You could nest the dictionaries as follows:

jsondata = {}
agent={}
content={}
agent['agentid'] = 'john'
content['eventType'] = 'view'
content['othervar'] = "new"

jsondata['agent'] = agent
jsondata['content'] = content
print(json.dumps(jsondata))

Output:

print {"content": {"eventType": "view", "othervar": "new"}, "agent": {"agentid": "john"}}

What @andrea-f looks good to me, here another solution:

Feel free to pick in both :)

import json

dic = {
        "bomber": [1, 2, 3, 4, 5],
        "irritation": [1, 3, 5, 7, 8]
      }

filename = "abc.pdf"

json_dict = {}
data = []

for k, v in dic.iteritems():
  tmp_dict = {}
  tmp_dict["keyword"] = k
  tmp_dict["term_freq"] = len(v)
  tmp_dict["lists"] = [{"occurrance": i} for i in v]
  data.append(tmp_dict)

json_dict["filename"] = filename
json_dict["data"] = data

with open("abc.json", "w") as outfile:
    json.dump(json_dict, outfile, indent=4, sort_keys=True)

It's the same idea, I first create a big json_dict to be saved directly in json. I use the with statement to save the json avoiding the catch of exception

Also, you should have a look to the doc of json.dumps() if you need future improve in your json output.

EDIT

And just for fun, if you don't like tmp var, you can do all the data for loop in a one-liner :)

json_dict["data"] = [{"keyword": k, "term_freq": len(v), "lists": [{"occurrance": i} for i in v]} for k, v in dic.iteritems()]

It could gave for final solution something not totally readable like this:

import json

json_dict = {
              "filename": "abc.pdf",
              "data": [{
                        "keyword": k,
                        "term_freq": len(v),
                        "lists": [{"occurrance": i} for i in v]
                       } for k, v in dic.iteritems()]
            }

with open("abc.json", "w") as outfile:
    json.dump(json_dict, outfile, indent=4, sort_keys=True)

EDIT 2

It looks like you don't want to save your json as the desired output, but be abble to read it.

In fact, you can also use json.dumps() in order to print your json.

with open('abc.json', 'r') as handle:
    new_json_dict = json.load(handle)
    print json.dumps(json_dict, indent=4, sort_keys=True)

There is still one problem here though, "filename": is printed at the end of the list because the d of data comes before the f.

To force the order, you will have to use an OrderedDict in the generation of the dict. Be careful the syntax is ugly (imo) with python 2.X

Here is the new complete solution ;)

import json
from collections import OrderedDict

dic = {
        'bomber': [1, 2, 3, 4, 5],
        'irritation': [1, 3, 5, 7, 8]
      }

json_dict = OrderedDict([
              ('filename', 'abc.pdf'),
              ('data', [ OrderedDict([
                                        ('keyword', k),
                                        ('term_freq', len(v)),
                                        ('lists', [{'occurrance': i} for i in v])
                                     ]) for k, v in dic.iteritems()])
            ])

with open('abc.json', 'w') as outfile:
    json.dump(json_dict, outfile)


# Now to read the orderer json file

with open('abc.json', 'r') as handle:
    new_json_dict = json.load(handle, object_pairs_hook=OrderedDict)
    print json.dumps(json_dict, indent=4)

Will output:

{
    "filename": "abc.pdf", 
    "data": [
        {
            "keyword": "bomber", 
            "term_freq": 5, 
            "lists": [
                {
                    "occurrance": 1
                }, 
                {
                    "occurrance": 2
                }, 
                {
                    "occurrance": 3
                }, 
                {
                    "occurrance": 4
                }, 
                {
                    "occurrance": 5
                }
            ]
        }, 
        {
            "keyword": "irritation", 
            "term_freq": 5, 
            "lists": [
                {
                    "occurrance": 1
                }, 
                {
                    "occurrance": 3
                }, 
                {
                    "occurrance": 5
                }, 
                {
                    "occurrance": 7
                }, 
                {
                    "occurrance": 8
                }
            ]
        }
    ]
}

But be carefull, most of the time, it is better to save a regular .json file in order to be cross languages.

Try this. Though your sample input and output data don't really give many clues as to where the "name" fields should come from. I've assumed you wanted the name of the original item in your list.

original_json = json.load(open('data/bricsinvestorsfirst.json'),'r')

response_json = {}
response_json["name"] = "analytics"

# where your children list will go
children = []

size = 500 # or whatever else you want

# For each item in your original list
for item in original_json:
    children.append({"name" : item["name"],
                     "size" : size})

response_json["children"] = children

print json.dumps(response_json,indent=2)

Your input JSON isn't proper, it should be something like:

{ "items": 
    [ 
       {
       "number": 
           {
           "value": 23,
           "label": "test"
           }
       } 
    ] 
}

Besides that it can get messy, but accessing the resultant dict is intuitive.

 jdict = json.loads(yourjson)
 jdict['items'] => [{"number":{...}}]
 jdict['items'][0] => {"number":{...}}
 jdict['items'][0]['number']['value'] => 23

Edit:

Think you actually just wanted this:

myjson.get('items').append({'number': d})

You have to append a dictionary, not entries of a dictionary to items.

