合并具有相同鍵值對的Json

我從以下格式的 API 中得到了一個結果 json

[{
        "Uid": "40cc6103-1cf0-4735-b882-d14d32018e58",
        "Id": "9e1a0057-4570-4a6e-8ff5-88b2facbaf4e",
        "Details": {
            "Name": "Kiran"
        }
    }, {
        "Uid": "40cc6103-1cf0-4735-b882-d14d32018e58",
        "Id": "9e1a0057-4570-4a6e-8ff5-88b2facbaf4e",
        "Details": {
            "Age": "24"
        }
    },
    {
        "Uid": "196f5865-e9fe-4847-86ae-97d0bf57b816",
        "Id": "84909ecb-c92e-48a7-bcaa-d478bf3a9220",
        "Details": {
            "Name": "Shreyas"
        }
    }
]

由于 Uid 和 Id 對于多個整數是相同的，我可以將它們與 Details 鍵作為逗號分隔鍵值對一起使用嗎？像下面提到的東西

[{
    "Uid": "40cc6103-1cf0-4735-b882-d14d32018e58",
    "Id": "9e1a0057-4570-4a6e-8ff5-88b2facbaf4e",
    "Details": {
        "Name": "Kiran",
        "Age": "24"
    }
},
{
    "Uid": "196f5865-e9fe-4847-86ae-97d0bf57b816",
    "Id": "84909ecb-c92e-48a7-bcaa-d478bf3a9220",
    "Details": {
        "Name": "Shreyas"
    }
}]

請在此指導我遵循的方法。謝謝

uj5u.com熱心網友回復：

您需要的是字典函式update()。下面是一個例子：

A = [{
    "Uid": "40cc6103-1cf0-4735-b882-d14d32018e58",
    "Id": "9e1a0057-4570-4a6e-8ff5-88b2facbaf4e",
    "Details": {
        "Name": "Kiran"
    }
}, {
    "Uid": "40cc6103-1cf0-4735-b882-d14d32018e58",
    "Id": "9e1a0057-4570-4a6e-8ff5-88b2facbaf4e",
    "Details": {
        "Age": "24"
    }
},
    {
        "Uid": "196f5865-e9fe-4847-86ae-97d0bf57b816",
        "Id": "84909ecb-c92e-48a7-bcaa-d478bf3a9220",
        "Details": {
            "Name": "Shreyas"
    }
}
]

B = []

def find(uid, id_):
    for i, d in enumerate(B):
        if d['Uid'] == uid and d['Id'] == id_:
            return i
    return -1

for d in A:
    if (i := find(d['Uid'], d['Id'])) < 0:
        B.append(d)
    else:
        B[i]['Details'].update(d['Details'])

print(B)

漂亮的輸出：

[
    {
        "Uid": "40cc6103-1cf0-4735-b882-d14d32018e58",
        "Id": "9e1a0057-4570-4a6e-8ff5-88b2facbaf4e",
        "Details": {
            "Name": "Kiran",
            "Age": "24"
        }
    },
    {
        "Uid": "196f5865-e9fe-4847-86ae-97d0bf57b816",
        "Id": "84909ecb-c92e-48a7-bcaa-d478bf3a9220",
        "Details": {
            "Name": "Shreyas"
        }
    }
]

筆記：

如果您的 API 回應包含大量字典，這可能會非常低效。您可能需要一種完全不同的方法

uj5u.com熱心網友回復：

您應該遍歷串列并與以 (Uid, Id) 為鍵的累加器合并：

from typing import Dict, List

l = [{
        "Uid": "40cc6103-1cf0-4735-b882-d14d32018e58",
        "Id": "9e1a0057-4570-4a6e-8ff5-88b2facbaf4e",
        "Details": {
            "Name": "Kiran"
        }
    }, {
        "Uid": "40cc6103-1cf0-4735-b882-d14d32018e58",
        "Id": "9e1a0057-4570-4a6e-8ff5-88b2facbaf4e",
        "Details": {
            "Age": "24"
        }
    },
    {
        "Uid": "196f5865-e9fe-4847-86ae-97d0bf57b816",
        "Id": "84909ecb-c92e-48a7-bcaa-d478bf3a9220",
        "Details": {
            "Name": "Shreyas"
        }
    }
]


def mergeItem(it: Dict, acc: Dict) -> Dict:
    uid = it["Uid"]
    id = it["Id"]
    if (uid, id) in acc:
        acc[(uid, id)] = {"Uid": uid, "Id": id, "Details": {**acc[(uid, id)]["Details"], **it["Details"]}}
    else:
        acc[(uid, id)] = {"Uid": uid, "Id": id, "Details": it["Details"]}
    return acc


def mergeList(a:List) -> Dict:
    acc = {}
    for v in a:
        acc = mergeItem(v, acc)
    return acc


print(list(mergeList(l).values()))

# [
#     {
#         'Uid': '40cc6103-1cf0-4735-b882-d14d32018e58',
#         'Id': '9e1a0057-4570-4a6e-8ff5-88b2facbaf4e',
#         'Details': {'Name': 'Kiran', 'Age': '24'}},
#     {
#         'Uid': '196f5865-e9fe-4847-86ae-97d0bf57b816',
#         'Id': '84909ecb-c92e-48a7-bcaa-d478bf3a9220',
#         'Details': {'Name': 'Shreyas'}
#     }
# ]

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