根据对象中的其他值对对象的值求和答案

【问题标题】：Sum values of objects based on other values in object根据对象中的其他值对对象的值求和
【发布时间】：2020-10-14 04:08:48
【问题描述】：

我有一个对象，格式是这样的：

let obj= {
  'Jan 01': [
             {dt: 'Jan 01', cat: 'abc', site: 'google', val1:10, val2:20, val3:30},
             {dt: 'Jan 01', cat: 'abc', site: 'bing', val1:23, val2:12, val3:14},
             {dt: 'Jan 01', cat: 'abc', site: 'jeeves', val1:67, val2:78, val3:12},
             {dt: 'Jan 01', cat: 'pqr', site: 'google', val1:10, val2:20, val3:30},
             {dt: 'Jan 01', cat: 'pqr', site: 'bing', val1:23, val2:12, val3:14},
             {dt: 'Jan 01', cat: 'pqr', site: 'jeeves', val1:67, val2:78, val3:12},
             {dt: 'Jan 01', cat: 'xyz', site: 'google', val1:10, val2:20, val3:30},
             {dt: 'Jan 01', cat: 'xyz', site: 'bing', val1:23, val2:12, val3:14},
             {dt: 'Jan 01', cat: 'xyz', site: 'jeeves', val1:67, val2:78, val3:12}
            ],
  'Feb 01': [
             {dt: 'Feb 01', cat: 'abc', site: 'google', val1:12, val2:11, val3:41},
             {dt: 'Feb 01', cat: 'abc', site: 'bing', val1:45, val2:67, val3:0},
             {dt: 'Feb 01', cat: 'abc', site: 'jeeves', val1:78, val2:34, val3:41},
             {dt: 'Feb 01', cat: 'pqr', site: 'google', val1:44, val2:88, val3:5},
             {dt: 'Feb 01', cat: 'pqr', site: 'bing', val1:56, val2:11, val3:99},
             {dt: 'Feb 01', cat: 'pqr', site: 'jeeves', val1:22, val2:34, val3:77},
             {dt: 'Feb 01', cat: 'xyz', site: 'google', val1:33, val2:99, val3:34},
             {dt: 'Feb 01', cat: 'xyz', site: 'bing', val1:77, val2:55, val3:14},
             {dt: 'Feb 01', cat: 'xyz', site: 'jeeves', val1:33, val2:23, val3:98}
            ],
  'Mar 01': [
             {dt: 'Mar 01', cat: 'abc', site: 'google', val1:11, val2:20, val3:6},
             {dt: 'Mar 01', cat: 'abc', site: 'bing', val1:22, val2:91, val3:89},
             {dt: 'Mar 01', cat: 'abc', site: 'jeeves', val1:33, val2:81, val3:12},
             {dt: 'Mar 01', cat: 'pqr', site: 'google', val1:44, val2:71, val3:33},
             {dt: 'Mar 01', cat: 'pqr', site: 'bing', val1:55, val2:61, val3:14},
             {dt: 'Mar 01', cat: 'pqr', site: 'jeeves', val1:66, val2:51, val3:43},
             {dt: 'Mar 01', cat: 'xyz', site: 'google', val1:77, val2:41, val3:98},
             {dt: 'Mar 01', cat: 'xyz', site: 'bing', val1:88, val2:31, val3:23},
             {dt: 'Mar 01', cat: 'xyz', site: 'jeeves', val1:99, val2:21, val3:4}
            ]
}

我想根据对象内的 2 个值对 val1 的值求和。我想对基于dt 和cat 的值求和。

我知道如何根据键对值求和：

    let group = Object.entries(obj).map(([key, group]) => ({
          ['dt']: key, // because it is similar to the key
          ['val1']: group.map(entry => entry[val1]).reduce((pv, cv) => {
                    return pv + (parseFloat(cv) || 0);,
          ['val2']: group.map(entry => entry[val2]).reduce((pv, cv) => {
                    return pv + (parseFloat(cv) || 0);,
          ['val3']: group.map(entry => entry[val3]).reduce((pv, cv) => {
                    return pv + (parseFloat(cv) || 0);,            
    }));

