如何在python中合并具有相同键的嵌套字典

我有一个这样的数据结构：

[ {'SNAPSHOT': {'SnapshotVersion': '304'}},

  {'SNAPSHOT': {'SnapshotCreationDate': '2015-06-21 17:33:41'}},


  {'CafeData': {'CafeVersion': '2807'}}, 

  {'CafeData': {'IsSoftwareOnly': '1'}}, 

  {'CafeData'{'IsPassportTCPIP': '1'}} 

  {'SNAPSHOT': {'SnapshotVersion': '777'}},

  {'SNAPSHOT': {'SnapshotCreationDate': '2017-07-27 17:37:47'}},]

输出应该是这样的：

 [ {'SNAPSHOT': {'SnapshotVersion': '304','SnapshotCreationDate': '2015-06-21 17:33:41'}},

   {'CafeData': {'CafeVersion': '2807','IsSoftwareOnly': '1','IsPassportTCPIP': '1'}} 
 
   {'SNAPSHOT': {'SnapshotVersion': '777','SnapshotCreationDate': '2017-07-27 17:37:47'}},
]

或者输出应该是这样的：

[ {'SNAPSHOT': {'SnapshotVersion': ['304','777'],
                'SnapshotCreationDate': ['2015-06-21 17:33:41','2017-07-27 17:37:47']}},

   {'CafeData': {'CafeVersion': '2807','IsSoftwareOnly': '1','IsPassportTCPIP': '1'}} 

]

可以使用 for 循环并跟踪最后合并的键。

from pprint import pprint

data = [
    {"SNAPSHOT": {"SnapshotVersion": "304"}},
    {"SNAPSHOT": {"SnapshotCreationDate": "2015-06-21 17:33:41"}},
    {"CafeData": {"CafeVersion": "2807"}},
    {"CafeData": {"IsSoftwareOnly": "1"}},
    {"CafeData": {"IsPassportTCPIP": "1"}},
    {"SNAPSHOT": {"SnapshotVersion": "777"}},
    {"SNAPSHOT": {"SnapshotCreationDate": "2017-07-27 17:37:47"}},
]

last_key = None
grouped = []
for value in data:
    # Easy way to get the key of a dict with one key
    curr_key = next(iter(value))
    # Decide if we should work on the next entry
    if last_key is None or curr_key != last_key:
        grouped.append(value)
    else:
        # update the last value in the group with the new data
        grouped[-1][curr_key].update(value[curr_key])
    # Move to the next item
    last_key = curr_key

pprint(grouped)

[{'SNAPSHOT': {'SnapshotCreationDate': '2015-06-21 17:33:41',
               'SnapshotVersion': '304'}},
 {'CafeData': {'CafeVersion': '2807',
               'IsPassportTCPIP': '1',
               'IsSoftwareOnly': '1'}},
 {'SNAPSHOT': {'SnapshotCreationDate': '2017-07-27 17:37:47',
               'SnapshotVersion': '777'}}]

选项 2 类似，但对于分组，我只使用字典。在这种情况下，您不需要知道最后一个密钥。

此外，如果嵌套字典中存在键冲突，您还需要对如何合并值做出一些假设。

grouped = {}
for value in data:
    curr_key = next(iter(value))
    curr_value = value[curr_key]
    group = grouped.setdefault(curr_key, {})

    for sub_key, sub_value in curr_value.items():
        # check if you need to merge
        if sub_key in group:
            # If the key is already present, but is not a list, make it one
            if not isinstance(group[sub_key], list):
                group[sub_key] = [group[sub_key]]
            # Add the new value to the list
            group[sub_key].append(sub_value)
        else:
            # Otherwise just copy it over
            group[sub_key] = sub_value

pprint(grouped)

{'CafeData': {'CafeVersion': '2807',
              'IsPassportTCPIP': '1',
              'IsSoftwareOnly': '1'},
 'SNAPSHOT': {'SnapshotCreationDate': ['2015-06-21 17:33:41',
                                       '2017-07-27 17:37:47'],
              'SnapshotVersion': ['304', '777']}}