在 javascript 中从扁平化的 JSON 创建嵌套的 json

Question

最近开始使用javascript。 我有一个扁平的 json，它看起来像：

[
{"timestamp":"2019-01-01T00:00:00.000Z","MarketShare":89905.0,"Territory":"PHILADELPHIA PA"},
{"timestamp":"2019-01-01T00:00:00.000Z","MarketShare":82991.0,"Territory":"SEATTLE WA"},
{"timestamp":"2019-02-01T00:00:00.000Z","MarketShare":73662.0,"Territory":"PHILADELPHIA PA"},
{"timestamp":"2019-02-01T00:00:00.000Z","MarketShare":67059.0,"Territory":"SEATTLE WA"},
{"timestamp":"2019-03-01T00:00:00.000Z","MarketShare":72898.0,"Territory":"PHILADELPHIA PA"},
{"timestamp":"2019-03-01T00:00:00.000Z","MarketShare":68013.0,"Territory":"SEATTLE WA"},
{"timestamp":"2019-04-01T00:00:00.000Z","MarketShare":74607.0,"Territory":"PHILADELPHIA PA"},
{"timestamp":"2019-04-01T00:00:00.000Z","MarketShare":66769.0,"Territory":"SEATTLE WA"},
{"timestamp":"2019-05-01T00:00:00.000Z","MarketShare":93932.0,"Territory":"PHILADELPHIA PA"},
{"timestamp":"2019-05-01T00:00:00.000Z","MarketShare":86514.0,"Territory":"SEATTLE WA"}
]

基本上，对于给定的时间戳，我想在结果键下收集 MarketShare 和 Territory。
预期的输出是：

    [
   {
      "timestamp":"2019-01-01T00:00:00.000Z",
      "result":[
         {
            "MarketShare":89905.0,
            "Territory":"PHILADELPHIA PA"
         },
         {
            "MarketShare":82991.0,
            "Territory":"SEATTLE WA"
         },

      ]
   },
   {
      "timestamp":"2019-02-01T00:00:00.000Z",
      "result":[
         {
            "MarketShare":73662.0,
            "Territory":"PHILADELPHIA PA"
         },
         {
            "MarketShare":67059.0,
            "Territory":"SEATTLE WA"
         }
      ]
   },
   {
      "timestamp":"2019-03-01T00:00:00.000Z",
      "result":[
         {
            "MarketShare":72898.0,
            "Territory":"PHILADELPHIA PA"
         },
         {
            "MarketShare":68013.0,
            "Territory":"SEATTLE WA"
         }
      ]
   },
   {
      "timestamp":"2019-04-01T00:00:00.000Z",
      "result":[
         {
            "MarketShare":74607.0,
            "Territory":"PHILADELPHIA PA"
         },
         {
            "MarketShare":66769.0,
            "Territory":"SEATTLE WA"
         }
      ]
   },
   {
      "timestamp":"2019-05-01T00:00:00.000Z",
      "result":[
         {
            "MarketShare":93932.0,
            "Territory":"PHILADELPHIA PA"
         },
         {
            "MarketShare":86514.0,
            "Territory":"SEATTLE WA"
         }
      ]
   }
]

我正在尝试创建一个通用实用程序，将任意数量的键值对放在给定的时间戳下。 目前取得的进展：

topNFormat=[]
for(var i = 0; i < firstResult.length; i++) {
    //res = [firstResult[i]["Territory"]
    topNFormat[timestamp] = firstResult[i][timestamp] || {};
    topNFormat[result] = firstResult[i][result] || [];
    topNFormat[result].push(firstResult[i]);
    //return collection;
}

Answer 1

这是一个通用的函数transform ，可以动态地对项目进行分组：

 const input = [{ "timestamp": "2019-01-01T00:00:00.000Z", "MarketShare": 89905.0, "Territory": "PHILADELPHIA PA" }, { "timestamp": "2019-01-01T00:00:00.000Z", "MarketShare": 82991.0, "Territory": "SEATTLE WA" }, { "timestamp": "2019-02-01T00:00:00.000Z", "MarketShare": 73662.0, "Territory": "PHILADELPHIA PA" }, { "timestamp": "2019-02-01T00:00:00.000Z", "MarketShare": 67059.0, "Territory": "SEATTLE WA" }, { "timestamp": "2019-03-01T00:00:00.000Z", "MarketShare": 72898.0, "Territory": "PHILADELPHIA PA" }, { "timestamp": "2019-03-01T00:00:00.000Z", "MarketShare": 68013.0, "Territory": "SEATTLE WA" }, { "timestamp": "2019-04-01T00:00:00.000Z", "MarketShare": 74607.0, "Territory": "PHILADELPHIA PA" }, { "timestamp": "2019-04-01T00:00:00.000Z", "MarketShare": 66769.0, "Territory": "SEATTLE WA" }, { "timestamp": "2019-05-01T00:00:00.000Z", "MarketShare": 93932.0, "Territory": "PHILADELPHIA PA" }, { "timestamp": "2019-05-01T00:00:00.000Z", "MarketShare": 86514.0, "Territory": "SEATTLE WA" } ]; const transform = (array, groupBy, groupUnder) => array.reduce((acc, object) => { if (groupBy in object) { if (object[groupBy] in acc) { acc[object[groupBy]][groupUnder] = [ ...acc[object[groupBy]][groupUnder], ...[Object.keys(object).filter((key) => key !== groupBy).reduce((acc, key) => { acc[key] = object[key]; return acc; }, {})] ] } else { acc[object[groupBy]] = { [groupUnder]: [Object.keys(object).filter((key) => key !== groupBy).reduce((acc, key) => { acc[key] = object[key]; return acc; }, {})] } } } return acc; }, {}); const groupBy = 'timestamp'; const groupUnder = 'result'; const transformedResult = transform(input, groupBy, groupUnder); console.log(transformedResult); console.log('-'.repeat(100)); console.log(Object.keys(transformedResult).map((key) => Object.assign({}, transformedResult[key], { [groupBy]: key })));

Answer 2

这是使用内置reduce函数的简单解决方案：

[].reduce((memo, item) => {
    memo[item.timestamp] = memo[item.timestamp] ||
                          { timestamp: item.timestamp, result: [] };
    memo[item.timestamp].result.push({
       MarketShare: item.MarketShare,
       Territory: item.Territory 
    });
    return memo;
 }, {});

更新 2（更通用）

[].reduce((memo, item) => {
    memo[item.timestamp] = memo[item.timestamp] ||
                          { timestamp: item.timestamp, result: [] };
    // push whole item into result list
    memo[item.timestamp].result.push({ ...item });
    return memo;
 }, {});