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"]
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
如果你想找出两个object数组之间的差异,你可以这样做:
let arrObj = [{id: 1},{id: 2},{id: 3}] let arrObj2 = [{id: 1},{id: 3}] let result = arrObj。filter(x => arrObj2.)每个(x2 => x2。Id == x.id) console.log(结果)
其他回答
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;
}
我一直在寻找一个不涉及使用不同库的简单答案,我想出了我自己的答案,我想这里没有提到过。 我不知道它的效率如何,但它确实有效;
function find_diff(arr1, arr2) {
diff = [];
joined = arr1.concat(arr2);
for( i = 0; i <= joined.length; i++ ) {
current = joined[i];
if( joined.indexOf(current) == joined.lastIndexOf(current) ) {
diff.push(current);
}
}
return diff;
}
对于我的代码,我也需要删除副本,但我想这并不总是可取的。
我想主要的缺点是它可能会比较许多已经被拒绝的选择。
如果你有两个对象列表
const people = [{name: 'cesar', age: 23}]
const morePeople = [{name: 'cesar', age: 23}, {name: 'kevin', age: 26}, {name: 'pedro', age: 25}]
let result2 = morePeople.filter(person => people.every(person2 => !person2.name.includes(person.name)))
数据:
var new_storage = JSON.parse('[{"id_order":"0003"},{"id_order":"0004"},{"id_order":"0006"}]');
var old_storage = JSON.parse('[{"id_order":"0001"},{"id_order":"0002"},{"id_order":"0003"},{"id_order":"0004"},{"id_order":"0005"}]');
使用过滤器:
var diff = new_storage
.filter(x => {if(!(old_storage.filter(y => y.id_order==x.id_order)).length){return x}})
.concat(old_storage
.filter(x => {if(!(new_storage.filter(y => y.id_order==x.id_order)).length){return x}})
)
console.log(JSON.stringify(diff))
两个数组的结果不同
[{"id_order":"0006"},{"id_order":"0001"},{"id_order":"0002"},{"id_order":"0005"}]
function diffArray(arr1, arr2) {
const newArr = [];
// arr1 match to arr2
arr1.map((item)=>{
if(arr2.indexOf(item)<0){
console.log(item)
newArr.push(item)
}
})
// arr2 match to arr1
arr2.map((item)=>{
if(arr1.indexOf(item)<0){
newArr.push(item)
}
})
return newArr;
}
差异([1,2,3,5],[1,2,3,4,5])
输出::[4]