python - Python 中联合多个嵌套 JSON

Question

我有多个包含需要合并的关系数据的json文件，每个文件都有一个带有commonkey的记录，它是所有文件中的公共(public)键，在下面的示例中a0，a1是公共(public)键。该值是一个嵌套的如下所示的 Key1、key2 等多个键的字典，我需要合并多个 json 文件并获取如 dboutput.json 中所示的输出，其中文件名充当合并操作中的索引。这样的问题是related question它合并丢失的信息，但在我的情况下，我不希望任何替换现有键或跳过更新的更新，如果击中现有键，则会创建另一个由文件名索引的嵌套字典，如下所示:

示例:

文件 db1.json:

"a0": {        "commonkey": [            "a1",             "parentkeyvalue1"        ],         "key1": "kvalue1",         "key2": "kvalue2"        "keyp": "kvalue2abc"    }, "a1": { ...}

File db2.json:

"a0": {        "commonkey": [            "a1",             "parentkeyvalue1"        ],         "key1": "kvalue1xyz",         "key2": "kvalue2",        "key3": "kvalue2"    }, "a1": { ...}

Desired Output

File dboutput.json

"a0": {        "commonkey": [            "a1",             "parentkeyvalue1"        ],         "key1": {"db1":"kvalue1","db2":"kvalue1xyz"} ,        "key2": {"db1":"kvalue2","db2":"kvalue2"} ,        "key3": {"db2":"kvalue2"}        "keyp": {"db1":"kvalue2abc"}    }, "a1": { ...}

So how to do such lossless merges? Note "key2": {"db1":"kvalue2","db2":"kvalue2"} even if the key\value pairs are same they need to be stored separately. In effect the output is a union of all input files and has all entries from all other files.

Also

"commonkey": [
            "a1", 
            "parentkeyvalue1"
        ],

所有文件都相同，因此无需重复

Answer 1

我终于成功了:

class NestedDict(collections.OrderedDict):
    """Implementation of perl's autovivification feature."""
    def __getitem__(self, item):
        try:
            return dict.__getitem__(self, item)
        except KeyError:
            value = self[item] = type(self)()
            return value

def mergejsons(jsns):
 ##use auto vification Nested Dict
    op=nesteddict.NestedDict()
    for j in jsns:
        jdata=json.load(open(j))
        jname=j.split('.')[0][-2:]
        for commnkey,val in jdata.items():
            for k,v in val.items():
                if k!='commonkey':
                    op[commnkey][k][jname]=v
                if  op[commnkey].has_key('commonkey'):
                    continue
                else:
                    op[commnkey][k][jname]=v