是否有一种方法可以在JavaScript中返回两个数组之间的差异?
例如:
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
// need ["c", "d"]
当前回答
这个怎么样:
Array.prototype.contains = function(needle){
for (var i=0; i<this.length; i++)
if (this[i] == needle) return true;
return false;
}
Array.prototype.diff = function(compare) {
return this.filter(function(elem) {return !compare.contains(elem);})
}
var a = new Array(1,4,7, 9);
var b = new Array(4, 8, 7);
alert(a.diff(b));
这样你就可以用array1。diff(array2)来得到它们的区别算法的时间复杂度很可怕- O(array1。长度(我相信)
其他回答
const dbData = [{name:'ally'},
{name:'James'}]
const localData = [{name:'James'}]
const diff = dbData.filter(a =>!localData.some(b => { return a.name === b.name}))
const difference = function (baseArray, arrayToCampare, callback = (a, b) => a!== b) {
if (!(arrayToCampare instanceof Array)) {
return baseArray;
}
return baseArray.filter(baseEl =>
arrayToCampare.every(compareEl => callback(baseEl, compareEl)));
}
function diffArray(newArr, oldArr) {
var newSet = new Set(newArr)
var diff = []
oldArr.forEach((a) => {
if(!newSet.delete(a))diff.push(a)
})
return diff.concat(Array.from(newSet))
}
function array_diff(a, b) {
let array = [];
for(let i = 0; i <a.length; i++) {
let k = 0;
for( let j = 0; j < b.length; j++) {
if(a[i]!==b[j]) {
k++;
}
if(k===b.length) {
array = array.concat(a[i]);
}
}
if(b.length ===0) {
array = array.concat(a[i]);
}
}
return array;
}
这个怎么样:
Array.prototype.contains = function(needle){
for (var i=0; i<this.length; i++)
if (this[i] == needle) return true;
return false;
}
Array.prototype.diff = function(compare) {
return this.filter(function(elem) {return !compare.contains(elem);})
}
var a = new Array(1,4,7, 9);
var b = new Array(4, 8, 7);
alert(a.diff(b));
这样你就可以用array1。diff(array2)来得到它们的区别算法的时间复杂度很可怕- O(array1。长度(我相信)
