将有序dict的列表转换为嵌套列表

Question

我列出了有序的dict,其中包括数据中的一些重复的ID ...这样的事情

[OrderedDict([('caseId', 20), ('userId', 1), ('emailStatus', 21)]), 
 OrderedDict([('caseId', 20), ('userId', 1), ('emailStatus', 20)]), 
 OrderedDict([('caseId', 18), ('userId', 4), ('emailStatus', 21)]), 
 OrderedDict([('caseId', 19), ('userId', 3), ('emailStatus', 21)]), 
 OrderedDict([('caseId', 18), ('userId', 1), ('emailStatus', 20)]),
 OrderedDict([('caseId', 20), ('userId', 3), ('emailStatus', 21)]),
 OrderedDict([('caseId', 18), ('userId', 4), ('emailStatus', 20)]), 
 OrderedDict([('caseId', 19), ('userId', 1), ('emailStatus', 20)])]

我想得到一个嵌套列表列表,就像这样;

[{
"caseId": "20",
"users": [
  {
    "userId": "1",
    "emailStatus": [
      {
      "emailStatus" : "20"
      },
      {
      "emailStatus" : "21"
      }

    ]
  },
    {
    "userId": "3",
    "emailStatus": [
      {
      "emailStatus" : "21"
      }

    ]

  }
]
},
{
"caseId": "19",
"users": [
  {
    "userId": "1",
    "emailStatus": [
      {
      "emailStatus" : "20"
      }
    ]

  },
    {
    "userId": "3",
    "emailStatus": [
      {
      "emailStatus" : "21"
      }

    ]

  }
]
},
{
"caseId": "18",
"users": [
  {
    "userId": "1",
    "emailStatus": [
      {
      "emailStatus" : "20"
      }
    ]

  },
    {
    "userId": "4",
    "emailStatus": [
      {
      "emailStatus" : "20"
      },
      {
      "emailStatus" : "21"
      }

    ]

  }
]
}
]

呈现这样的嵌套列表;

我试图通过迭代两个列表来实现这一点,但无法知道如何记录上一个和下一个记录以及相同的数据..这太令人困惑了......如果有人能给我一个开头,我可以如何迭代我的列表,你真好.

很多问候..

更新的问题

这里有更详细的问题

Answer 1

首先,您可以使用循环并将dict.setdefault数据分组到嵌套的dict中:

temp = {}
for d in lst:
    temp.setdefault(d["caseId"], {}).setdefault(d["userId"], []).append(d["emailStatus"])
print(temp)
# {18: {1: [20], 4: [21, 20]}, 19: {1: [20], 3: [21]}, 20: {1: [21, 20], 3: [21]}}

或者使用collections.defaultdict:

temp = defaultdict(lambda: defaultdict(list))
for d in lst:
    temp[d["caseId"]][d["userId"]].append(d["emailStatus"])

然后,使用嵌套的混合字典和列表推导来聚合最终结果:

res = [{"caseId": case, "users": [{"userId": user, "emailStatus": [{"emailStatus": s} for s in status]} 
                                  for user, status in users.items()]} 
       for case, users in temp.items()]
print(res)
# [{'caseId': 18, 'users': [{'userId': 1, 'emailStatus': [{'emailStatus': 20}]}, {'userId': 4, 'emailStatus': [{'emailStatus': 21}, {'emailStatus': 20}]}]},
#  {'caseId': 19, 'users': [{'userId': 1, 'emailStatus': [{'emailStatus': 20}]}, {'userId': 3, 'emailStatus': [{'emailStatus': 21}]}]},
#  {'caseId': 20, 'users': [{'userId': 1, 'emailStatus': [{'emailStatus': 21}, {'emailStatus': 20}]}, {'userId': 3, 'emailStatus': [{'emailStatus': 21}]}]}]