[英]How do I convert dict of dicts/nested dicts into dict of list
这是原始字典
uid_coins = {'141632864': {'username': 'Guest130679138', 'coins': 0},
'141632884': {'username': 'Guest130679156', 'coins': 39123441},
'141632886': {'username': 'Guest130679158', 'coins': 44006638}}
我想得到什么
d = {'uid':[141632864, 141632884, 141632886],
'username': ['Guest130679138', 'Guest130679156', 'Guest130679158'],
'coins': [0, 39123441, 44006638]}
原始dict中的键代表uid。
这是我所做的:
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)
但我宁愿寻找一种通用方法,无需再次手动声明键即可实现,因此它应该与原始数据中的新键一起使用。
尝试:
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)
印刷:
{'uid': ['141632864', '141632884', '141632886'],
'username': ['Guest130679138', 'Guest130679156', 'Guest130679158'],
'coins': [0, 39123441, 44006638]}
与@Andrej Kesely 类似的版本,但带有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]})
概括(基于 Andrej Kesely 的原始版本):
d = {}
for k, v in uid_coins.items():
d.setdefault('uid', []).append(k)
for i in v.keys():
d.setdefault(i, []).append(v[i])
这就是您如何从给定字典中以所需格式(如问题所述)获取 output 的方式。
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())
将返回您所有的 uid 作为列表。
[i['username'] for i in list(uid_coins.values())]
将返回 'username' 值作为列表。
[i['coins'] for i in list(uid_coins.values())]
将返回 'coins' 值作为列表。
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