如何根据dt 和cat 求和？我需要我的最终值是：

let obj= {
  'Jan 01': [
             {dt: 'Jan 01', cat: 'abc', val1:100, val2:110, val3:56},
             {dt: 'Jan 01', cat: 'pqr', val1:100, val2:110, val3:56},
             {dt: 'Jan 01', cat: 'xyz', val1:100, val2:110, val3:56},   
             ],
  'Feb 01': [
             {dt: 'Feb 01', cat: 'abc', val1:135, val2:112, val3:82},
             {dt: 'Feb 01', cat: 'pqr', val1:122, val2:133, val3:181},
             {dt: 'Feb 01', cat: 'xyz', val1:143, val2:177, val3:146}
            ],
  'Mar 01': [
             {dt: 'Mar 01', cat: 'abc', val1:66, val2:192, val3:107},
             {dt: 'Mar 01', cat: 'pqr', val1:165, val2:183, val3:90},
             {dt: 'Mar 01', cat: 'xyz', val1:264, val2:93, val3:125}
            ]
}

甚至dt 和site。

我希望能够基于 2 个值进行分组。这可能吗？

【问题讨论】：

使用 dt+cat 作为查找表键
@user120242，你是怎么做到的？抱歉，我有点不熟悉

标签： javascript jquery arrays object ecmascript-6

【解决方案1】：

使用dt+','+cat 作为Map 中的键来跟踪骗子。总结所有其他键。
假定除站点 dt 和 cat 之外的任何键都是 val。

for(const key of Object.keys(obj)) {
  const catMap = new Map()
  for(const {site, dt, cat, ...vals} of obj[key]) {
    const row = catMap.get(dt+','+cat)||{dt, cat}
    Object.entries(vals).forEach(([k,v])=>row[k]=(row[k]||0)+v)
    catMap.set(dt+','+cat, row)
  }
  obj[key] = [...catMap.values()]
}

console.log(
  obj
)

<script>
let obj= {
  'Jan 01': [
             {dt: 'Jan 01', cat: 'abc', site: 'google', val1:10, val2:20, val3:30},
             {dt: 'Jan 01', cat: 'abc', site: 'bing', val1:23, val2:12, val3:14},
             {dt: 'Jan 01', cat: 'abc', site: 'jeeves', val1:67, val2:78, val3:12},
             {dt: 'Jan 01', cat: 'pqr', site: 'google', val1:10, val2:20, val3:30},
             {dt: 'Jan 01', cat: 'pqr', site: 'bing', val1:23, val2:12, val3:14},
             {dt: 'Jan 01', cat: 'pqr', site: 'jeeves', val1:67, val2:78, val3:12},
             {dt: 'Jan 01', cat: 'xyz', site: 'google', val1:10, val2:20, val3:30},
             {dt: 'Jan 01', cat: 'xyz', site: 'bing', val1:23, val2:12, val3:14},
             {dt: 'Jan 01', cat: 'xyz', site: 'jeeves', val1:67, val2:78, val3:12}
            ],
  'Feb 01': [
             {dt: 'Feb 01', cat: 'abc', site: 'google', val1:12, val2:11, val3:41},
             {dt: 'Feb 01', cat: 'abc', site: 'bing', val1:45, val2:67, val3:0},
             {dt: 'Feb 01', cat: 'abc', site: 'jeeves', val1:78, val2:34, val3:41},
             {dt: 'Feb 01', cat: 'pqr', site: 'google', val1:44, val2:88, val3:5},
             {dt: 'Feb 01', cat: 'pqr', site: 'bing', val1:56, val2:11, val3:99},
             {dt: 'Feb 01', cat: 'pqr', site: 'jeeves', val1:22, val2:34, val3:77},
             {dt: 'Feb 01', cat: 'xyz', site: 'google', val1:33, val2:99, val3:34},
             {dt: 'Feb 01', cat: 'xyz', site: 'bing', val1:77, val2:55, val3:14},
             {dt: 'Feb 01', cat: 'xyz', site: 'jeeves', val1:33, val2:23, val3:98}
            ],
  'Mar 01': [
             {dt: 'Mar 01', cat: 'abc', site: 'google', val1:11, val2:20, val3:6},
             {dt: 'Mar 01', cat: 'abc', site: 'bing', val1:22, val2:91, val3:89},
             {dt: 'Mar 01', cat: 'abc', site: 'jeeves', val1:33, val2:81, val3:12},
             {dt: 'Mar 01', cat: 'pqr', site: 'google', val1:44, val2:71, val3:33},
             {dt: 'Mar 01', cat: 'pqr', site: 'bing', val1:55, val2:61, val3:14},
             {dt: 'Mar 01', cat: 'pqr', site: 'jeeves', val1:66, val2:51, val3:43},
             {dt: 'Mar 01', cat: 'xyz', site: 'google', val1:77, val2:41, val3:98},
             {dt: 'Mar 01', cat: 'xyz', site: 'bing', val1:88, val2:31, val3:23},
             {dt: 'Mar 01', cat: 'xyz', site: 'jeeves', val1:99, val2:21, val3:4}
            ]
}
</script>

