Get Multiple Keys From Json In Python

I have the following json structure in a file: [ { 'Date': 'timevalue', 'Org': 'b4256282', 'Referer': 'somevalue' }, { 'Date': 'timevalue', 'Org': 'b4257482', 'Referer': 'somevalue

Solution 1:

You need to iterate over each dictionary in the list and then index dict in turn with dict indexing. For each dictionary in the list,

1. get the i dictionary
2. get the value associated with Org in the i dictionary
3. print said value from (2)

In code, this is

for dict_ in jdata:
    org = dict_['Org']
    print(org)

However, we have the power of list comprehension at our disposal, so the code above can more succinctly be represented by,

jdata = json.loads(str_of_above_json)
orgs = [x['Org'] for x in jdata]      
print(orgs)       

Why isn't your current code working? You do jdata['Org'], but [...] is a dict indexing operation, and this will throw errors because jdata is a list.

Solution 2:

Remember, your data once loaded in as a Python structure is going to be a list of dictionaries. To keep it simple, it's just:

[{...}, {...}, {...}]

Keyword here is a list. So, your for loop will/should iterate over a list, giving you each dictionary at every iteration.

At this point you can then access the 'Org' key.

So when you do:

for orgs in jdata:
    print(orgs) # each dictionary

At that point, you can now access the Org key in the dictionary:

for orgs in jdata:
    print(orgs) # each dictionaryprint(orgs['Org']) # The value of 'Org' in each dictionary

Solution 3:

import json

jdata = json.loads(str_of_above_json)
for orgs in jdata:
    print(orgs["Org"])

You need to iterate over the list [] then print the "org" part of each data.

Solution 4:

All the other answers are correct and will solve your current problem.

But if you manipulate a lot with structures like this, you might find the package plucky helpful (full disclosure: I'm the author of that package).

For example (assuming your jdata is loaded):

>>> import plucky
>>> plucky.plucks(data, '.Org')
['b4256282', 'b4257482', 'b4253345']

or:

>>> plucky.pluckable(data).Org['b4256282', 'b4257482', 'b4253345']