 print(values)
87 # ChainMap({'x': 2}, {'x': 1})
88 values = values.new_child()
89 values['x'] = 3
90 print(values)
91 # ChainMap({'x': 3}, {'x': 2}, {'x': 1})
92 
93 # 刪除第一個字典
94 values = values.parents
95 print(values)
96 # ChainMap({'x': 2}, {'x': 1})
97 values = values.parents
98 print(values)
99 # ChainMap({'x': 1})

參考資料：
　　Python Cookbook, 3rd edition, by David Beazley and Brian K. Jones (O’Reilly).