JS Array Reduce 以计算具有多个匹配键的对象值答案

【问题标题】：JS Array Reduce to count Object Values with a Multiple Matching KeysJS Array Reduce 以计算具有多个匹配键的对象值
【发布时间】：2021-11-19 19:35:04
【问题描述】：

我有一个这样的数组：

var objects = [{
    cat: "Fixtures"
    name: "brilliance_denali_(best)"
    shadowColor: "warmWhite"
    subCat: "Spot & Bullet Lights"
    title: "Brilliance Denali (Best)"
    wattage: {upPrice: 10, value: '10W'}
},
{
    cat: "Fixtures"
    name: "brilliance_denali_(best)"
    shadowColor: "warmWhite"
    subCat: "Spot & Bullet Lights"
    title: "Brilliance Denali (Best)"
    wattage: {upPrice: 10, value: '10W'}
},
{
    cat: "Fixtures"
    name: "brilliance_denali_(best)"
    shadowColor: "red"
    subCat: "Spot & Bullet Lights"
    title: "Brilliance Denali (Best)"
    wattage: {upPrice: 12, value: '12W'}
},
{
    cat: "Fixtures"
    name: "bullet_lights"
    shadowColor: "blue"
    subCat: "Lights"
    title: "Bullet Lights"
    wattage: {upPrice: 28, value: '10W'}
}]

我想根据“cat”减少这个数组，然后在 cat 内部是“subCat”，然后在我想用额外的“COUNT”键减少它们的对象内部，如果它们具有相同的“名称”，“ shadowColor”和“功率值”。所以总的输出数组应该是这样的：

var formattedObjects = [
    {
        cat: 'Fixtures',
        children: [
            {
                subCat: 'Lights',
                childres: [
                    {
                        cat: "Fixtures"
                        name: "bullet_lights"
                        shadowColor: "blue"
                        subCat: "Lights"
                        title: "Bullet Lights"
                        wattage: {upPrice: 28, value: '10W'},
                        count: 1     <-- COUNT BASED ON SAME : NAME && SHADOWCOLOR && WATTAGE VALUE
                    }
                ]
            }, {
                subCat: 'Spot & Bullet Lights',
                childres: [
                    {
                        {
                            cat: "Fixtures"
                            name: "brilliance_denali_(best)"
                            shadowColor: "warmWhite"
                            subCat: "Spot & Bullet Lights"
                            title: "Brilliance Denali (Best)"
                            wattage: {upPrice: 10, value: '10W'},
                            count: 2         <-- COUNT BASED ON SAME : NAME && SHADOWCOLOR && WATTAGE VALUE
                        },
                        {
                            cat: "Fixtures"
                            name: "brilliance_denali_(best)"
                            shadowColor: "red"
                            subCat: "Spot & Bullet Lights"
                            title: "Brilliance Denali (Best)"
                            wattage: {upPrice: 12, value: '12W'},
                            count: 1        <-- COUNT BASED ON SAME : NAME && SHADOWCOLOR && WATTAGE VALUE
                        }

                    }
                ]
            }
        ]
    }
]

到目前为止，我能够根据“subCat”和“cat”来减少它，每个都有嵌套的孩子，但我不确定我应该如何在其中添加计数并删除具有相同“名称”、“sahdowColor”的对象和“瓦数”。这就是我的减速器在不计数时的样子：

const formatedObjects = objects.reduce((list, item) => {
    const { cat, subCat } = item;
    const hasList = !!list?.[cat];
    const subList = list?.[cat]?.children.find(child => child.subCat === subCat)
    const value = { 
      name: item.name,
      title: item.title,
      shadowColor: item.shadowColor ? item.shadowColor : null ,
      wattage: item.wattage ? item.wattage : null
    }; // the element in data property
    if (!hasList) {
      list[cat] = { cat, children: [
        {
          subCat,
          children: [value]
        }
      ]}
    } else if (!subList){
      list[cat].children.push({ subCat, children: [value]})
    } else {
      subList.children.push(value)     
    }
    return list;
  }, {})

知道我应该如何在减速器中添加计数吗？？？谢谢。

【问题讨论】：

标签： javascript arrays reduce

【解决方案1】：

不确定，但这是您要找的吗？

let objects = [{ cat: "Fixtures", name: "brilliance_denali_(best)", shadowColor: "warmWhite", subCat: "Spot & Bullet Lights", title: "Brilliance Denali (Best)", wattage: { upPrice: 10, value: '10W' } }, { cat: "Fixtures", name: "brilliance_denali_(best)", shadowColor: "warmWhite", subCat: "Spot & Bullet Lights", title: "Brilliance Denali (Best)", wattage: { upPrice: 10, value: '10W' } }, { cat: "Fixtures", name: "brilliance_denali_(best)", shadowColor: "red", subCat: "Spot & Bullet Lights", title: "Brilliance Denali (Best)", wattage: { upPrice: 12, value: '12W' } }, { cat: "Fixtures", name: "bullet_lights", shadowColor: "blue", subCat: "Lights", title: "Bullet Lights", wattage: { upPrice: 28, value: '10W' } }]

let result = [], counter = {};

for (let item of objects) {
    let countId = item.name + item.shadowColor + item.wattage.value
    if (counter[countId]) {
      counter[countId].count++;
      continue
    }
    let cat = result.find(o => o.cat == item.cat);
    if (!cat) result.push(cat = { cat: item.cat, children: [] });
    let subCat = cat.children.find(o => o.subCat == item.subCat);
    if (!subCat) cat.children.push(subCat = { subCat: item.subCat, childres: [] });
    item.count = 1
    subCat.childres.push(counter[countId] = item);
}

console.log(result)

【讨论】：

这很接近，但它仍然没有删除重复的。一旦我们在计数中添加相同的对（使用 item.name + item.shadowColor + item.wattage.value），我需要在结果中删除重复的一对。
在您的示例中，名称为“brilliance_denali_(best)”，shadowColor“warmWhite”，功率值“10W”打印两次。一个计数：1，一个计数：2。我只需要 count:2 的总计数。