Question

我有一个如下所示的数组

const data = [
    {
        name:'AAAA',
        group: 'A',
        age:10
    },
    {
        name:'ABCD',
        group: 'A',
        age:10
    },
    {
        name:'DJSHDJH',
        group: 'B',
        age:15
    },
    {
        name:'FJHF',
        group: 'B',
        age:20
    }
]

我希望group和age es6找到不同的数据

预期结果

[
    {
        group: 'A',
        age: 10
    },
    {
        group: 'B',
        age: 15
    },
    {
        group: 'B',
        age: 20
    }
]

我尝试使用[... new Set(data.map(x => {x.group, g.age}))]，但没有机会

我知道如何遍历数据并查找唯一值，但我想使用es6函数

Answer 1

您可以将Object.values()与reduce()一起使用，以返回值数组。

const data = [{"name":"AAAA","group":"A","age":10},{"name":"ABCD","group":"A","age":10},{"name":"DJSHDJH","group":"B","age":15},{"name":"FJHF","group":"B","age":20}]

var result = Object.values(data.reduce((r, e) => (r[e.group + '|' + e.age] = {group: e.group, age: e.age}, r), {}))
console.log(result)

编写@Jeremy Thille建议的相同代码的另一种方法是使用模板文字和属性值缩写

const data = [{"name":"AAAA","group":"A","age":10},{"name":"ABCD","group":"A","age":10},{"name":"DJSHDJH","group":"B","age":15},{"name":"FJHF","group":"B","age":20}]

var result = Object.values(data.reduce(function(r, e) {
  let group = e.group, age = e.age;
  return r[`${group}|${age}`] = {group, age}, r
}, {}))

console.log(result)

您也可以使用@gyre建议的ES6参数解构。

const data = [{"name":"AAAA","group":"A","age":10},{"name":"ABCD","group":"A","age":10},{"name":"DJSHDJH","group":"B","age":15},{"name":"FJHF","group":"B","age":20}]

var result = Object.values(data.reduce((r, {group, age}) => (r[group+'|'+age] = {group, age}, r), {}))
console.log(result)

Answer 2

使用像Nenad的代码一样的查找表是完成这项工作的正确方法，但是对于各种各样的我来说，通过排序和减少包含一个明显效率较低的解决方案。

＆＃13;

var data   = [{"name":"AAAA","group":"A","age":10},
              {"name":"DJSHDJH","group":"B","age":15},
              {"name":"ABCD","group":"A","age":10},
              {"name":"FJHF","group":"B","age":20},
              {"name":"ABCD","group":"A","age":12},
              {"name":"SOLUS","group":"B","age":15},],
    result = data.sort((a,b) => a.group.localeCompare(b.group) || a.age - b.age)
                 .reduce((r,c,i) => i ? r[r.length-1].group == c.group && r[r.length-1].age == c.age ? r 
                                                                                                     : r.concat({group:c.group, age:c.age})
                                      : [{group:c.group, age:c.age}], []);
console.log(result);

＆＃13;

Answer 3

另一种这样做的方法，可读性更高但更长的方法是：

var filteredData = [];
data.forEach((item)=>{
    if(!filteredData.some(x=>x.group === item.group && x.age === item.age)){
        filteredData.push(item);
    }
});

// then you can change your result array if you want by using map
var result = filteredData.map(x=>{name:x.name,age:x.age});

通过es6在数组中查找不同的对象

3 个答案: