先易后難,先對最簡單的一維數字數組求交並補
var a=[1,2,3,5,6]; var b=[1,2,6,7,8] // 交集 a.filter(item=>b.includes(item)); // 並集 [...new Set(a.concat(b))] // 補集 a.filter(item=>!b.includes(item));
對象數組也相差無幾,每個對象都有一個唯一的標識的id
var a=[ {id:'001',name:'zhangsan'}, {id:'002',name:'tom'}, {id:'003',name:'jack'}, {id:'004',name:'fei'}, {id:'005',name:'ming'} ]; var b=[ {id:'003',name:'jack'}, {id:'005',name:'ming'}, {id:'007',name:'hello'}, {id:'008',name:'milk'}, ]; var obj={}; var arr=a.concat(b); // 交集:定義一個對象,通過其屬性值是否出現多次判斷交集 arr.reduce(function(pre,cur){ obj.hasOwnProperty(cur.id)?pre.push(cur):obj[cur.id]=true; return pre; },[]); // 並集:每次遍歷將還未出現的項進行收集 arr.reduce(function(pre,cur){ if(!obj.hasOwnProperty(cur.id)){ pre.push(cur); } obj[cur.id]=true; return pre; },[]) //補集:a中每一項都不在b中 let test=a.reduce(function(pre,cur){ if(b.every(item=>item.id!==cur.id)){ pre.push(cur) } return pre; },[])