如何将字典/嵌套字典的字典转换为列表字典

Question

Heres the original dict

uid_coins = {'141632864': {'username': 'Guest130679138', 'coins': 0}, 
        '141632884': {'username': 'Guest130679156', 'coins': 39123441}, 
        '141632886': {'username': 'Guest130679158', 'coins': 44006638}}

What I am trying to get

d = {'uid':[141632864, 141632884, 141632886], 
'username': ['Guest130679138', 'Guest130679156', 'Guest130679158'], 
'coins': [0, 39123441, 44006638]}

The keys in original dict represent uid.

This is my what I have done:

uid = list(uid_coins.keys())
username = [u_data['username'] for u_data in uid_coins.values()]
coins = [[u_data['coins'] for u_data in uid_coins.values()]]

d = {"uid":uid, "username":username, "coins":coins}


dict((key,d[key]) for d in data for key in d)

But I am rather looking for a generalized approach, to achieve without manually declaring the keys again, just so it should work with new keys in the original data.

Answer 1

Try:

uid_coins = {
    "141632864": {"username": "Guest130679138", "coins": 0},
    "141632884": {"username": "Guest130679156", "coins": 39123441},
    "141632886": {"username": "Guest130679158", "coins": 44006638},
}

out = {}
for k, v in uid_coins.items():
    out.setdefault("uid", []).append(k)
    out.setdefault("username", []).append(v["username"])
    out.setdefault("coins", []).append(v["coins"])

print(out)

Prints:

{'uid': ['141632864', '141632884', '141632886'], 
 'username': ['Guest130679138', 'Guest130679156', 'Guest130679158'], 
 'coins': [0, 39123441, 44006638]}

Answer 2

A similar version as @Andrej Kesely but with defaultdict

from collections import defaultdict

uid_coins = {
    '141632864': {'username': 'Guest130679138', 'coins': 0}, 
    '141632884': {'username': 'Guest130679156', 'coins': 39123441}, 
    '141632886': {'username': 'Guest130679158', 'coins': 44006638}
}

d = defaultdict(list)
for key, value in uid_coins.items():
    d['uid'].append(key)
    d['username'].append(value['username'])
    d['coins'].append(value['coins'])
    

Output:

defaultdict(<class 'list'>, {'uid': ['141632864', '141632884', '141632886'], 'username': ['Guest130679138', 'Guest130679156', 'Guest130679158'], 'coins': [0, 39123441, 44006638]})

Answer 3

Generalized (based on original version by Andrej Kesely):

d = {}
for k, v in uid_coins.items():
    d.setdefault('uid', []).append(k)
    for i in v.keys():
        d.setdefault(i, []).append(v[i])

Answer 4

This is how you can get the output in desired format (as mentioned in question) from the given dictionary.

d = {'uid': list(uid_coins.keys()), 'username': [i['username'] for i in list(uid_coins.values())], 'coins': [i['coins'] for i in list(uid_coins.values())]}

list(uid_coins.keys()) will return all your uids as list.

[i['username'] for i in list(uid_coins.values())] will return 'username' values as list.

[i['coins'] for i in list(uid_coins.values())] will return 'coins' values as list.