Question

我有这样的json响应

order_response = {
"orders": [
    {
        "id": '1',
        "email": "b@mail.com",
        "location_id": 9,
        "line_items": [
            {
                "id": 5,
                "product_id": 6,
            }, {
                "id": 7,
                "product_id": 8,
            }
        ]
    }, {
        "id": '2',
        "email": "b@mail.com",
        "location_id": 10,
        "line_items": {
            "id": 3,
            "product_id": 4,
        }
    },
]

}

我想要这样的输出

id email      location_id line_items_id line_items_product_id
1  b@mail.com 9           5             6
1  b@mail.com 9           7             8
1  b@mail.com 10          3             4

我想根据line_items中对象的数量来拆分行。所以我的方法是使用Pandas的json_normalize功能如果我在代码中指定了列名，就会出现如下情况。

pd.io.json.json_normalize(report_json, ['line_items'], ['id', 'email'], record_prefix='line_items_')

，但除id, email之外可能还有其他列。我希望这是动态的，即它应该能够处理提供的任何数量的对象而无需显式定义在这方面的任何帮助都将受到高度赞赏。

Answer 1

首先将列表添加到一个元素字典，然后提取字典的所有键：

    val password = EditText(this)
    password.inputType = InputType.TYPE_CLASS_TEXT or InputType.TYPE_TEXT_VARIATION_PASSWORD
    password.hint = "Password"

L = []
keys = []
for x in report_json['orders']:
    d = {}
    for k, v in x.items():
        if isinstance(v, dict) and k =='line_items':
            d[k] = [v]
        else:
            d[k] = v
        if k !='line_items':
            keys.append(k)
    L.append(d)

print (L)

[
    {
        "id": '1',
        "email": "b@mail.com",
        "location_id": 9,
        "line_items": [
            {
                "id": 5,
                "product_id": 6,
            }, {
                "id": 7,
                "product_id": 8,
            }
        ]
    }, {
        "id": '2',
        "email": "b@mail.com",
        "location_id": 10,
        "line_items": [{
            "id": 3,
            "product_id": 4,
        }]
    }
]

json_normalize拆分多个json对象：python

1 个答案: