如何使用jq创建转换后的嵌套JSON元素？

Question

我有一个包含嵌套元素的JSON文件，我正试图将其操作为一个未嵌套的JSON文件。我该怎么做呢？

使用js，我尝试分离百分比，这是我能够做到的。我不知道如何重命名百分比字段。我见过一些实例，看起来value.gender或value.grade应该可以工作，但我也不确定如何组合它们。

jq '.data[] | .id as $id | (.demographics[] | .percentage as $percentage | .gender as $gender | .grade as $grade | {"id":$id, "percentage":$percentage})' test2.json

在这里，我希望能够将percentage字段重命名为性别和成绩值。然后我想按id分组。

以下是原始JSON文件(test2.json)：

{
   "data": [{
         "id": "abc",
         "students": "elementary",
         "demographics": [{
               "grade": "K-2",
               "percentage": "0.1",
               "gender": "unspecified"
            },
            {
               "grade": "K-2",
               "gender": "male",
               "percentage": "0.5"
            },
            {
               "gender": "female",
               "percentage": "0.4",
               "grade": "K-2"
            },
            {
               "grade": "3-6",
               "percentage": "0.3",
               "gender": "male"
            },
            {
               "percentage": "0.2",
               "gender": "unspecified",
               "grade": "3-6"
            },
            {
               "grade": "3-6",
               "gender": "female",
               "percentage": "0.5"
            }
         ],

         "neighborhood_name": [{
               "percentage": "0.5",
               "neighborhood": "atwood"
            },
            {
               "region": "bluff",
               "percentage": "0.5"
            }
         ]
      },
      {
         "id": "def",
         "students": "midhigh",
         "demographics": [{
               "grade": "7-9",
               "percentage": "0.2",
               "gender": "unspecified"
            },
            {
               "grade": "7-9",
               "gender": "male",
               "percentage": "0.2"
            },
            {
               "gender": "female",
               "percentage": "0.6",
               "grade": "7-9"
            },
            {
               "grade": "10-12",
               "percentage": "0.1",
               "gender": "male"
            },
            {
               "percentage": "0.1",
               "gender": "unspecified",
               "grade": "10-12"
            },
            {
               "grade": "10-12",
               "gender": "female",
               "percentage": "0.8"
            }
         ],

         "neighborhood_name": [{
               "percentage": "0.2",
               "neighborhood": "atwood"
            },
            {
               "region": "bluff",
               "percentage": "0.8"
            }
         ]

      }
   ]
}

以下是我所期望的：

{
         "id": "abc",
         "students": "elementary",
         "demo_K-2_unspecified": "0.1",
         "demo_K-2_male": "0.5",
         "demo_K-2_female": "0.4",
         "demo_3-6_male": "0.3",
         "demo_3-6_unspecified": "0.6",
         "demo_3-6_female": "0.5",
            },
      {
         "id": "def",
         "students": "midhigh",
         "demo_7-9_unspecified": "0.2",
         "demo_7-9_male": "0.2",
         "demo_7-9_female": "0.6",
         "demo_10-12_male": "0.1",
         "demo_10-12_unspecified": "0.1",
         "demo_10-12_female": "0.8",

      }

Answer 1

对于示例数据，以下筛选器将生成所需的输出：

.data[]
| {id, students} as $ix
| .demographics
| map( {"demo_\(.grade)_\(.gender)": .percentage} )
| $ix + add

这里的主要思想是使用map创建键-值对的列表，这样就可以使用add轻松地创建复合对象。

作为一行程序

jq '.data[] | {id,students} + (.demographics | map( {"demo_\(.grade)_\(.gender)": .percentage} ) | add)' test2.json