將字典元組轉換為嵌套字典

Question

我的代碼有問題

s1 = ({'server_id':'myserver1','cloud':'google','time':'1hr'},
      {'server_id':'myserver2','cloud':'aws','time':'2hr'},
      {'server_id':'myserver3','cloud':'google','time':'1hr'}
      {'server_id':'myserver4','cloud':'aws','time':'3hr'})
mydict = {'mine':
          {data[i]:
           {key:value for key,value in data.iteritems() if key!='cloud'}
          for data in s1 for i in data if i=='cloud'}}
print mydict

結果：

{'mine': {'aws': {'server_id': 'myserver3','time':'3hr'}}}

但我期待這樣的結果

{'mine':
 {'aws':
  {'server_id': 'myserver1','time':'1hr'},
  {'server_id': 'myserver3','time':'3hr'}},
 {
 'google':
  {'server_id': 'myserver2','time':'2hr'},
  {'server_id': 'myserver4','time':'4hr'}}}

任何人都可以幫我這個嗎？

Answer 1

考慮初始化字典並在循環內附加到它。 使用dict.pop負責擺脫cloud密鑰，而不必像現在這樣迭代整個事物。

from collections import defaultdict
d = defaultdict(list)  # {}

for s in s1:
    d[s.pop('cloud')].append(s)  # d.setdefault(s.pop('cloud'), []).append(s)

{'mine' : dict(d)}

{'mine': {'aws': [{'server_id': 'myserver1', 'time': '1hr'},
   {'server_id': 'myserver2', 'time': '2hr'},
   {'server_id': 'myserver3', 'time': '3hr'}]}}

免責聲明！ 這會就地修改s1 （ pop改變子句，返回該鍵的值（如果存在）。

要處理KeyErrors，您可以提供合適的默認值 - s.pop('cloud', 'default') ，其中default是cloud密鑰不存在時返回的值。

Answer 2

問題是您正在嘗試將三個詞典映射到相同的鍵。 如果你把它作為字典列表，它就可以了。

s1 = ({'server_id':'myserver1','cloud':'aws','time':'1hr'},
      {'server_id':'myserver2','cloud':'aws','time':'2hr'},
      {'server_id':'myserver3','cloud':'aws','time':'3hr'})
mydict = {'mine':
          {data['cloud']:
           [{key:value for key, value in data.iteritems() if key != "cloud"}
            for data in s1]}}
print mydict

輸出：

{'mine': {'aws': [{'server_id': 'myserver1', 'time': '1hr'},
   {'server_id': 'myserver2', 'time': '2hr'},
   {'server_id': 'myserver3', 'time': '3hr'}]}}