jsobj["a"]["b"]["e"].append({"f":var3, "g":var4, "h":var5})
jsobj["a"]["b"]["e"].append({"f":var6, "g":var7, "h":var8})

Python includes SimpleJSON via the json module. Use it to serialize/de-serialize from JSON strings and python dicts:

myJSON_d = json.loads(myJSON)

myJSON_d.['DataSet'].update({'values': values, 'timeStamps': timeStamps})

myJSON = json.dumps(myJSON_d)

model.fruits = {"id": 1, "name": "apple"}* defines a dictionary, you cannot append to a dictionary, you can add keys to it

You can append elements into a JSON array though, e.g. if model.fruits was an array (or a list, as it is called in Python), e.g.

model.fruits = [{"id": 1, "name": "apple"},]

which defines an array with 1 element in it, then you can append to it

model.fruits.append({"id": 2, "name": "banana"})

Some references:

https://docs.djangoproject.com/en/1.8/ref/contrib/postgres/fields/#hstorefield https://docs.djangoproject.com/en/1.8/ref/contrib/postgres/fields/#arrayfield

* To be honest it doesn't make sense to name something in the plural like fruits to hold a single object, should be a container, like a list

First, accidents is a dictionary, and you can't write to a dictionary; you just set values in it.

So, what you want is:

for accident in accidents:
    accident['Turn'] = 'right'

The thing you want to write out is the new JSON—after you've finished modifying the data, you can dump it back to a file.

Ideally you do this by writing to a new file, then moving it over the original:

with open('sanfrancisco_crashes_cp.json') as json_file:
    json_data = json.load(json_file)
accidents = json_data['accidents']
for accident in accidents:
    accident['Turn'] = 'right'
with tempfile.NamedTemporaryFile(dir='.', delete=False) as temp_file:
    json.dump(temp_file, json_data)
os.replace(temp_file.name, 'sanfrancisco_crashes_cp.json')

But you can do it in-place if you really want to:

# notice r+, not rw, and notice that we have to keep the file open
# by moving everything into the with statement
with open('sanfrancisco_crashes_cp.json', 'r+') as json_file:
    json_data = json.load(json_file)
    accidents = json_data['accidents']
    for accident in accidents:
        accident['Turn'] = 'right'
    # And we also have to move back to the start of the file to overwrite
    json_file.seek(0, 0)
    json.dump(json_file, json_data)
    json_file.truncate()

If you're wondering why you got the specific error you did:

In Python—unlike many other languages—assignments aren't expressions, they're statements, which have to go on a line all by themselves.

But keyword arguments inside a function call have a very similar syntax. For example, see that tempfile.NamedTemporaryFile(dir='.', delete=False) in my example code above.

So, Python is trying to interpret your accident['Turn'] = 'right' as if it were a keyword argument, with the keyword accident['Turn']. But keywords can only be actual words (well, identifiers), not arbitrary expressions. So its attempt to interpret your code fails, and you get an error saying keyword can't be an expression.

I would do this:

data["list"].append({'b':'2'})

so simply you are adding an object to the list that is present in "data"

You could do a json.loads to be able to work with Python data structures, then do the mapping into the target structure, and afterwards do a json.dumps to write back to the target JSON file.

import json

with open(file_name) as file:
    string = file.read()
    obj = json.loads(string)

target = map_to_target(obj)

with open(target_file_name, 'w') as out_file:
    out = json.dumps(target)
    out_file.write(out)

As mentioned in the comments, once loaded the json is a python dictionary. You should now iterate on the value of the 'data' key, extract the information and store them in a list.

Something like this should do the job:

json_obj = json.load(json1)

ips = []
for piece in json_obj['data']:
    this_ip = [piece['custom3'], piece['modemId']]
    ips.append(this_ip)

One line form:

ips = [[piece['custom3'], piece['modemId']] for piece in json_obj['data']]