var arr = [{x:1}, {x:2}, {x:3}];
arr.map(function(a) {return a.x;})
.reduce(function(a, b) {return a + b});
console.log(arr);
//I tried using the following code and the result is the data array
//result = [{x:1}, {x:2}, {x:3}];
var arr2 = [{x:1}, {x:2}, {x:3}]
.reduce((total, thing) => total + thing.x, 0);
console.log(arr2);
// and I changed the code to like this and it worked.
// result = 6

只是我对用object literal设置默认值的看法。

让arr = [{ 持续时间:1 },{ 持续时间:3 },{ 持续时间:5 },{ 持续时间:6 }); Const out = arr。Reduce ((a, b) => { 返回{ 持续时间:a.duration + b.duration }; },{ 持续时间:0 })； console.log(出);

我在ES6中做了一点改进:

arr.reduce((a, b) => ({x: a.x + b.x})).x

返回数

我们可以使用数组减少方法来创建新的对象，我们可以使用这个选项来求和或过滤

const FRUITS = ["apple"， "orange"] const fruitBasket ={香蕉:{数量:10,kg:3}，苹果:{数量:30,kg:10}，橙子:{数量:1,kg:3}} const newFruitBasket =水果。Reduce ((acc, fruit) =>({…acc， [fruit]: fruitBasket[fruit]})， {}) console.log (newFruitBasket)

泛型typescript函数:

const sum = <T>(array: T[], predicate: (value: T, index: number, array: T[]) => number) => {
      return array.reduce((acc, value, index, array) => {
        return acc + predicate(value, index, array);
      }, 0);
    };

例子:

const s = sum(arr, (e) => e.x);

//fill创建包含n元素的数组 //reduce需要2个参数，第3个参数作为长度 var fibonacci = (n) =>数组(n).fill()。Reduce ((a, b, c) => { 返回a.concat(c < 2 ?C: a[C - 1] + a[C - 2]) },[]) console.log(斐波纳契(8))