Array.reduce
2021-06-07 本文已影响0人
diviner_杨
reduce() 方法对数组中的每个元素执行一个由您提供的reducer函数(升序执行),将其结果汇总为单个返回值。
reducer 函数接收four个参数:
- Accumulator (acc) (累计器)
- Current Value (cur) (当前值)
- Current Index (idx) (当前索引)
- Source Array (src) (源数组)
arr.reduce(callback(accumulator, currentValue[, index[, array]])[, initialValue])
数组累加等
const res1 = [0, 1, 2, 3, 4].reduce((prev, curr) => prev + curr )
const res2 = [0, 1, 2, 3, 4].reduce((prev, curr, currentIndex, array) => {
return prev+ curr
}, 0)
console.log(res1 === res2 , res1, res2) // true 10 10
数组去重
const arr = [1, 1, 2, 2, 6 , 5 ,66]
const newArr= arr.reduce((pre, cur) => {
if(!pre.includes(cur)){
return pre.concat(cur)
}else{
return pre
}
}, [])
console.log(newArr) // [1, 2, 6, 5, 66]
console.log(newArr.sort()) // [1, 2, 5, 6, 66]
数组计数
const arr = ['星期一', '星期二', '星期三', '星期四', '星期四', '星期五', '星期二', '星期一']
const newObj= arr.reduce((pre, cur ) => {
if(cur in pre){
pre[cur] ++
}else{
pre[cur] = 1
}
return pre
}, {}) //初始传入空对象
console.log(newObj) // {星期一: 2, 星期二: 2, 星期三: 1, 星期四: 2, 星期五: 1}
数组扁平化
const arr = [1, [2,3],[4,[7,8]]]
no.flat
console.log(arr.flat(Infinity)) //[1, 2, 3, 4, 7, 8]
no.正则
const res2 = JSON.stringify(arr).replace(/\[|\]/g, '')
console.log(res2.split(',')) //["1", "2", "3", "4", "7", "8"]
const res3 = JSON.parse('[' + res2 + ']')
console.log(res3) // [1, 2, 3, 4, 7, 8]
no.for
const array = []
const fn = (arr) =>{
for (let i = 0; i < arr.length; i++) {
if(Array.isArray(arr[i])){
fn(arr[i])
}else{
array.push(arr[i])
}
}
}
fn(arr)
console.log(array) //[1, 2, 3, 4, 7, 8]
no.reduce
const newArr = (arr) =>{
return arr.reduce((pre, cur)=>{
return pre.concat(Array.isArray(cur) ? newArr(cur) : cur)
}, [])
}
console.log(newArr(arr)) //[1, 2, 3, 4, 7, 8]
手写一个reduce
//放在数组原型下
Array.prototype.reduceArr = function(fn, initValue){
//判断是否为function
if(Object.prototype.toString.call(fn) !== '[object Function]'){
throw new Error('不是一个function')
}
let arr = this;//当前数组
let initIndex;//初始下标
let acc;//累加器
//判断是否有第二个初始参数
initIdnex = arguments.length === 0 ? 1 : 0;
acc = arguments.length === 0 ? arr[0] : initValue;
//遍历回调并返回出去
for (let i = initIdnex; i < arr.length; i++) {
acc = fn(acc,arr[i],i,arr)
}
return acc;
}
//实验
let arr = [1,2,3]
let arrRes1 = arr.reduceArr((pre, cur) => {
return pre + cur
}, 0)
console.log(arrRes1) // 6
let arrRes2 = arr.reduceArr((pre, cur) => {
return pre + cur
}, 1)
console.log(arrRes2) // 7
let arrRes3 = arr.reduceArr((pre, cur) => {
return pre + cur
}, 10)
console.log(arrRes3) // 16