【讨论】：

【解决方案2】：

这是使用reduce的另一种方法

  let objx= {
    'Jan 01': [
               {dt: 'Jan 01', cat: 'abc', site: 'google', val1:10, val2:20, val3:30},
               {dt: 'Jan 01', cat: 'abc', site: 'bing', val1:23, val2:12, val3:14},
               {dt: 'Jan 01', cat: 'abc', site: 'jeeves', val1:67, val2:78, val3:12},
               {dt: 'Jan 01', cat: 'pqr', site: 'google', val1:10, val2:20, val3:30},
               {dt: 'Jan 01', cat: 'pqr', site: 'bing', val1:23, val2:12, val3:14},
               {dt: 'Jan 01', cat: 'pqr', site: 'jeeves', val1:67, val2:78, val3:12},
               {dt: 'Jan 01', cat: 'xyz', site: 'google', val1:10, val2:20, val3:30},
               {dt: 'Jan 01', cat: 'xyz', site: 'bing', val1:23, val2:12, val3:14},
               {dt: 'Jan 01', cat: 'xyz', site: 'jeeves', val1:67, val2:78, val3:12}
              ],
              'Feb 01': [
                {dt: 'Feb 01', cat: 'abc', site: 'google', val1:12, val2:11, val3:41},
                {dt: 'Feb 01', cat: 'abc', site: 'bing', val1:45, val2:67, val3:0},
                {dt: 'Feb 01', cat: 'abc', site: 'jeeves', val1:78, val2:34, val3:41},
                {dt: 'Feb 01', cat: 'pqr', site: 'google', val1:44, val2:88, val3:5},
                {dt: 'Feb 01', cat: 'pqr', site: 'bing', val1:56, val2:11, val3:99},
                {dt: 'Feb 01', cat: 'pqr', site: 'jeeves', val1:22, val2:34, val3:77},
                {dt: 'Feb 01', cat: 'xyz', site: 'google', val1:33, val2:99, val3:34},
                {dt: 'Feb 01', cat: 'xyz', site: 'bing', val1:77, val2:55, val3:14},
                {dt: 'Feb 01', cat: 'xyz', site: 'jeeves', val1:33, val2:23, val3:98}
               ],
     'Mar 01': [
                {dt: 'Mar 01', cat: 'abc', site: 'google', val1:11, val2:20, val3:6},
                {dt: 'Mar 01', cat: 'abc', site: 'bing', val1:22, val2:91, val3:89},
                {dt: 'Mar 01', cat: 'abc', site: 'jeeves', val1:33, val2:81, val3:12},
                {dt: 'Mar 01', cat: 'pqr', site: 'google', val1:44, val2:71, val3:33},
                {dt: 'Mar 01', cat: 'pqr', site: 'bing', val1:55, val2:61, val3:14},
                {dt: 'Mar 01', cat: 'pqr', site: 'jeeves', val1:66, val2:51, val3:43},
                {dt: 'Mar 01', cat: 'xyz', site: 'google', val1:77, val2:41, val3:98},
                {dt: 'Mar 01', cat: 'xyz', site: 'bing', val1:88, val2:31, val3:23},
                {dt: 'Mar 01', cat: 'xyz', site: 'jeeves', val1:99, val2:21, val3:4}
               ]
            }

fo={}
  for(let o of Object.entries(objx)){
    fo[o[0]]=[]
     o[1].reduce((acc,curr,i)=>{
      return curr.dt==acc.dt && acc.cat==curr.cat? (acc= {...acc,...{...curr,['val1']:acc.val1+curr.val1, ['val2']:acc.val2+curr.val2, ['val3']:acc.val3+curr.val3}} 
           ,i!=o[1].length-1 ? acc : (delete acc.site  ,fo[o[0]].push(acc))): (delete acc.site,fo[o[0]].push(acc) , acc=curr)
    })
  }
 console.log(fo)

【讨论】：