js數組交集、并集、差集

數組交集

1. 利用filter + indexOf

var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN], arr2 = [1, 3, 5, NaN];
var intersection = arr1.filter(function (val) {
    return arr2.indexOf(val) > -1;
})
console.log(intersection);   // [1, 3, 5]
           

但這種方法針對是value值比較簡單的Number(不包括NaN)、String、Boolean類型， 因為indexOf 本質上還是使用 === 來判斷。

2. 利用ES6中 Array.from方法及Set

function intersect (a, b) {
    var bSet = new Set(b);
    return Array.from(a.filter(function (val) {
        return bSet.has(val);
    }));
}
var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN], arr2 = [1, 3, 5, NaN];
var intersection = intersect (arr1, arr2);
console.log(intersection);   // [1, 3, 5, NaN]
           

 可以解決NaN

3. 利用ES7中數組新增的includes + filter

var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN], arr2 = [1, 3, 5, NaN];
var intersection = arr1.filter(function (val) {
    return arr2.includes(val);
})
console.log(intersection);   // [1, 3, 5, NaN]
           

 數組并集

1. 利用filter + indexOf

var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN], arr2 = [1, 3, 5, NaN];
var union= arr1.concat(arr2.filter(function (val) {
    return arr1.indexOf(val) === -1;
}))
console.log(union);  // [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN, NaN]
           

2. 利用ES6中 Array.from方法及Set

var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN], arr2 = [1, 3, 5, NaN];
var union= Array.from(new Set(arr1.concat(arr2)));
console.log(union);   // [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN]
           

3. 利用ES7中新增的includes + filter

var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN], arr2 = [1, 3, 5, NaN];
var union= arr1.concat(arr2.filter(function (val) {
    return !arr1.includes(val)
}));
console.log(union);   // [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN]
           

 數組差集

1. 利用filter + indexOf

var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN], arr2 = [1, 3, 5, NaN];
var difference = arr1.filter(function (val) {
    return arr2.indexOf(val) === -1;
}).concat(arr2.filter(function (val) { 
    return arr1.indexOf(val) === -1 
}));
console.log(difference);  //  [2, 4, 6, 7, 8, 9, NaN, NaN]
           

2. 利用ES6中 Array.from方法及Set

function difference (a, b) {
    var aSet = new Set(a), bSet = new Set(b);
    return Array.from(new Set(a.concat(b).filter(function (val) {
        return !aSet.has(val) || !bSet.has(val)
    })))
}
var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN], arr2 = [1, 3, 5, NaN];
var differenceVal = difference (arr1, arr2);
console.log(differenceVal );  //  [2, 4, 6, 7, 8, 9]
           

3. 利用ES7中新增的includes + filter

var arr1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, NaN], arr2 = [1, 3, 5, NaN];
var difference = arr1.concat(arr2).filter(function (val) {
    return  !a.includes(val) || !b.includes(val)
});
console.log(difference);  // [2, 4, 6, 7, 8, 9]