You simply put them in another dictionary, and use a new list. So:

import json

list1 = ['key1 value1', 'key2 value2', 'key3 value3']
dict1 = {'Name1': [dict(item.split(" ",1) for item in list1)] }
#       ^ dict    ^ list with 1 element            end list ^ ^ end dict
json.dumps(dict1)

And this produces:

>>> print(json.dumps(dict1))
{"Name1": [{"key2": "value2", "key3": "value3", "key1": "value1"}]}

Notes:

1. A dictionary can only contain different keys (both in JSON and Python);
2. You better split with .split(" ",1) since if the value contains spaces, these are all seen as still a single value.
3. dictionaries are unordered, so the order of th keys can be shuffled.

Maybe I'm missing something, but I'd start with the json docs. json.dumps and json.loads specifically.

The csv module will handle the CSV reading nicely - including handling line breaks that are within quotes.

import csv
with open('my_csv.csv') as csv_file:
   for row in csv.reader(csv_file):
       # do work

The csv.reader object is an iterator - you can iterate through the rows in the CSV by using a for loop. Each row is a list, so you can get each field as row[0], row[1], etc. Be aware that this will load the first row (which just contains field names in your case).

As we have field names given to us in the first row, we can use csv.DictReader so that fields in each row can be accessed as row['id'], row['name'], etc. This will also skip the first row for us:

import csv
with open('my_csv.csv') as csv_file:
   for row in csv.DictReader(csv_file):
       # do work

For the JSON export, use the json module. json.dumps() will take Python data structures such as lists and dictionaries and return the appropriate JSON string:

import json
my_data = {'id': 123, 'name': 'Test User', 'emails': ['[email protected]', '[email protected]']}
my_data_json = json.dumps(my_data)

If you want to generate JSON output exactly as you posted, you'd do something like:

output = {'persons': []}
with open('my_csv.csv') as csv_file:
    for person in csv.DictReader(csv_file):
        output['persons'].append({
            'type': 'config.profile',
            'id': person['id'],
            # ...add other fields (email etc) here...
        })

        # ...do similar for config.pictures, config.status, etc...

output_json = json.dumps(output)

output_json will contain the JSON output that you want.

However, I'd suggest you carefully consider the structure of the JSON output that you're after - at the moment, you're defining an outer dictionary that serves no purpose, and you're adding all your 'config' data directly under 'persons' - you may want to reconsider this.

The rgp loop can be made more compact, and a bit faster, with:

def foo2(rgp):
    alist = []
    for n, g in rgp:
        temp2 = {"name": n}
        values = g.T.to_dict().values()
        n = len(values)
        def size(t): 
            return (t['Rating'] * t['Count'] * 10000) / n
        temp3 = [{'name': t['Name'], 'size': size(t)} for t in values]
        temp2['children'] = temp3
        alist.append(temp2)
    return alist

I don't have enough experience with Pandas to know whether it is possible to improve on the groupby. For example, it would be possible to perform a 2 level grouping with one call - ie. group on 'Count' and within that 'Rating'?

Considering that json is a string version of a dict, and you have a specific dictionary layout in mind, I don't see how you can organize the code in any other way. With the exception of update, all dictionary additions are key by key. So you have to have these 2 loops over groups.

You're getting error because your json file is incorrectly formatted and thus calling json.load() will raise an JSONDecodeError.

Your json structure should look like,

{
    "companies": {
        "company1": [
            {
                "path": "C:/USER/Path/Company1/",
                "files": [
                    {
                        "_CO": {
                            "ID": "ID",
                            "Report Number": "Report_Number"
                        }
                    },
                    {
                        "_TR": {
                            "ID": "Trade_Ident",
                            "Report Number": "Number of Report"
                        }
                    }
                ]
            }
        ],
        "company2": [
            {
                "path": "C:/USER/Path/Company2/",
                "files": [
                    {
                        "_CO": {
                            "ID": "Identification",
                            "Report Number": "Report-Number"
                        }
                    },
                    {
                        "_TR": {
                            "ID": "Ident",
                            "Report Number": "NumberReport"
                        }
                    }
                ]
            }
        ]
    }
}

Hope it helps you!