使用map,reduce或filter获取属性的唯一值数组javascript

问题描述 投票:-1回答:4

我有这个数组,并希望只获得独特的课程和更新的DateStart。

const classes = [{
    Course: {id: 1, Name: 'JS'},
    Name: 'JS Morning',DateStart: "2018/09/01"
}, {
    Course: {id: 1, Name: 'JS'},
Name: 'JS Afternoon',DateStart: "2018/10/15"
}, {
    Course: {id: 1, Name: 'JS'},
    Name: 'JS Night',DateStart: "2018/10/01"
}, {
    Course: {id: 2, Name: 'Jquery'},
    Name: 'JQ Morning',DateStart: "2018/10/01"
}, {
    Course: {id: 2, Name: 'Jquery'},
    Name: 'JQ Night',DateStart: "2018/09/15"
}];

每个例子,我的结果数组应该是:

const results = [{
    DataStart:"2018/09/01",Course{id: 1, Name: "JS"}
},{
    DataStart:"2018/09/15",Course{id: 2, Name: "Jquery"}

我知道使用foreach这样做,但我想使用MAP或REDUCE或FILTER。有人能帮助我吗?

谢谢!!!亚历克斯

javascript ecmascript-6
4个回答
0
投票

我们可以按日期先做一个sort(),在这种情况下更容易比较日期作为数字,取代/20180901,一旦我们有排序数组我们可以做filter()只保留Course.Name的第一个记录。

var classes = [{
    Course: {id: 1, Name: 'JS'},
    Name: 'JS Morning',DateStart: "2018/09/01"
}, {
    Course: {id: 1, Name: 'JS'},
Name: 'JS Afternoon',DateStart: "2018/10/15"
}, {
    Course: {id: 1, Name: 'JS'},
    Name: 'JS Night',DateStart: "2018/10/01"
}, {
    Course: {id: 2, Name: 'Jquery'},
    Name: 'JQ Morning',DateStart: "2018/10/01"
}, {
    Course: {id: 2, Name: 'Jquery'},
    Name: 'JQ Night',DateStart: "2018/09/15"
}];


var diff = {}

results = classes.sort((a,b)=>{

    return a.DateStart.replace(/\//g, "") - b.DateStart.replace(/\//g, ""); 
 
}).filter(a=>{
    delete a.Name
    if(a.Course.Name in diff){
        return false;
    }else{
        diff[a.Course.Name] = true;
        return true;
    }
    
})

console.log(results)

0
投票

var classes = [{
    Course: {id: 1, Name: 'JS'},
    Name: 'JS Morning',DateStart: "2018/09/01"
}, {
    Course: {id: 1, Name: 'JS'},
Name: 'JS Afternoon',DateStart: "2018/09/01"
}, {
    Course: {id: 1, Name: 'JS'},
    Name: 'JS Night',DateStart: "2018/09/01"
}, {
    Course: {id: 2, Name: 'Jquery'},
    Name: 'JQ Morning',DateStart: "2018/09/15"
}, {
    Course: {id: 2, Name: 'Jquery'},
    Name: 'JQ Night',DateStart: "2018/09/15"
}];

var result = classes.map(x => { return { DateStart: x.DateStart, Course: x.Course } }).filter((x,i,a)=> i == a.findIndex(y => y.DateStart == x.DateStart && y.Course.id == x.Course.id && y.Course.Name == x.Course.Name));

console.log(result);

0
投票

您可以使用array#reduce使用id获取独特的课程。如果你得到一个重复的课程,那么比较DateStart并存储先前的值。获得对象累加器中的所有唯一课程后,使用Object.values()获取所有值。

const classes = [{ Course: {id: 1, Name: 'JS'}, Name: 'JS Morning',DateStart: "2018/09/01" }, { Course: {id: 1, Name: 'JS'}, Name: 'JS Afternoon',DateStart: "2018/10/15" }, { Course: {id: 1, Name: 'JS'}, Name: 'JS Night',DateStart: "2018/10/01" }, { Course:{id: 2, Name: 'Jquery'}, Name: 'JQ Morning',DateStart: "2018/10/01" }, { Course: {id: 2, Name: 'Jquery'}, Name: 'JQ Night',DateStart: "2018/09/15" }],
    result = Object.values(classes.reduce((r,{Course, DateStart}) => {
      if(r[Course.id]) {
        if(r[Course.id].DateStart > DateStart)
          r[Course.id] = {Course, DateStart};
      } else 
        r[Course.id] = {Course, DateStart};
      return r;
    },{}));
console.log(result);

-1
投票

您可以使用以下解决方案使用map通过属性从数组中获取唯一值:

classes.map( c => ({ DataStart: c.DateStart, Course: c.Course }))
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