对来自 DJango QuerySet 的 JSON 数据进行分组

Question

我的 Django QuerySet 有这个结果（JSON）：

{
    "high": [{
        "counthigh": 27,
        "brgy_locat": "Barangay 6"
    }, {
        "counthigh": 3,
        "brgy_locat": "Mabini"
    }],
    "medium": [{
        "brgy_locat": "Barangay 6",
        "countmedium": 31
    }, {
        "brgy_locat": "Tolosa",
        "countmedium": 9
    }],
    "low": [{
        "brgy_locat": "Barangay 12",
        "countlow": 29
    }, {
        "brgy_locat": "Mabini",
        "countlow": 25
    }, {
        "brgy_locat": "Barangay 9",
        "countlow": 35
    }]
}

我想按 brgy_locat 及其值对其进行分组：

brgy_locat | countlow | countmedium | counthigh

主要是因为，我使用的是数据表。

顺便说一句，这是我的观点.py：

response_data = {} 
response_data["medium"] = list(BuildingStructure.objects.filter(geom__intersects=getgeom_medium).values( 'brgy_locat').annotate(countmedium=Count('brgy_locat'))) 
response_data["high"] = list(BuildingStructure.objects.filter(geom__intersects=getgeom).values('brgy_locat').annotate( counthigh=Count('brgy_locat'))) 
response_data["low"] = list(BuildingStructure.objects.filter(geom__intersects=getgeom_low).values('brgy_locat').annotate( countlow=Count('brgy_locat'))) 
return HttpResponse(json.dumps(response_data), content_type='application/json')

Answer 1

您无法在 QuerySet 中执行您要查找的内容。您需要在 python 中操作数据。这是一种方法。它假定每个元素不会有冲突的键。意思是high、medium、low 的字典中只有counthigh 等键。当然除了brgy_locat。

data = {
    "high": [{
        "counthigh": 27,
        "brgy_locat": "Barangay 6"
    }, {
        "counthigh": 3,
        "brgy_locat": "Mabini"
    }],
    "medium": [{
        "brgy_locat": "Barangay 6",
        "countmedium": 31
    }, {
        "brgy_locat": "Tolosa",
        "countmedium": 9
    }],
    "low": [{
        "brgy_locat": "Barangay 12",
        "countlow": 29
    }, {
        "brgy_locat": "Mabini",
        "countlow": 25
    }, {
        "brgy_locat": "Barangay 9",
        "countlow": 35
    }]
}

result = {}

for category in data.values():
    for element in category:
        key = element.pop('brgy_locat')
        if key not in result:
            result[key] = {}
        result[key].update(element)

那么结果应该是：

{
    "Barangay 6": {
        "counthigh": 27,
        "countmedium": 31
    }, 
    "Mabini": {
        "counthigh": 3,
        "countlow": 25
    },
    "Tolosa": {
        "countmedium": 9
    },
    "Barangay 12":{
        "countlow": 29
    },
    "Barangay 9": {
        "countlow": 35
    }
}