Question

我有一本字典，我想根据子字典列表中所有设备的会话总数对父字典进行排序/重新排序

编辑，我使用python 3.6

使用以下示例数据，我想订购

曼彻斯特...
爱丁堡...
伦敦...

我如何实现这种排序，这种数据结构可能吗？

{
    "London": {
        "EDGE-01": [
            {
                "bgp_peer_as": "1",
                "bgp_session": "3:35",
                "bgp_routes": "0",
                "service_status": "Down",
                "circuit_name": "MPLS",
                "subnet_name": null,
                "last_updated": "2019-05-08 17:45:43",
                "time_error": true,
                "service_name": "MPLS"
            }
        ]
    },
    "Manchester": {
        "EDGE-01": [
            {
                "bgp_peer_as": "1",
                "bgp_session": "1w1d",
                "bgp_routes": "2",
                "service_status": "Up",
                "circuit_name": null,
                "subnet_name": "Live ",
                "last_updated": "2019-05-08 17:45:13",
                "time_error": true,
                "service_name": "Live Primary"
            },
            {
                "bgp_peer_as": "2",
                "bgp_session": "1w6d",
                "bgp_routes": "140",
                "service_status": "Up",
                "circuit_name": null,
                "subnet_name": "Dev Primary",
                "last_updated": "2019-05-08 17:45:08",
                "time_error": true,
                "service_name": "Dev Primary"
            }
        ],
        "INT-GW-01": [
            {
                "bgp_peer_as": "10",
                "bgp_session": "1d2h",
                "bgp_routes": "10",
                "service_status": "Up",
                "circuit_name": null,
                "subnet_name": "PUBLIC",
                "last_updated": "2019-05-08 17:41:58",
                "time_error": true,
                "service_name": "PUBLIC"
            }
        ]
    },
    "Edinburgh": {
        "EDGE-01": [
            {
                "bgp_peer_as": "1",
                "bgp_session": "3:35",
                "bgp_routes": "0",
                "service_status": "Down",
                "circuit_name": "MPLS",
                "subnet_name": null,
                "last_updated": "2019-05-08 17:45:43",
                "time_error": true,
                "service_name": "MPLS"
            },
            {
                "bgp_peer_as": "65001",
                "bgp_session": "1w6d",
                "bgp_routes": "140",
                "service_status": "Up",
                "circuit_name": null,
                "subnet_name": "SV Primary",
                "last_updated": "2019-05-08 17:45:08",
                "time_error": true,
                "service_name": "SV Primary"
            }
        ]
    }
}

Answer 1

按照@masasa的建议，您可以在集合模块中使用for(const [j, k, l, m] of loop(1, 2, 3, 4)) { //... }：

OrderedDict

对于python3.7，您可以只使用标准的from collections import OrderedDict # You can use sorted to get the keys of the dict based on the len of all devices # and reverse=True will order from longest to shortest sorted_keys = sorted([(k, sum(len(dev) for name, dev in v.items())) for k,v in y.items()], key=lambda x: x[1], reverse=True) # [('Manchester', 3), ('Edinburgh', 2), ('London', 1)] # So to put this into OrderedDict z = OrderedDict((k, y[k]) for k, _ in sorted_keys) z.keys() odict_keys(['Manchester', 'Edinburgh', 'London'])，因为@buran已经回答了，其他人已经发表了评论

Answer 2

import json
spam = json.loads("""{
    ...
}""")

eggs = dict(sorted(spam.items(), key=lambda item: sum(len(value2) for value2 in item[1].values()), reverse=True))
print(eggs)

请注意，您所拥有的不是有效的字典（例如，null，true之类的值），显然是json。

我的解决方案假定使用python3.7 +命令。如果不是这种情况，那么鸡蛋应该是collections.OrderedDict

的实例。

python-按嵌套项的总数对嵌套字典排序？

2 个答案: