CODEWITHSUNDEEP
pythonjsonapidata-extraction
Brzozova
Brzozova

Reputation: 382

Python: how to assign multiple values to one key

I extract data using API and retrieve a list of servers and backups. Some servers have more than one backup. This is how I get list of all servers with backaup IDs.

bkplist = requests.get('https://heee.com/1.2/storage/backup')
bkplist_json = bkplist.json()
backup_list = bkplist.json()
backupl = backup_list['storages']['storage']

Json looks like this:

{
   "storages": {
      "storage": [
         {
            "access": "",
            "created": "",
            "license": ,
            "origin": "01165",
            "size": ,
            "state": "",
            "title": "",
            "type": "backup",
            "uuid": "01019",
            "zone": ""
         },

Firstly I create a dictionary to store this data:

backup = {}
for u in backup_list['storages']['storage']:
    srvuuidorg = u['origin']
    backup_uuid = u['uuid']
    backup[srvuuidorg] = backup_uuid

But then I find out there is more than one value for every server. As dictionary can have just one value assigned to one key I wanted to use some hybrid of list and dictionary, but I just can't figure it out how to do this with this example.

Servers are nested in storages->storage and I need to assign a couple of uuid which is backup ID to one origin which is server ID.

I know about collections module and with a simple example it is quite understandable, but I have a problem how to use this in my example with extracting data through API.

How extract origin and assign to this key other values stored in json uuid?

What's more it is a massive amount of data so I cannot add every value manually.

Upvotes: 3

Views: 156

Answers (4)

Aryamaan Goswamy
Aryamaan Goswamy

Reputation: 319

You can use a list for multiple values.

dict = {"Greetings": ["hello", "hi"]}

Upvotes: 1

Max Voitko
Max Voitko

Reputation: 1629

You could store uuid list for origin key.

I sugget the following 2 ways:

  1. Creating empty list for first accessing origin, and then appending to it:
backup = {}
for u in backup_list['storages']['storage']:
    srvuuidorg = u['origin']
    backup_uuid = u['uuid']
    if not backup.get(srvuuidorg):
        backup[srvuuidorg] = []
    backup[srvuuidorg].append(backup_uuid)
  1. Using defaultdict collection, which basically does the same for you under the hood:
from collections import defaultdict


backup = defaultdict(list)
for u in backup_list['storages']['storage']:
    srvuuidorg = u['origin']
    backup_uuid = u['uuid']
    backup[srvuuidorg].append(backup_uuid)

It seems to me that the last way is more elegant. If you need to store uuid unique list you should use the saem approach with set instead of list.

Upvotes: 2

Florian Bernard
Florian Bernard

Reputation: 2569

You can do something like this.

from collections import defaultdict

backup = defaultdict(list)
for u in backup_list['storages']['storage']: 
    srvuuidorg = u['origin'] 
    backup_uuid = u['uuid'] 
    backup[srvuuidorg].append(backup_uuid)

Note that you can simplify your loop like this.

from collections import defaultdict

backup = defaultdict(list)
for u in backup_list['storages']['storage']:
    backup[u['origin']].append(u['uuid'])

But this may be considering as less readable.

Upvotes: 2

Raul Perez
Raul Perez

Reputation: 45

A json allows to contain an array in a key:

var= {
  "array": [
    {"id": 1, "value": "one"},
    {"id": 2, "value": "two"},
    {"id": 3, "value": "three"}
  ]
}

print var
{'array': [{'id': 1, 'value': 'one'}, {'id': 2, 'value': 'two'}, {'id': 3, 'value': 'three'}]}

var["array"].append({"id": 4, "value": "new"})

print var

{'array': [{'id': 1, 'value': 'one'}, {'id': 2, 'value': 'two'}, {'id': 3, 'value': 'three'}, {'id': 4, 'value': 'new'}]}

Upvotes: 1

Related Questions