CODEWITHSUNDEEP
pythonjsonpython-3.x
John Doe
John Doe

Reputation: 89

Nested JSON to Flat JSON using Python

I have a Nested JSON as shown below -

sample4 = { "a": 1, "b": 2, "c": 3, "d": [{"a": 5, "b": 6}, {"a": 7, "b": 8}], "e": [{"a": 1}, {"a": 2}], "f": 9, "g": [{"a": 5, "b": 6}, {"a": 7, "b": 8}], "i": {"a": 5, "b": 6}, "j": {} }

I want to convert this into a flat JSON file.

Currently, I am using this code - 

def count_steps(dictionary):
    """counts the needed steps from the longest list inside the dictionary"""
    return max((len(value) for value in dictionary.values() if isinstance(value, list)))

def flatten(dictionary, name=''):
    steps = count_steps(dictionary)
    return_out = []
    for step in range(0, steps):
        out = {}
        for key, value in dictionary.items():
            if isinstance(value, list):
                for key_inner, value_inner in value[step].items():
                    combined_key = key + '_' + key_inner
                    if combined_key not in out:
                        out[combined_key] = []
                    out[combined_key] = value_inner
            else:
                out[key] = value
        return_out.append(out)
    return return_out

When I am using this code, I get the following output -

[{'a': 1,
  'b': 2,
  'c': 3,
  'd_a': 5,
  'd_b': 6,
  'e_a': 1,
  'f': 9,
  'g_a': 5,
  'g_b': 6,
  'i': {'a': 5, 'b': 6},
  'j': {}},
 {'a': 1,
  'b': 2,
  'c': 3,
  'd_a': 7,
  'd_b': 8,
  'e_a': 2,
  'f': 9,
  'g_a': 7,
  'g_b': 8,
  'h_a': 7,
  'h_b': 8,
  'i': {'a': 5, 'b': 6},
  'j': {}}]

But I want the following output - 

[{'a': 1,
  'b': 2,
  'c': 3,
  'd_a': 5,
  'd_b': 6,
  'e_a': 1,
  'f': 9,
  'g_a': 5,
  'g_b': 6,
  'i_a': 5, 
  'i_b': 6,
  'j': {}},
 {'a': 1,
  'b': 2,
  'c': 3,
  'd_a': 7,
  'd_b': 8,
  'e_a': 2,
  'f': 9,
  'g_a': 7,
  'g_b': 8,
  'h_a': 7,
  'h_b': 8,
  'i_a': 5, 
  'i_b': 6,
  'j': {}}]

The code here first counts the maximum number of elements in all the lists present in the JSON.

Upvotes: 0

Views: 208

Answers (2)

sam
sam

Reputation: 1896

data = { "a": 1, "b": 2, "c": 3, "d": [{"a": 5, "b": 6}, {"a": 7, "b": 8}], "e": [{"a": 1}, {"a": 2}], "f": 9, "g": [{"a": 5, "b": 6}, {"a": 7, "b": 8}], "i": {"a": 5, "b": 6}, "j": {} }

def flatten(dictionary):
    """counts the needed steps from the longest list inside the dictionary"""
    bag = [] # keys to be deleted
    new_dict = dict() # new keys to be added
    for key, value in dictionary.items():
        if type(value) is list:
          bag.append(key)
          for _value in value:
            if type(_value) is dict:  
                for key1, value2 in _value.items():
                    new_key = key + '_' + key1
                    new_dict[new_key] = value2
                    print((new_key, value2))
        else:
            print((key, value))
    for key in bag:
        del dictionary[key]
    for key, value in new_dict.items():
        dictionary[key] = value
    return dictionary

print(flatten(data))

Upvotes: 1

ladofa
ladofa

Reputation: 667

I thought there is more beautiful way to solve it, but I tried to follow your way and do smallest modifying.

The key is to care about type(dict).

sample4 = { 
    "a": 1, 
    "b": 2, 
    "c": 3, 
    "d": [{"a": 5, "b": 6}, {"a": 7, "b": 8}], 
    "e": [{"a": 1}, {"a": 2}], 
    "f": 9, 
    "g": [{"a": 5, "b": 6}, {"a": 7, "b": 8}], 
    "i": {"a": 5, "b": 6}, 
    "j": {} }


def count_steps(dictionary):
    """counts the needed steps from the longest list inside the dictionary"""
    return max((len(value) for value in dictionary.values() if isinstance(value, list)))

def merge_dict(outer_dict, inner_dict, key):
    for key_inner, value_inner in inner_dict.items():
        combined_key = key + '_' + key_inner
        outer_dict[combined_key] = value_inner

def flatten(dictionary, name=''):
    steps = count_steps(dictionary)
    return_out = []
    for step in range(0, steps):
        out = {}
        for key, value in dictionary.items():
            if isinstance(value, list):
                merge_dict(out, value[step], key)
                # for key_inner, value_inner in value[step].items():
                #     combined_key = key + '_' + key_inner
                #     if combined_key not in out:
                #         out[combined_key] = []
                #     out[combined_key] = value_inner
            elif isinstance(value, dict):
                #exception for "j"
                if len(value) == 0:
                    out[key] = {}
                else:
                    merge_dict(out, value, key)
            else:
                out[key] = value

        return_out.append(out)
    return return_out

sample5 = flatten(sample4)
print(sample5)

Upvotes: 1

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