Question

var developers = [
{ name: "Joe", age: 23, overallLevel: "high", date: "Aug 14, 2015" },
{ name: "Sue", age: 28, overallLevel: "advanced", date: "Aug 11, 2015" },
{ name: "Jon", age: 32, overallLevel: "high", date: "Aug 10, 2015" },
{ name: "Bob", age: 24, overallLevel: "high", date: "Sept 07, 2015" },
{ name: "Johnson", age: 23, overallLevel: "advanced", date: "Aug 20, 2015" },
{ name: "Paul", age: 25, overallLevel: "basic", date: "Jan 30, 2016" },
{ name: "Jane", age: 27, overallLevel: "intermediate", date: "Aug 30, 2016" },
{ name: "Chris", age: 20, overallLevel: "basic", date: "Aug 1, 2016" },
{ name: "Susan", age: 25, overallLevel: "basic", date: "Dec 12, 2015" },
{ name: "Jenny", age: 23, overallLevel: "intermediate", date: "Aug 30, 2016" },
{ name: "Stone", age: 25, overallLevel: "basic", date: "June 13, 2016" },
]

以下是从数组中提取的数据，表示级别后跟日期。

＆＃34;高＆＃34; ： 2015年8月10日， 2015年8月14日， 2015年9月7日

＆＃34;高级＆＃34; ： 2015年8月11日， 2015年8月20日

＆＃34;碱性＆＃34; ： 2016年1月30日， 2016年8月1日， 2015年12月12日， 2016年6月13日，

＆＃34;中间＆＃34; ： 2016年8月30日， 2016年8月30日

预期输出：要求最小和最大日期范围，即进一步显示

{ 高：2015年8月10日 - 2015年9月7日，高级：2015年8月11日 - 2015年8月20日，基本：2015年12月12日 - 2016年8月1日，中级：2016年1月30日 }

这是我能够获得用于分组目的的overallLevel计数。

var developers = [
{ name: "Joe", age: 23, overallLevel: "high", date: "Aug 14, 2015" },
{ name: "Sue", age: 28, overallLevel: "advanced", date: "Aug 11, 2015" },
{ name: "Jon", age: 32, overallLevel: "high", date: "Aug 10, 2015" },
{ name: "Bob", age: 24, overallLevel: "high", date: "Sept 07, 2015" },
{ name: "Johnson", age: 23, overallLevel: "advanced", date: "Aug 20, 2015" },
{ name: "Paul", age: 25, overallLevel: "basic", date: "Jan 30, 2016" },
{ name: "Jane", age: 27, overallLevel: "intermediate", date: "Aug 30, 2016" },
{ name: "Chris", age: 20, overallLevel: "basic", date: "Aug 1, 2016" },
{ name: "Susan", age: 25, overallLevel: "basic", date: "Dec 12, 2015" },
{ name: "Jenny", age: 23, overallLevel: "intermediate", date: "Aug 30, 2016" },
{ name: "Stone", age: 25, overallLevel: "basic", date: "June 13, 2016" },
],
    overallLevel = developers.reduce(function (r, a) {
        r[a.overallLevel] = (r[a.overallLevel] || 0) + 1;
        return r;
    }, {});

console.log(overallLevel);

请访问： https://jsfiddle.net/Saily/6nzxd6ss/

Answer 1

我也很喜欢使用reduce。它对很多案件来说非常方便。

＆＃13;

var developers = [
{ name: "Joe", age: 23, overallLevel: "high", date: "Aug 14, 2015" },
{ name: "Sue", age: 28, overallLevel: "advanced", date: "Aug 11, 2015" },
{ name: "Jon", age: 32, overallLevel: "high", date: "Aug 10, 2015" },
{ name: "Bob", age: 24, overallLevel: "high", date: "Sept 07, 2015" },
{ name: "Johnson", age: 23, overallLevel: "advanced", date: "Aug 20, 2015" },
{ name: "Paul", age: 25, overallLevel: "basic", date: "Jan 30, 2016" },
{ name: "Jane", age: 27, overallLevel: "intermediate", date: "Aug 30, 2016" },
{ name: "Chris", age: 20, overallLevel: "basic", date: "Aug 1, 2016" },
{ name: "Susan", age: 25, overallLevel: "basic", date: "Dec 12, 2015" },
{ name: "Jenny", age: 23, overallLevel: "intermediate", date: "Aug 30, 2016" },
{ name: "Stone", age: 25, overallLevel: "basic", date: "June 13, 2016" },
],
reduced = developers.reduce((p,c) => {var d = new Date(c.date);
                                      p[c.overallLevel] ? d < p[c.overallLevel][0] && d < p[c.overallLevel][1] ? p[c.overallLevel][0] = d
       	                                                                                                       : p[c.overallLevel][1] = d
                                                        : p[c.overallLevel] = [d,d];
                                      return p},{});
reduced = Object.keys(reduced).reduce((p,c) => (p[c] = p[c].map(e => e.toDateString(2)), p),reduced);
       
console.log(reduced);

＆＃13;

JavaScript代码，使用array.reduce（）根据数组的一个属性从对象数组中获取日期范围

1 个答案: