将具有相同键值的数组中的JavaScript对象组合在一起

Question

我一直在尝试将具有相同id值的数组内的对象组合在一起。 我拥有的数组是这样的：

[
{"id" : "abcd","val1" : 1,"val2": 1, "val3" : 0},
{"id" : "abcd","val1" : 1,"val2": 1, "val3" : 1},
{"id" : "efgh","val1" : 0,"val2": 0, "val3" : 1}
]

我一直试图找到一种方法来组合这些，以便结果如下：

[
{"id" : "abcd","val1" : 2,"val2": 2, "val3" : 1},
{"id" : "efgh","val1" : 0,"val2": 0, "val3" : 1}
]

有没有办法用underscore.js做到这一点？

Answer 1

这应该通过下划线的功能方法来实现：

_.map(_.groupBy(arr, "id"), function(vals, id) {
    return _.reduce(vals, function(m, o) {
        for (var p in o)
            if (p != "id")
                m[p] = (m[p]||0)+o[p];
        return m;
    }, {id: id});
});

Answer 2

标准的方法是使用一个对象作为map（这里是b ）：

var b = {}, arr = [];
for (var id in a) {
  var oa = a[id], ob = b[oa.id];
  if (!ob) arr.push(ob = b[oa.id] = {}); 
  for (var k in oa) ob[k] = k==='id' ? oa.id : (ob[k]||0)+oa[k];
}
console.log(arr)

Answer 3

请参考以下示例

 var array = [{ "sequence" : 1, "locationId" : "332228", "lat" : 25.246511, "lng" : 55.293837, "stopName" : "332228", "locationType" : "service", "serviceType" : "Delivery", "arrivalTime" : 37666, "departureTime" : 37830, "travelTime" : 1593, "travelDistance" : 20985, "travelCost" : 0, "serviceTime" : 30, "serviceTimeCost" : 0, "tripNumber" : 0, "orders" : [ { "orderId" : "AQ137O1701240", "SIZE1" : "28", "SIZE2" : "520", "SIZE3" : "52" } ], "stopId" : "SkhirG2ioZ" }, { "sequence" : 2, "locationId" : "332228", "lat" : 25.236407, "lng" : 55.272403, "stopName" : "332228", "locationType" : "service", "serviceType" : "Delivery", "arrivalTime" : 38575, "departureTime" : 38605, "travelTime" : 1593, "travelDistance" : 20985, "travelCost" : 0, "serviceTime" : 30, "serviceTimeCost" : 0, "tripNumber" : 0, "orders" : [ { "orderId" : "AQ137O1701233", "SIZE1" : "23", "SIZE2" : "402", "SIZE3" : "30" } ], "stopId" : "H1iirfhisW" }, { "sequence" : 3, "locationId" : "332228", "lat" : 25.221368, "lng" : 55.265915, "stopName" : "332228", "locationType" : "service", "serviceType" : "Delivery", "arrivalTime" : 39137, "departureTime" : 39167, "travelTime" : 974, "travelDistance" : 5717, "travelCost" : 0, "serviceTime" : 30, "serviceTimeCost" : 0, "tripNumber" : 0, "orders" : [ { "orderId" : "AQ110O1705036", "SIZE1" : "60", "SIZE2" : "1046", "SIZE3" : "68" } ], "stopId" : "H1csHM3jjb" }] var arrOut = []; array.forEach(function(value) { var existing = arrOut.filter(function(v, i) { return v.locationId == value.locationId; }); if (existing.length) { var existingIndex = arrOut.indexOf(existing[0]); arrOut[existingIndex].orders = arrOut[existingIndex].orders.concat(value.orders); } else { if(Array.isArray(value.orders)){ value.orders = value.orders arrOut.push(value); } } }); document.write('<pre>' + JSON.stringify(arrOut, 0, 4) + '</pre>'); 

Answer 4

这是我乏味的方式：小提琴： http ： //jsfiddle.net/hN8b6/5/

var a = [{
    "id": "abcd",
        "val1": 1,
        "val2": 1,
        "val3": 0
}, {
    "id": "abcd",
        "val1": 1,
        "val2": 1,
        "val3": 1
}, {
    "id": "efgh",
        "val1": 0,
        "val2": 0,
        "val3": 1
}];

// clone a
a2 = JSON.parse(JSON.stringify(a));

// make sure elements with the same id are next to each other
a2.sort(function (x, y) {
    if (x['id'] < y['id']) {
        return -1;
    }
    if (x['id'] > y['id']) {
        return 1;
    }
    return 0;
});

// iterate over each one, if this one has the same id as the previous one, accumulate
// else add to b
var lastId;
var b = [];
for (var i = 0; i < a2.length; i++) {
    if (lastId == a2[i]['id']) {
        b[b.length-1]['val1'] += a2[i]['val1'];
        b[b.length-1]['val2'] += a2[i]['val2'];
        b[b.length-1]['val3'] += a2[i]['val3'];
    } else {
        b[b.length] = (a2[i]);
        lastId = a2[i]['id'];
    }
}

console.log(b);