Python -将字典的列表重新组合成两个嵌套的字典列表？
Python -将字典的列表重新组合成两个嵌套的字典列表？
提问于 2019-05-08 10:04:06
我有一个字典列表，其中包含匹配的站点和匹配的设备，我想按站点和设备重新组合这些字典。
我添加了一个示例输出字典和一个所需的字典。
我想我可以用迭代工具来做多个有效的组--我确实有这些组--但是我不知道如何将它们合并，或者这是否是最有效的方法
迭代工具尝试：
site_groups = itertools.groupby(bgp_data_query, lambda i: i['location'])
for key, site in site_groups:
device_groups = itertools.groupby(site, lambda i: i['device_name'])
for key, device in site_groups:原始数据
[
{
"bgp_peer_as": "1",
"bgp_session": "3:35",
"bgp_routes": "0",
"service_status": "Down",
"location": "London",
"circuit_name": "MPLS",
"device_name": "LON-EDGE",
"timestamp" : "2019-5-8 12:30:00"
},
{
"bgp_peer_as": "3",
"bgp_session": "4:25",
"bgp_routes": "100",
"service_status": "UP",
"location": "London",
"circuit_name": "MPLS 02",
"device_name": "LON-EDGE",
"timestamp" : "2019-5-8 12:30:00"
},
{
"bgp_peer_as": "18",
"bgp_session": "1:25",
"bgp_routes": "1",
"service_status": "UP",
"location": "London",
"circuit_name": "INTERNET",
"device_name": "LON-INT-GW",
"timestamp" : "2019-5-8 12:31:00"
},
{
"bgp_peer_as": "20",
"bgp_session": "1:25",
"bgp_routes": "1",
"service_status": "UP",
"location": "Manchester",
"circuit_name": "INTERNET",
"device_name": "MAN-INT-GW",
"timestamp" : "2019-5-8 12:20:00"
},
{
"bgp_peer_as": "20",
"bgp_session": "1:25",
"bgp_routes": "1",
"service_status": "UP",
"location": "Manchester",
"circuit_name": "INTERNET 02",
"device_name": "MAN-INT-GW",
"timestamp" : "2019-5-8 12:20:00"
},
{
"bgp_peer_as": "45",
"bgp_session": "1:25",
"bgp_routes": "1",
"service_status": "UP",
"location": "Manchester",
"circuit_name": "MPLS 01",
"device_name": "MAN-EDGE",
"timestamp" : "2019-5-8 12:21:00"
},
]期望的dict
[
{
"London": {
"LON-EDGE": {
"bgp_peer_as": "1",
"bgp_session": "3:35",
"bgp_routes": "0",
"service_status": "DOWN",
"circuit_name": "MPLS",
},
{
"bgp_peer_as": "1",
"bgp_session": "4:25",
"bgp_routes": "100",
"service_status": "UP",
"circuit_name": "MPLS 02",
}
},
{
"LON-INT-GW" : {
"bgp_peer_as": "18",
"bgp_session": "1:25",
"bgp_routes": "1",
"service_status": "UP",
"circuit_name": "INTERNET",
}
}
}
],
[
{
"Manchester": {
"MAN-EDGE": {
"bgp_peer_as": "45",
"bgp_session": "1:25",
"bgp_routes": "1",
"service_status": "UP",
"circuit_name": "MPLS 01",
}
},
{
"MAN-INT-GW": {
"bgp_peer_as": "20",
"bgp_session": "1:25",
"bgp_routes": "1",
"service_status": "UP",
"circuit_name": "INTERNET",
},
{
"bgp_peer_as": "20",
"bgp_session": "1:25",
"bgp_routes": "1",
"service_status": "UP",
"circuit_name": "INTERNET 02",
}
}
}
]回答 2
Stack Overflow用户
回答已采纳
发布于 2019-05-08 10:14:09
为此，使用双collections.defaultdict，其中包含一个最高级的列表，并在项目上循环，弹出“键”，这样它们就不会出现在最终数据中：
result = collections.defaultdict(lambda :collections.defaultdict(list))
for d in raw_dict:
location = d.pop("location")
device_name = d.pop("device_name")
result[location][device_name].append(d)结果与您的数据(转储为json以消除特殊dicts的表示)：
import json
print(json.dumps(result,indent=4))
{
"Manchester": {
"MAN-INT-GW": [
{
"bgp_routes": "1",
"service_status": "UP",
"bgp_peer_as": "20",
"circuit_name": "INTERNET",
"bgp_session": "1:25"
},
{
"bgp_routes": "1",
"service_status": "UP",
"bgp_peer_as": "20",
"circuit_name": "INTERNET 02",
"bgp_session": "1:25"
}
],
"MAN-EDGE": [
{
"bgp_routes": "1",
"service_status": "UP",
"bgp_peer_as": "45",
"circuit_name": "MPLS 01",
"bgp_session": "1:25"
}
]
},
"London": {
"LON-EDGE": [
{
"bgp_routes": "0",
"service_status": "Down",
"bgp_peer_as": "1",
"circuit_name": "MPLS",
"bgp_session": "3:35"
},
{
"bgp_routes": "100",
"service_status": "UP",
"bgp_peer_as": "3",
"circuit_name": "MPLS 02",
"bgp_session": "4:25"
}
],
"LON-INT-GW": [
{
"bgp_routes": "1",
"service_status": "UP",
"bgp_peer_as": "18",
"circuit_name": "INTERNET",
"bgp_session": "1:25"
}
]
}
}请注意，itertools.groupby-based解决方案也可以工作，但只有在相同的键是连续的情况下才能工作。否则，它会创建几个组，而不是您想要的。
Stack Overflow用户
发布于 2019-05-08 10:23:11
可以使用defaultdict和itertools.groupby
import itertools
from collections import defaultdict
res = defaultdict(dict)
for x, g in itertools.groupby(bgp_data_query, key=lambda x: x["location"]):
for d, f in itertools.groupby(g, key=lambda x: x["device_name"]):
res[x][d] = [{k:v} for z in f for k, v in z.items() if k not in {"location", "device_name"}]
print(dict(res))输出：
{'London': {'LON-EDGE': [{'bgp_peer_as': '1'},
{'bgp_routes': '0'},
{'circuit_name': 'MPLS'},
{'bgp_session': '3:35'},
{'service_status': 'Down'},
{'bgp_peer_as': '3'},
{'bgp_routes': '100'},
{'circuit_name': 'MPLS 02'},
{'bgp_session': '4:25'},
{'service_status': 'UP'}],
'LON-INT-GW': [{'bgp_peer_as': '18'},
{'bgp_routes': '1'},
{'circuit_name': 'INTERNET'},
{'bgp_session': '1:25'},
{'service_status': 'UP'}]},
'Manchester': {'MAN-EDGE': [{'bgp_peer_as': '45'},
{'bgp_routes': '1'},
{'circuit_name': 'MPLS 01'},
{'bgp_session': '1:25'},
{'service_status': 'UP'}],
'MAN-INT-GW': [{'bgp_peer_as': '20'},
{'bgp_routes': '1'},
{'circuit_name': 'INTERNET'},
{'bgp_session': '1:25'},
{'service_status': 'UP'},
{'bgp_peer_as': '20'},
{'bgp_routes': '1'},
{'circuit_name': 'INTERNET 02'},
{'bgp_session': '1:25'},
{'service_status': 'UP'}]}}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56038162
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