The brackets denote a JSON array, containing one element in your example. In Python, simply pick out the first element of the root array and convert back to JSON.

import json
data = json.loads('[...]')
str = json.dumps(data[0])

why?

the whole point of the json format is that once you parse it you can easily access whatever info you need.

what info do you need?

Per discussion in the comments, the problem was that the original code was using commas at the end of every value passed to json_object:

Wrong code:

json_object['date_value'] = date_value,
json_object['boolean_value'] = boolean_value,
json_object['string_value'] = string_value,
json_object['integer_value'] = integer_value,

Correct code:

json_object['date_value'] = date_value
json_object['boolean_value'] = boolean_value
json_object['string_value'] = string_value
json_object['integer_value'] = integer_value

Python interprets a value with a comma as a tuple with one item which then gets converted to a list when written out as JSON.

[] indicates a list, so you need to get into the first element in the list first.

data = data["source_contents"][0]
df = pd.DataFrame.from_dict(data["fields"], orient="index")

Output:

                                 type original_field_name      mode
employee_id  {'field_type': 'string'}         Employee ID  required

If there's always exactly one "extra" set of array brackets:

for result in page['blah']:
    that = result[0]['this']
    list.append(that)

You don't need to "remove" anything. You have a dictionary, each of whose values is a list. Calling str on a list will show the repr of the elements in it. If you don't want that, print the elements of the list:

for item in data:
    print ', '.join(item['brand'])

Your desired output isn't a valid JSON format. You can manipulate the string representation of the JSON easily though.

import json

json.dumps(json_formatted).strip('[]').replace('},', '}')

strip removes all of the characters in the argument from the beginning and end of the string. The replace will just remove all of the commas following an end brace.

grades.append(allgrades)

allgrades is a list, which you're appending to grades - another list. This creates a nested list which is where the extra ] is from.

What you want is to extend the existing list with the new one:

grades.extend(allgrades)

Which adds all the entries in allgrades to grades - without any nesting.

Use this when you return:

return properties[0];

Or

var data = [

{
    "Name": "TEST",
    "deviceId": "",
    "CartId": "",
    "timestamp": 1383197265540,
    "FOOD": [],
    "City": "LONDON CA"
 }

]; // Or whatever the Json is
data = data[0];

Or if you're accessing the json via another object

var data = jsonObj[0];

You should use ', '.join:

'value': ', '.join(df['email'])

e.g

df = pd.DataFrame({'email': ['[email protected]', '[email protected]']})
print(', '.join(df['email']))
# [email protected], [email protected]

You can load the json into a python object, and then convert the python object to YAML. The other solution is to simply iterate over the dictionaries and format it however you want.

Here's an example of converting it to YAML. It doesn't give you precisely what you want, but it's pretty close. There are lots of ways to customize the output, this is just a quick hack to show the general idea:

import json
import yaml

data = json.loads('''
   {
        "A": {
            "A1": 1,
            "A2": 2
        },
        "B": {
            "B1": {
                "B11": {
                    "B111": 1,
                    "B112": 2
                },
                "B12": {
                    "B121": 1,
                    "B122": 2
                }
            },
            "B2": {
                "B21": [1,2],
                "B22": [1,2]
            }
        },
        "C": "CC"
    }
''')

print yaml.safe_dump(data, allow_unicode=True, default_flow_style=False)

This is the output I get:

A:
  A1: 1
  A2: 2
B:
  B1:
    B11:
      B111: 1
      B112: 2
    B12:
      B121: 1
      B122: 2
  B2:
    B21:
    - 1
    - 2
    B22:
    - 1
    - 2
C: CC