如何基於三個鍵對數組對象進行分組

Question

基於這個問題Group array items using object

我引用了這個答案，由https://stackoverflow.com/users/1094311/1983回答

所以我從這個 jquery 獲取數據然后我解析它，之后我將它推送到變量

我從 myurl 得到的數據是這樣的

{"pname":"some1","datax":2.0643278,"prov":"11","datay":1},{"pname":"some1","datax":3.2142857142857144,"prov":"11","datay":1},{"pname":"some2","datax":1.125,"prov":"12","datay":1},{"pname":"some2","datax":1.6666666666666667,"prov":"12","datay":2}

這就是我的陣列開始的方式

$(document).ready(function(){
    $.ajax({
                        url: "myurl",
                        method: "GET",
                        success: function(data) {

var parsedJSON = JSON.parse(data);

var result1 = [
  (function(data) {
    let tmp = [];

    data.forEach(e => {
      tmp.push({
        type: "firstid",
        prov: e.prov,
        name: e.pname,
        showInLegend:false,
        dataPoints:[{x:e.datax, y:e.datay}],
      })
    });

    return tmp;
  })(parsedJSON)
];

從那里我得到了 result1[0]，如果我控制台記錄它，它會是這樣的

[        { type: "firstid",
           prov: 11,
           name: some1,
           showInLegend:false,
           dataPoints:[{x:2.0643278, y:1}] 
         },
         { type: "firstid",
           prov: 11,
           name: some1,
           showInLegend:false,
           dataPoints:[{x:3.2142857142857144, y:1}] 
         },
         { type: "firstid",
           prov: 12,
           name: some2,
           showInLegend:false,
           dataPoints:[{x:1.125, y:1}] 
         },
         { type: "firstid",
           prov: 12,
           name: some2,
           showInLegend:false,
           dataPoints:[{x:1.6666666666666667, y:1}] 
         }]

當我控制台記錄它時，我想得到一些數組是這樣的：

[        { type: "firstid",
           prov: 11,
           name: some1,
           showInLegend:false,
           dataPoints:[ {x:2.0643278, y:1}, 
                        {x:3.2142857142857144, y:1}] 
         },
         { type: "firstid",
           prov: 12,
           name: some2,
           showInLegend:false,
           dataPoints:[ {x:1.125, y:1}, 
                        {x:1.6666666666666667, y:1}] 
         }]

我花了我的時間試圖通過數組合並來解決它，但似乎我無法弄清楚，任何幫助都會很好

Answer 1

根據您的方法，如果元素不存在，我將嘗試僅推送，如果存在，則推送到現有元素上的 dataPoints 數組。 我沒有試過這個，但它可以工作。

data.forEach(e => {
    const tempIndex = tmp.findIndex(el => el.prov === e.prov);
    if (tempIndex) {
        tmp[tempIndex].dataPoints.push({ x: e.datax, y: e.datay });
    } else {
        tmp.push({
            type: "firstid",
            prov: e.prov,
            name: e.pname,
            showInLegend: false,
            dataPoints: [{ x: e.datax, y: e.datay }],
        })
    }
});

請讓我知道這種編輯方法是否更好

Answer 2

使用Array.prototype.reduce和Array.prototype.find應該可以解決問題。

reduce循環遍歷數組中的每個元素。 它以一個空數組開始： [] 。 然后代碼使用find檢查元素是否已經在新數組中。 如果是這樣，將dataPoints添加到現有數組中。 如果它不存在，則將對象推送到新數組。

 const obj = [{"pname":"some1","datax":2.0643278,"prov":"11","datay":1},{"pname":"some1","datax":3.2142857142857144,"prov":"11","datay":1},{"pname":"some2","datax":1.125,"prov":"12","datay":1},{"pname":"some2","datax":1.6666666666666667,"prov":"12","datay":2}]; objReduced = obj.reduce((acc, cur) => { const dbItem = acc.find( element => element.prov === cur.prov); if (dbItem) { dbItem.dataPoints.push({x: cur.datax, y: cur.datay}); } else { acc.push({ type: "firstid", prov: cur.prov, name: cur.pname, showInLegend: false, dataPoints: [{x: cur.datax, y: cur.datay}] }); } return acc; }, []); console.log(objReduced);

 obj = [{ type: "firstid", prov: 11, name: "some1", showInLegend: false, dataPoints: [{ x: 2.0643278, y: 1 }]

Answer 3

在合並它們之前，這還將檢查 dataPoints 數組中的重復項：

 const data = [{ type: "firstid", prov: 11, name: "some1", showInLegend: false, dataPoints: [{ x: 2.0643278, y: 1 }] }, { type: "firstid", prov: 11, name: "some1", showInLegend: false, dataPoints: [{ x: 3.2142857142857144, y: 1 }] }, { type: "firstid", prov: 12, name: "some2", showInLegend: false, dataPoints: [{ x: 1.125, y: 1 }] }, { type: "firstid", prov: 12, name: "some2", showInLegend: false, dataPoints: [{ x: 1.6666666666666667, y: 1 }] } ]; const res = Object.values(data.reduce((acc, curr) => { const groupKey = ['type', 'prov', 'name', 'showInLegend'].map(x => curr[x]).join('-'); if (!acc[groupKey]) { acc[groupKey] = curr; } else { for (const { x: currX, y: currY } of curr.dataPoints) { for (const { x: accX, y: accY } of acc[groupKey].dataPoints) { if ((currX + '-' + currY) !== (accX + '-' + accY)) { acc[groupKey].dataPoints.push({ x: currX, y: currY }); } } } } return acc; }, {})); console.log(res);