在python中从字典列表创建分层json转储

Question

表格：

categories = Table("categories", metadata,
                   Column("id", Integer, primary_key=True),
                   Column("name", String),
                   Column("parent_id", Integer, ForeignKey("categories.id"),
                          CheckConstraint('id!=parent_id'), nullable=True),

)

一个类别可以有多个子项，但只有一个父项。我使用CTE得到了如下字典值列表：例如，对于id:14，parent是13，并从parent 8->；10->；12->；13->；14遍历，其中parent 8没有父id

^{pr2}$

我希望父对象的属性也作为子类别嵌入到列表中，如下所示：

{
  "id": 14, 
  "name": "cat14", 
  "parent_id": 13, 
  "subcats": [
       {
         "id: 8", 
         "name": "cat8", 
         "parent_id":null
       }, 
       {
         "id: 10", 
         "name": "cat10", 
         "parent_id":8
       },  
       {
         "id: 12", 
         "name": "cat12", 
         "parent_id":10
       },   
      and similarly for ids 13 and 14..... 
     ]
}, 
{
  "id": 15, 
  "name": "cat15", 
  "parent_id": 13, 
  "subcats": [
       {
         "id: 8", 
         "name": "cat8", 
         "parent_id":null
       }, 
       {
         "id: 10", 
         "name": "cat10", 
         "parent_id":8
       },  
       {
         "id: 12", 
         "name": "cat12", 
         "parent_id":10
       },   
       and similarly for ids 13, 14, 15..... 
     ]
}

] 请注意，“path_info”已从字典中删除，并且每个id都显示了其详细信息。我想要上面缩进格式的json转储。怎么走？使用Flask0.10，Python2.7

Answer 1

可以用python代码执行：

假设我们有一个json对象。我稍微修改了一下—添加了缺少的节点，并按照规范的要求将其包装到一个对象中：

    {
      "array": [
          {
            "id": 14, 
            "name": "cat14", 
            "parent_id": 13, 
            "path_info": [
              8, 
              10, 
              12, 
              13, 
              14
            ]
          }, 
          {
            "id": 15, 
            "name": "cat15", 
            "parent_id": 13, 
            "path_info": [
              8, 
              10, 
              12, 
              13, 
              15
            ]
          },
          {
            "id": 13, 
            "name": "cat13", 
            "parent_id": 12, 
            "path_info": [
              8, 
              10, 
              12, 
              13
            ]
          },
         {
            "id": 12, 
            "name": "cat12", 
            "parent_id": 10, 
            "path_info": [
              8, 
              10, 
              12
            ]
          }, 
          {
            "id": 10, 
            "name": "cat10", 
            "parent_id": 8, 
            "path_info": [
              8, 
              10
            ]
          },
          {
            "id": 8, 
            "name": "cat8", 
            "parent_id": null, 
            "path_info": [
              8
            ]
          }
        ]
    }

然后您可以使用以下代码：

^{pr2}$

为了确保你得到你需要的东西：

    >>> print(json.dumps(a, indent=True))
    [
     {
      "name": "cat14",
      "subcats": [
       {
        "name": "cat10",
        "id": 10,
        "parent_id": 8
       },
       {
        "name": "cat12",
        "id": 12,
        "parent_id": 10
       },
       {
        "name": "cat13",
        "id": 13,
        "parent_id": 12
       },
       {
        "name": "cat14",
        "id": 14,
        "parent_id": 13
       }
      ],
      "id": 14,
      "parent_id": 13
     },
     {
      "name": "cat15",
      "subcats": [
       {
        "name": "cat10",
        "id": 10,
        "parent_id": 8
       },
       {
        "name": "cat12",
        "id": 12,
        "parent_id": 10
       },
       {
        "name": "cat13",
        "id": 13,
        "parent_id": 12
       },
       {
        "name": "cat15",
        "id": 15,
        "parent_id": 13
       }
      ],
      "id": 15,
      "parent_id": 13
     },
     {
      "name": "cat13",
      "subcats": [
       {
        "name": "cat10",
        "id": 10,
        "parent_id": 8
       },
       {
        "name": "cat12",
        "id": 12,
        "parent_id": 10
       },
       {
        "name": "cat13",
        "id": 13,
        "parent_id": 12
       }
      ],
      "id": 13,
      "parent_id": 12
     },
     {
      "name": "cat12",
      "subcats": [
       {
        "name": "cat10",
        "id": 10,
        "parent_id": 8
       },
       {
        "name": "cat12",
        "id": 12,
        "parent_id": 10
       }
      ],
      "id": 12,
      "parent_id": 10
     },
     {
      "name": "cat10",
      "subcats": [
       {
        "name": "cat10",
        "id": 10,
        "parent_id": 8
       }
      ],
      "id": 10,
      "parent_id": 8
     },
     {
      "name": "cat8",
      "subcats": [],
      "id": 8,
      "parent_id": null
     }
    ]
    >>>

Answer 2

有一种可以接受的方法，只需理解一些列表/dict。在

lst = [{"id": 14, "name": "cat14", "parent_id": 13, "path_info": [8, 10, 12, 13, 14]}, {"id": 15, "name": "cat15", "parent_id": 13, "path_info": [8, 10, 12, 13, 15]}]

master_dct = { d['id'] : d for d in lst}
for d in lst:
    d['subcats'] = [{field : master_dct[i][field] for field in ['id', 'name', 'parent_id']} \
        for i in d['path_info'] if i in master_dct]

import json
with open('out.json', 'w') as f:
    json.dump(lst, f)

Answer 3

对此的python代码：简单明了

import json
categories = [] #input
def transform(category, child_node_id):
    category['subcats'].append({
       'id': child_node_id,
       'name': 'cat%s' % child_node_id,
       'parent_id': category['id']
    })


for category in categories:
    category['subcats'] = []
    [transform(category, child_node_id) for child_node_id in category['path_info']]
    category.pop('path_info', None)

print(json.dumps(categories, indent=4))