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如何在ES6中使用reduce()计算平均值?

ecmascript-6 javascript

stone rock · 技术社区 · 7 年前

我正在尝试从JSON数据中提取 econ 和 season 我想找出每个季节的平均经济率


   
    
     
      例如:
      
       season=2018 economy rate=10,20,30 so avg economy rate=(10+20+30)/3=20
      
     
     
      
       
        season=2017 economy rate=30,40,50 so avg economy rate=(30+40+50)/3=40
       
      
      
       因此,我要计算特定季节的所有经济率之和,然后取平均值,即将特定季节的总金额除以该季节的比赛数。
      
      
      
      
       下面是我尝试使用ES6中
       
        reduce()的代码:
       
      
      
       const economy=bowldata.reduce((a,season,econ)=>。{
A[季节]=A[季节]+ParseInt(Econ);
返回A;
},);
console.log(经济型);
我得到的是2018:nanbut I want my output to be 2018:22,2017:50,etc.I.E I want key value pairs of season and average economy rate for that season.


屏幕截图:




假设我有json数据:

var bowldata=[
{
季节:2018
经济:10
},请
季节:2018
经济:20
},请
季节:2018
经济:30
},请
季节:2017年
经济:40
},请
季节:2017年
经济:50
},请
季节:2016年
经济:10
},请
]
现在,对象whichreduce()should return is key value pairs of year and average economy rate.


结果:[2018:20,2017:45,2016:10]
如:season = 2018 economy rate = 10,20,30 so avg economy rate = (10+20+30)/3 = 20


season = 2017 economy rate = 30,40,50 so avg economy rate = (30+40+50)/3 = 40

所以我要计算出特定季节的所有经济率之和,然后取平均值,即用特定季节的总金额除以该季节的比赛数量。

下面是我试图利用的代码reduce()在ES6中:

const economy = bowldata.reduce( (a,{season, econ})  => {
                    a[season] = a[season] + parseInt(econ);
                    return a; 
                }, {});
console.log(economy);


我得到了2018:NaN但是,我希望我的产出是2018:22、2017:50等,也就是说,我希望那个季节的关键价值对和平均经济率。

截图:



假设我有JSON数据:

var bowldata = [
    {
        season:2018
        economy:10
    },
    {   season:2018
        economy:20
    },
    {   season:2018
        economy:30
    },
    {   season:2017
        economy:40
    },
    {   season:2017
        economy:50
    },
    {   season:2016
        economy:10
    },
]


现在这个物体减少()应回报是关键价值对年平均经济率。

结果:[ 2018:20 , 2017:45 , 2016:10 ]

5 回复 | 直到 7 年前

Nina Scholz 7 年前

var bowldata = [{ season: 2001, econ: '30' }, { season: 2001, econ: '40' }, { season: 2001, econ: '50' }, { season: 2001, econ: '60' }, { season: 2002, econ: '30' }, { season: 2002, econ: '40' }, { season: 2002, econ: '50' }, { season: 2003, econ: '60' }, { season: 2003, econ: '30' }, { season: 2003, econ: '40' }, { season: 2003, econ: '50' }, { season: 2003, econ: '60' }],
    economy = bowldata.reduce((a, { season, econ }) => {
        a[season] = a[season] || { sum: 0, count: 0 };
        a[season].sum += +econ;
        a[season].average = a[season].sum / ++a[season].count;
        return a;
    }, Object.create(null)),
    nice = Object.assign(...Object
        .entries(economy)
        .map(([k, { average }]) => ({ [k]: average }))
    );

console.log(nice);
console.log(economy);

.as-console-wrapper { max-height: 100% !important; top: 0; }

charlietfl 7 年前

更改

a[season] = a[season] + parseInt(econ);

a[season] = (a[season] || 0 ) + parseInt(econ);

Sergio Mazzoleni 7 年前

第一次执行 a[season] = a[season] + parseInt(econ); a[season] 0 第一次访问它。

const economy = bowldata.reduce( (a,{season, econ})  => {
    if (!a[season]) {
        a[season] = 0;
    }
    a[season] = a[season] + parseInt(econ);
    return a; 
}, {});

Lionel Rowe 7 年前

平均值

const totals = bowldata.reduce((acc, cur)  => {

  if (!acc[cur.season]) {
    acc[cur.season] = {count: 0, total: 0};
  }

  acc[cur.season].count++;
  acc[cur.season].total += parseInt(cur.econ);

  return acc; 
}, {});

const averages = {};

Object.keys(totals).forEach(key => {

  averages[key] = totals[key].total / totals[key].count;

});

godpanrupesh 7 年前

hi您可以通过以下减少和映射链来解决问题:
ii)获得季节和平均经济的地图

bowldata.reduce((acc,b)=>。{
},[])

},[])


)