在列表理解中访问不同的 JSON 键/值对

Question

不确定我是否正在尝试实现不可能的目标？我有这个 JSON 字符串：

dot= [{"type": 1, "date": "2018-12-02T00:40:03.2186792+00:00", "device": 
[{"id": "20165cf4e596", "deviceName": "17", "records": [{"timestamp": "2018- 
12-02T00:40:00.499+00:00", "grp": "undefined", "val": 887}]}, {"id": 
"5f401a6a6f66", "deviceName": "18", "records": [{"timestamp": "2018-12- 
02T00:42:00.499+00:00", "grp": "undefined", "val": 1063}, {"timestamp": 
"2018-12-02T00:41:00.498+00:00", "grp": "undefined", "val": 907}]}, {"id": 
"569bb0147a72", "deviceName": "19", "records": [{"timestamp": "2018-12- 
02T00:44:00.499+00:00", "grp": "undefined", "val": 817}, {"timestamp": 
"2018-12-02T00:43:00.498+00:00", "grp": "undefined", "val": 1383}]}, {"id": 
"ef829aa3", "deviceName": "2", "records": [{"timestamp": "2018-12- 
02T00:46:00.499+00:00", "grp": "undefined", "val": 1173}]}, {"id": 
"388ae8f2fa64", "deviceName": "17", "records": [{"timestamp": "2018-12- 
02T00:41:00.499+00:00", "grp": "undefined", "val": 866}, {"timestamp": 
"2018-12-02T00:32:00.492+00:00", "grp": "undefined", "val": 1080}]}, {"id": 
"01f874b30b55", "deviceName": "19", "records": [{"timestamp": "2018-12- 
02T00:43:00.499+00:00", "grp": "undefined", "val": 1050}, {"timestamp": 
"2018-12-02T00:42:00.498+00:00", "grp": "undefined", "val": 1084}]}]}]

我想实现以下目标：

[{'id': '20165cf4e596','deviceName': '17','timestamp': '2018-12- 
02T00:40:00.499+00:00','grp': 'undefined','val': 887},
{'id': '5f401a6a6f66','deviceName': '18','timestamp': '2018-12- 
02T00:42:00.499+00:00','grp': 'undefined','val': 1063},
{'id': '5f401a6a6f66','deviceName': '18','timestamp': '2018-12- 
02T00:41:00.498+00:00','grp': 'undefined','val': 907},...]

我使用了以下代码：

for i in dot:
    for k in i['device']:
        d2= [[{l:m},{'value':v}] for l,m in k.items() for p in m if 
        isinstance(p,list) for v in p]
        print(d2)

得到了空列表：

[]
[]
[]
[]

提前致谢

Answer 1

您需要遍历每个设备的records 元素。然后将设备中的字段与每条记录结合起来。

result = []
for i in dot:
    for k in i['device']:
        for r in k['records']:
            result.append({"id": k["id"], "deviceName": k["deviceName"], "timestamp": r["timestamp"], "grp": r["grp"], "val": r["val"]})
print(result)

Answer 2

Barmar's answer 可能是正确的 - 但我相信，如果您通过创建代表数据结构中项目的对象层次结构来对代码进行去混淆处理，您将会轻松得多：

class SomeObject:
    def __init__(self, some_object_type, date, devices):
        self.type = some_object_type
        self.date = date
        self.devices = []

        for d in devices:
            some_device = SomeDevice(d["id"], d["deviceName"], d["records"])
            self.devices.append(some_device)

class SomeDevice:
    def __init__(self, some_device_id, name, records):
        self.id = some_device_id
        self.name = name
        self.records = []

        for r in records:
            some_record = SomeDeviceRecord(r["timestamp"], r["grp"], r["val"])
            self.records.append(some_record)

class SomeDeviceRecord:
    def __init__(self, timestamp, group, value):
        self.timestamp = timestamp
        self.group = group
        self.value = value

如果你走这条路，你可以很容易地将整个 JSON 解析成强类型对象：

some_object = SomeObject(dot[0]["type"], dot[0]["date"], dot[0]["device"])

然后报告数据结构的特定部分应该相当容易/直接：

for device in some_object.devices:
    print(device.id, device.name, device.records[0].timestamp, device.records[0].group, device.records[0].value)