最效率获得俩个数组的相同值

Question

letarr1=['a','b','c','d','e']letarr2=['c','d','g','h','a']找到arr1和arr2中相同的值，并在arr2中删除对应的值怎么写才更简单效率，我只能想到套个双层循环，就比如这样arr1.forEach((val)=>{arr2.forEach((v,i)=>{val===v&&arr2.splice(i,1)})})然后现在问题变了arr2=[{'id':'c'},{'id':'d'},{'id':'g'},{'id':'h'},{'id':'a'}]我的只要改一个地方就能继续运行arr1.forEach((val)=>{arr2.forEach((v,i)=>{val===v.id&&arr2.splice(i,1)})})你们代码需要怎么改才能实现原来的目的，改动最少最简单的，我认为是最合理的就采纳了

Answer 1

letarr1=['a','b','c','d','e']
letarr2=['c','d','g','h','a']
letres=[]
vara=arr2.filter(v=>{
if(arr1.indexOf(v)>=0){
res.push(v)
}
returnarr1.indexOf(v)<0
})
res//["c","d","a"]
a//["g","h"]
                            

Answer 2

array转dict
letarr1=['a','b','c','d','e']
letarr2=['c','d','g','h','a']
letobj=arr1.reduce((iter,val)=>{
iter[val]=true
returniter
},{})
letresult=arr2.filter(v=>!obj[v])
console.log(result)