Python：解析Json以查找与几个键匹配的匹配行，并将每个匹配集作为单个Json记录返回

Question

我有一个JSON，如下所示

{"processId":"p1","userId":"user1","reportName":"report1","threadId": "12234", "some_other_keys":"respective values"}
{"userId":"user1","processId":"p1","reportName":"report1","threadId":"12335", "some_other_keys":"respective values"}
{"reportName":"report2","processId":"p1","userId":"user1","threadId":"12434", "some_other_keys":"respective values"}
{"threadId":"12734", "some_other_keys":"respective values", "processId":"p1","userId":"user2","reportName":"report1"}
{"processId":"p1","reportName":"report1","threadId":"12534", "some_other_keys":"respective values","userId":"user2"}
{"processId":"p1","userId":"user1","reportName":"report2","threadId":"12934", "some_other_keys":"respective values"}
{"processId":"p1","userId":"user1","reportName":"report1","threadId":"12834", "some_other_keys":"respective values"}
{"processId":"p1","userId":"user1","reportName":"report2","threadId":"12634", "some_other_keys":"respective values"}

目标：编写一个函数，返回所有不同的行集，这些行的值“processId”、“userId”、“reportName”与单个JSON记录相同，每个匹配记录的键名都经过修改，如下所示

在上面的示例中，有三个匹配集

Set1（对于“processId”：“p1”，“userId”：“user1”，“reportName”：“report1”）：

{"processId":"p1","userId":"user1","reportName":"report1","threadId":"12234", "some_other_keys":"respective values"}
{"userId":"user1","processId":"p1","reportName":"report1","threadId":"12335", "some_other_keys":"respective values"}
{"processId":"p1","userId":"user1","reportName":"report1","threadId":"12834", "some_other_keys":"respective values"}

Set2（“processId”：“p1”，“userId”：“user1”，“reportName”：“report2”）：

{"reportName":"report2","processId":"p1","userId":"user1","threadId":"12434", "some_other_keys":"respective values"}
{"processId":"p1","userId":"user1","reportName":"report2","threadId":"12934", "some_other_keys":"respective values"}
{"processId":"p1","userId":"user1","reportName":"report2","threadId":"12634", "some_other_keys":"respective values"}

Set3（“processId”：“p1”，“userId”：“user2”，“reportName”：“report2”）：

{"threadId":"12734", "some_other_keys":"respective values", "processId":"p1","userId":"user2","reportName":"report1"}
{"processId":"p1","reportName":"report1","threadId":"12534", "some_other_keys":"respective values","userId":"user2"}

因此，在这个特定的示例中，函数应该返回三个不同的集合，如下所示

Set1（对于“processId”：“p1”，“userId”：“user1”，“reportName”：“report1”）： {"processId":"p1","userId":"user1","reportName":"report1","threadId_1":"12234", "some_other_keys_1":"respective values", "threadId_2":"12335", "some_other_keys_2":"respective values", "threadId_3":"12834", "some_other_keys_3":"respective values"}

Set2（“processId”：“p1”，“userId”：“user1”，“reportName”：“report2”）： {"processId":"p1","userId":"user1","reportName":"report2","threadId_1":"12934", "some_other_keys_1":"respective values","threadId_2":"12434", "some_other_keys_2":"respective values","threadId_3":"12634", "some_other_keys_3":"respective values"}

Set3（“processId”：“p1”，“userId”：“user2”，“reportName”：“report2”）： {"threadId_1":"12734", "some_other_keys_1":"respective values", "processId":"p1","userId":"user2","reportName":"report1""threadId_2":"12534", "some_other_keys_2":"respective values"}

因此，一个函数返回三个集合（这可能或多或少也取决于匹配集合的数量）

我需要一个解决上述问题的方案，作为（a）性能高效的代码（b）行数较少的代码，因为我将处理大量的行。因此，我希望我的代码运行得更快，而且代码的行数应该更少

Answer 1

import json

f = open('data.json')
data = json.load(f)
f.close()

sets_of_procces = dict()

for item in data:
    set_id = processId, userId, reportName = item['processId'], item['userId'], item['reportName']
    if set_id not in sets_of_procces.keys():
        sets_of_procces[set_id] = []
    thread_number = len(sets_of_procces[set_id]) + 1
    thread_data = { f'threadId_{thread_number}' : item['threadId'], f'some_other_keys_{thread_number}' : item['some_other_keys'] }
    sets_of_procces[set_id].append(json.dumps(thread_data))

for i, procces_set in enumerate(sets_of_procces):
    print(f'Set {i+1} : \n')
    processId, userId, reportName = procces_set
    json_dict = { 'processId' : processId, 'userId' : userId, 'reportName' : reportName }
    for item in sets_of_procces[procces_set]:
        json_dict = {**json_dict, **json.loads(item)}
    print(json.dumps(json_dict))