Python如何将字典列表按UserName分组为嵌套字典结构

现有如下字典列表：

[
    {'UserName': 'aaa', 'AccessKeyId': 'AKIAYWQTISJD6X27YVK', 'Status': 'Active', 'CreateDate': datetime.datetime(2022, 9, 8, 15, 56, 39, tzinfo=tzutc())},
    {'UserName': 'eee', 'AccessKeyId': 'AKIAYWQTISJD6QXMAKY', 'Status': 'Active', 'CreateDate': datetime.datetime(2023, 1, 24, 12, 30, 59, tzinfo=tzutc())}, 
    {'UserName': 'eee', 'AccessKeyId': 'AKIAYWQTISJDUARK6FV', 'Status': 'Active', 'CreateDate': datetime.datetime(2023, 1, 24, 16, 58, 38, tzinfo=tzutc())}
]

需要转换为以UserName为键、对应值为该用户相关字典（移除UserName字段）列表的嵌套字典：

{
  "aaa": [
    {'AccessKeyId': 'AKIAYWQTISJD6X27YVK', 'Status': 'Active', 'CreateDate': datetime.datetime(2022, 9, 8, 15, 56, 39, tzinfo=tzutc())}],
  "eee": [
    {'AccessKeyId': 'AKIAYWQTISJD6QXMAKY', 'Status': 'Active', 'CreateDate': datetime.datetime(2023, 1, 24, 12, 30, 59, tzinfo=tzutc())}, 
    {'AccessKeyId': 'AKIAYWQTISJDUARK6FV', 'Status': 'Active', 'CreateDate': datetime.datetime(2023, 1, 24, 16, 58, 38, tzinfo=tzutc())}
   ]
}

尝试了以下代码，但同一用户的多条数据仅能保留最后一条，无法生成列表：

import copy

list_per_user = {i['UserName']: copy.deepcopy(i) for i in key_list} 
for obj in list_per_user:     
   del list_per_user[obj]['UserName'] 

问题出在字典推导式的特性：当键重复时，后面的键值对会覆盖前面的，所以同一用户的多条记录最后只留下了最后一条。要实现需求，需要为每个用户维护一个列表，把对应记录添加进去。

方法一：普通循环实现 #

from datetime import datetime, timezone
import copy

key_list = [
    {'UserName': 'aaa', 'AccessKeyId': 'AKIAYWQTISJD6X27YVK', 'Status': 'Active', 'CreateDate': datetime(2022, 9, 8, 15, 56, 39, tzinfo=timezone.utc)},
    {'UserName': 'eee', 'AccessKeyId': 'AKIAYWQTISJD6QXMAKY', 'Status': 'Active', 'CreateDate': datetime(2023, 1, 24, 12, 30, 59, tzinfo=timezone.utc)}, 
    {'UserName': 'eee', 'AccessKeyId': 'AKIAYWQTISJDUARK6FV', 'Status': 'Active', 'CreateDate': datetime(2023, 1, 24, 16, 58, 38, tzinfo=timezone.utc)}
]

result = {}
for item in key_list:
    username = item['UserName']
    # 深拷贝原字典并移除UserName字段
    user_data = copy.deepcopy(item)
    del user_data['UserName']
    # 如果用户已在结果中，追加数据；否则创建新列表
    if username in result:
        result[username].append(user_data)
    else:
        result[username] = [user_data]

print(result)

方法二：使用collections.defaultdict简化代码 #

defaultdict可以自动为不存在的键生成默认值（这里是空列表），让代码更简洁：

from datetime import datetime, timezone
import copy
from collections import defaultdict

key_list = [
    {'UserName': 'aaa', 'AccessKeyId': 'AKIAYWQTISJD6X27YVK', 'Status': 'Active', 'CreateDate': datetime(2022, 9, 8, 15, 56, 39, tzinfo=timezone.utc)},
    {'UserName': 'eee', 'AccessKeyId': 'AKIAYWQTISJD6QXMAKY', 'Status': 'Active', 'CreateDate': datetime(2023, 1, 24, 12, 30, 59, tzinfo=timezone.utc)}, 
    {'UserName': 'eee', 'AccessKeyId': 'AKIAYWQTISJDUARK6FV', 'Status': 'Active', 'CreateDate': datetime(2023, 1, 24, 16, 58, 38, tzinfo=timezone.utc)}
]

result = defaultdict(list)
for item in key_list:
    user_data = copy.deepcopy(item)
    del user_data['UserName']
    result[item['UserName']].append(user_data)

# 如果需要转为普通字典，可执行：result = dict(result)
print(result)

方法三：不使用深拷贝（如果原数据不需要保留） #

如果不需要保留原列表中的字典，可以直接修改原字典（注意这会改变原数据），省去深拷贝的开销：

from datetime import datetime, timezone
from collections import defaultdict

key_list = [
    {'UserName': 'aaa', 'AccessKeyId': 'AKIAYWQTISJD6X27YVK', 'Status': 'Active', 'CreateDate': datetime(2022, 9, 8, 15, 56, 39, tzinfo=timezone.utc)},
    {'UserName': 'eee', 'AccessKeyId': 'AKIAYWQTISJD6QXMAKY', 'Status': 'Active', 'CreateDate': datetime(2023, 1, 24, 12, 30, 59, tzinfo=timezone.utc)}, 
    {'UserName': 'eee', 'AccessKeyId': 'AKIAYWQTISJDUARK6FV', 'Status': 'Active', 'CreateDate': datetime(2023, 1, 24, 16, 58, 38, tzinfo=timezone.utc)}
]

result = defaultdict(list)
for item in key_list:
    username = item.pop('UserName')
    result[username].append(item)

print(result)

内容的提问来源于stack exchange，提问作者opti2k4