es6 set取兩個數組的交集

const difference= (a, b) => {
const sA = new Set(a), sB = new Set(b);
return [...a.filter(x => !sB.has(x)), ...b.filter(x => !sA.has(x))];
}
// difference([1,2,3], [1,2,4]) -> [3,4]

 

let arr1 = [
　　{"__row":12,"name":"甲"},

　　{"__row":3,"name":"乙"},

　　{"__row":6,"name":"丙"},

　　{"__row":17,"name":"丁"},

];
let arr2 = [
　　

　　{"__row":12,"name":"張"},

　　{"__row":23,"name":"李"},

　　{"__row":6,"name":"王"},

　　{"__row":57,"name":"陳"},

];
                        
const id_arr1 = new Set(arr1.map(o => o.__row));
const id_arr2 = new Set(arr2.map(o => o.__row));
//id的交集
const allIds = new Set([...id_arr1].filter(x => id_arr2.has(x)))
console.log(allIds)
arr1.forEach(o => {
   //如果有交集
   if (allIds.has(o.__row)) {
        console.log(o)
　　　　//結果                        
   } else {
       
   }
});