You can do this.

data[0]['f'] = var

Sounds like you want to load a dictionary from json, add new key values and write it back. If that's the case, you can do this:

with open('python_dictionary.json','r+') as f:
    dic = json.load(f)
    dic.update(new_dictionary)
    json.dump(dic, f)

(mode is 'r+' for reading and writing, not appending because you're re-writing the entire file)

If you want to do the append thing, along with json.dumps, I guess you'd have to remove the first { from the json.dumps string before appending. Something like:

with open('python_dictionary.json','a') as f:
    str = json.dumps(new_dictionary).replace('{', ',', 1)
    f.seek(-2,2)
    f.write(str)

There are several questions here. The main points worth mentioning:

• Use can use a list to hold your arguments and use *args to unpack when you supply them to add_entry.
• To check / avoid duplicates, you can use set to track items already added.
• For writing to JSON, now you have a list, you can simply iterate your list and write in one function at the end.

Putting these aspects together:

import json

res = []
seen = set()

def add_entry(res, name, element, type):

    # check if in seen set
    if (name, element, type) in seen:
        return res

    # add to seen set
    seen.add(tuple([name, element, type]))

    # append to results list
    res.append({'name': name, 'element': element, 'type': type})

    return res

args = ['xyz', '4444', 'test2']

res = add_entry(res, *args)  # add entry - SUCCESS
res = add_entry(res, *args)  # try to add again - FAIL

args2 = ['wxy', '3241', 'test3']

res = add_entry(res, *args2)  # add another - SUCCESS

Result:

print(res)

[{'name': 'xyz', 'element': '4444', 'type': 'test2'},
 {'name': 'wxy', 'element': '3241', 'type': 'test3'}]

Writing to JSON via a function:

def write_to_json(lst, fn):
    with open(fn, 'a', encoding='utf-8') as file:
        for item in lst:
            x = json.dumps(item, indent=4)
            file.write(x + '\n')

#export to JSON
write_to_json(res, 'elements.json')

I would do this:

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

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

Dictionaries in Python (and most other languages) have no order to them - they are a set (unordered collection) of key:value pairs. Hence the argument that they are in the wrong order is incorrect as the premise that the data structure has an order is false.

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

You have to read your JSON file and then convert it to list instead of dict. Then you just need to append to that list and overwrite your JSON file.

import json

data = json.load(open('data.json'))

# convert data to list if not
if type(data) is dict:
    data = [data]

# append new item to data lit
data.append({
  "user": "user2",
  "id": "21780"
})

# write list to file
with open('data.json', 'w') as outfile:
    json.dump(data, outfile)

The assignment statement x = y = z implies that both x and y will take on the value of z.

As an example, look at the byte code for the assignment a = b = 2:

In [45]: import dis; dis.dis(compile('a = b = 2', '', 'exec'))
  1           0 LOAD_CONST               0 (2)
              3 DUP_TOP
              4 STORE_NAME               0 (a)
              7 STORE_NAME               1 (b)
             10 LOAD_CONST               1 (None)
             13 RETURN_VALUE

With 4 STORE_NAME, a is assigned first to 2, followed by 7 STORE_NAME where b is then assigned to the same value, 2.

So, with

sal = json.loads(salaries)["Hritik"] = 0

sal receives the value 0. Also, a temporary variable is created when you call json.loads and that is modified, following which the reference is lost.

In order to get this to work, you'll need to break this up into 2 parts, as you have already done.

sal = json.loads(salaries)
sal['Hritik'] = 0