Modifying Json Data using Python

Question

Below is my data.

input = {'key1': {'key1.1': 'value1.1',
                  'key1.2': 'value1.2',
                  'key1.3': {'key1.3.1': 'value1.3.1', 'key1.3.2': [{'key1.3.2.1': 'value1.3.2.1'},
                                                                    {'key1.3.2.2': 'value1.3.2.2'}
                                                                    ]
                             }
                  },
         'key2': 'value2'}

I want to recursively convert inner inner key/values into a list of sub dictionaries with Name and Value as their keys. Below is my desired output.

output = {"key1": [{"NAME": "key1.1", "VALUE": "value1.1"},
                   {"NAME": "key1.2", "VALUE": "value1.2"},
                   {"NAME": "key1.3", "VALUE": [{"NAME": "key1.3.1", "VALUE": "value1.3.1"},
                                                {"NAME": "key1.3.2", "VALUE": [{"NAME": "key1.3.2.1", "VALUE": "value1.3.2.1"},
                                                                               {"NAME": "key1.3.2.2", "VALUE": "value1.3.2.2"}]
                                                 }]
                    }],
          "key2": "value2"}

Below is what is tried but it is not working correctly for list of dictionaries (ex: "key1.3.2.1": "value1.3.2.1"). Is there a way to achieve this output?

def my_func(d):
    new_dict = {}
    for k, v in d.items():
        a = []
        if isinstance(v, dict):
            for key, value in my_func(v).items():
                x = {}
                y = {}
                x['NAME'] = key
                y['VALUE'] = value
                z = {**x, **y}
                a.append(z)
        elif isinstance(v, list):
            a = [my_func(item) if isinstance(item, dict) else item for item in v]
        else:
            a = v
        new_dict[k] = a
    return new_dict


output = json.dumps(my_func(input), indent=2)
print(output)

Btw, do not use input as variable name, you're shadowing built-in function. — Olvin Roght
– Olvin Roght, Commented Sep 5, 2020 at 14:53
Your expected output is really bad, cause it violates own logic. Once you convert dict to list of dicts, but then you want dict to be converted to dict (not list of single dict). To generate such as output you need to handle lot of different exceptions, also you will need to do same when you will try to parse it back. — Olvin Roght
– Olvin Roght, Commented Sep 5, 2020 at 15:20

deadshot · Accepted Answer · 2020-09-05 15:30:40Z

1

You can use recursive function

def fun(d, res=None):
    if res is None:
        res = []
    for k, v in d.items():
        if isinstance(v, dict):
            res.append({'NAME': k, 'VALUE': fun(v)})
        elif isinstance(v, list):
            temp = {'NAME': k, 'VALUE': []}
            for x in v:
                temp['VALUE'].append(fun(x)[0])
            res.append(temp)
        else:
            res.append({'NAME': k, 'VALUE': v})
    return res

d = {k: fun(v) if isinstance(v, dict) else v  for k, v in d.items()}
print(d)

Output:

{'key1': [{'NAME': 'key1.1', 'VALUE': 'value1.1'},
          {'NAME': 'key1.2', 'VALUE': 'value1.2'},
          {'NAME': 'key1.3',
           'VALUE': [{'NAME': 'key1.3.1', 'VALUE': 'value1.3.1'},
                     {'NAME': 'key1.3.2',
                      'VALUE': [{'NAME': 'key1.3.2.1', 'VALUE': 'value1.3.2.1'},
                                {'NAME': 'key1.3.2.2',
                                 'VALUE': 'value1.3.2.2'}]}]}],
 'key2': 'value2'}

edited Sep 5, 2020 at 15:30

answered Sep 5, 2020 at 15:03

deadshot

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4 Comments

Olvin Roght Over a year ago

This produce output different from example in question

Sid Over a year ago

[{'key1.3.2.1': 'value1.3.2.1'}, {'key1.3.2.2': 'value1.3.2.2'}] also needs to be converted to dictionary of Name and Value keys such as [{"NAME": "key1.3.2.1", "VALUE": "value1.3.2.1"}, {"NAME": "key1.3.2.2", "VALUE": "value1.3.2.2"}]

Olvin Roght Over a year ago

@deadshot, don't waste your time, this output has no strict logic, so you can't write universal solution to generate it.

Olvin Roght Over a year ago

@deadshot, but what you'll do if value of key 'key1.3.2' will be list of lists? How to proceed if somewhere appeared just list of strings? Too much exceptional cases, you can't generate object without strict rules. +1 for efforts, btw, it's not your fault.

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