获取数组中每一天的平均值

Question

我正在尝试在 Javascript/jQuery 中找到一种有效的方法来执行以下操作：

我得到了一个包含对象的数组，每个对象都包含一个日期和一个评级。许多条目具有相同的日期。我想创建另一个对象数组，每个对象都包含日期（唯一）和当天所有评分的平均值。

示例数据：

var array = [{
    date: "07-08-2018"
    rating: 3
    },{
    date: "07-08-2018"
    rating: 5
    },{
    date: "08-08-2018"
    rating: 2
    },{
    date: "08-08-2018"
    rating: 1
    },{
    date: "09-08-2018"
    rating: 3
    },{
    date: "09-08-2018"
    rating: 5
    }
]

我要做什么：

var averageArray = [{
    date: "07-08-2018",
    rating: 4
    },{
    date: "08-08-2018",
    rating: 1.5
    },{
    date: "09-08-2018",
    rating: 4
    }]

我当前的代码（只返回 2 个值而不是 3 个）

var oldDate = new Date(array[0].date);
var sum = 0;
var counter = 0
var newArray = []

for(var i = 0; i < array.length; i++){
    var date = new Date(array[i].date);

    if(date > oldDate){
        console.log("new day");
        var avg = (sum/counter);
        //console.log(avg);
        var newObject = new Object({
            date: date,
            rating: avg
        });
        newArray.push(newObject);
        counter = 0;
        sum = 0;
        oldDate = date;
    }
    counter++;
    sum = sum + array[i].rating;
    console.log(counter + " " + sum)
}

console.log(newArray)

Answer 1

对事物进行分组的最佳方法是创建一个使用公共值作为属性键的对象。

然后将它们分组到该对象中后，将结果映射到新数组

var tmp ={};

data.forEach(function(item){
  // if property for current date already exists  we update existing otherwise start new one
  var obj =  tmp[item.date] = tmp[item.date] || {count:0, total: 0};
  // increment count and total of all the ratings
  obj.count ++;
  obj.total += item.rating

});

var res = Object.entries(tmp).map(function(entry){
    return { date: entry[0], ave: entry[1].total/entry[1].count}
})

console.log( res)

<script>
var data = [{
    date: "07-08-2018",
    rating: 3
    },{
    date: "07-08-2018",
    rating: 5
    },{
    date: "08-08-2018",
    rating: 2
    },{
    date: "08-08-2018",
    rating: 1
    },{
    date: "09-08-2018",
    rating: 3
    },{
    date: "09-08-2018",
    rating: 5
    }
]
</script>