1.forEach(),遍历数组的每个元素

 let arrFor = ['muzi','digbig','muzidigbig','lucky']
 //forEach(),遍历数组的每个元素
 arrFor.forEach((item,index) => {
 console.log(`${index}---${item}`)
 })

2.map(参数为回调函数)函数，遍历数组每个元素，并回调操作，需要返回值，返回值组成新数组，原数组不变；

 let arr = [{id:1,name:'muzi'},{id:2,name:'digbig'},{id:3,name:'muzidigbig'},{id:4,name:'lucky'}]
 const arrMap = arr.map((item,index) => {
 return {
 id:item.id,
 name:item.name,
 sex:'男'
 }
 })
 console.log(arrMap)

3.filter(参数为回调函数)函数：过滤通过条件的元素组成一个新数组，原数组不变；

 let arr = [{id:1,name:'muzi'},{id:2,name:'digbig'},{id:3,name:'muzidigbig'},{id:4,name:'lucky'}]
 const arrFilter = arr.filter((item,index) => {
 return item.id >= 2;
 })
 console.log(arrFilter)

4.some(参数为回调函数)函数，遍历数组中是否有符合条件的函数，返回布尔值；

 let arr = [{id:1,name:'muzi'},{id:2,name:'digbig'},{id:3,name:'muzidigbig'},{id:4,name:'lucky'}]
 const arrSome = arr.some((item,index) => {
 return item.id === 5
 })
 console.log(arrSome)

5.every(参数为回调函数)函数，遍历数组是否每个元素都符合条件，返回布尔值；

 let arr = [{id:1,name:'muzi'},{id:2,name:'digbig'},{id:3,name:'muzidigbig'},{id:4,name:'lucky'}]
 const arrEvery = arr.every((item,index) => {
 return item.id >= 1
 })
 console.log(arrEvery)

6.find()函数，数组中的每个元素都执行这个回调函数；返回第一个满足条件的元素 之后的元素就不在调用；没有符合的返回undefined；并没有改变数组的原始值。

 let arr = [{id:1,name:'muzi'},{id:2,name:'digbig'},{id:3,name:'muzidigbig'},{id:4,name:'lucky'}]
 const arrFind = arr.find((item,index) => {
 return item.id === 1
 })
 console.log(arrFind)

7.reduce(),合并二维数组

 var twoArr = [['mu','zi'],['dig','big'],['lucky','jiji']];
 var oneArr = twoArr.reduce(function(total,currentValue){
 // console.log(total)
 return total.concat(currentValue);
 },[])
 console.log(oneArr);//["mu", "zi", "dig", "big", "lucky", "jiji"]

总结