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javascript - 如何在javascript中的对象数组下获取值重复的计数

问题描述

我有一个有一些重复计数的数组。我想知道那个重复的次数。我的数组是这样的：

var array = [
    { asset: "A", sector: "Hospital" },
    { asset: "B", sector: "Hardware" },
    { asset: "C", sector: "Networking" },
    { asset: "D", sector: "Networking" },
    { asset: "E", sector: "Food" },
    { asset: "F", sector: "Hospital" },
    { asset: "G", sector: "Hardware" },
    { asset: "H", sector: "Industrial" },
    { asset: "I", sector: "Transport" },
    { asset: "J", sector: "Hardware" },
    { asset: "K", sector: "Networking" },
    { asset: "L", sector: "Transport" }
]

现在我希望重复计数也像这样排序，如下所示：

final_array = [
    { sector: 'Hardware', count: 3 },
    { sector: 'Networking', count: 3 },
    { sector: 'Hospital', count: 2 },
    { sector: 'Transport', count: 2 },
    { sector: 'Food', count: 1 },
    { sector: 'Industrial', count: 1 }
]

我不知道从哪里可以做到这一点。我有很多链接，但它们解决了数组下的重复，而不是对象数组上的重复。

我使用一种方法，但这不能解决我的问题

var finalD = [];
c.forEach(x => {
    if (isSectorExists(x.sector, finalD) == true) {
        //Here I don't know how I will increase the counter.
    } else {
        finalD.push({ sector: x.sector, count: 1 });
    }
});

var isSectorExists = (sector, arr) => {
    return arr.some(function(el) {
        return el.sector === sector;
    });
};

我知道我的方法很长。是否有人有最好和最简单的方法来完成这种任务。任何帮助真的很感激。提前致谢

标签： javascriptarraysobjectecmascript-6

解决方案

您可以构建一个counts以扇区名称作为键的对象和一个将计数存储为值的对象，遍历您的数组并更新这些计数，然后最终获取该counts对象的值。

var array = [
      { asset: "A", sector: "Hospital" },
      { asset: "B", sector: "Hardware" },
      { asset: "C", sector: "Networking" },
      { asset: "D", sector: "Networking" },
      { asset: "E", sector: "Food" },
      { asset: "F", sector: "Hospital" },
      { asset: "G", sector: "Hardware" },
      { asset: "H", sector: "Industrial" },
      { asset: "I", sector: "Transport" },
      { asset: "J", sector: "Hardware" },
      { asset: "K", sector: "Networking" },
      { asset: "L", sector: "Transport" }
];
            
            
var counts = array.reduce((m, c) => {
    if (c.sector in m) m[c.sector].count += 1;
    else m[c.sector] = { sector: c.sector, count: 1};
    return m;
}, {});

let finalArray = Object.values(counts).sort((a, b) => b.count - a.count);

console.log(finalArray);

这将在线性时间内获得计数（而不是调用isSectorExists每个元素使其成为O(n^2)。

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