有没有一种方法可以在 JavaScript 中返回两个数组的差异?
例如:
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
// need ["c", "d"]
有没有一种方法可以在 JavaScript 中返回两个数组的差异?
例如:
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
// need ["c", "d"]
function diff(o, n) {
// deal with empty lists
if (o == undefined) o = [];
if (n == undefined) n = [];
// sort both arrays (or this won't work)
o.sort(); n.sort();
// don't compare if either list is empty
if (o.length == 0 || n.length == 0) return {added: n, removed: o};
// declare temporary variables
var op = 0; var np = 0;
var a = []; var r = [];
// compare arrays and add to add or remove lists
while (op < o.length && np < n.length) {
if (o[op] < n[np]) {
// push to diff?
r.push(o[op]);
op++;
}
else if (o[op] > n[np]) {
// push to diff?
a.push(n[np]);
np++;
}
else {
op++;np++;
}
}
// add remaining items
if( np < n.length )
a = a.concat(n.slice(np, n.length));
if( op < o.length )
r = r.concat(o.slice(op, o.length));
return {added: a, removed: r};
}
var
,让我感觉有点…… - Ian Grainger这是有效的:基本上合并两个数组,寻找重复项,并将未重复的内容推入一个新数组中,这就是差异。
function diff(arr1, arr2) {
var newArr = [];
var arr = arr1.concat(arr2);
for (var i in arr){
var f = arr[i];
var t = 0;
for (j=0; j<arr.length; j++){
if(arr[j] === f){
t++;
}
}
if (t === 1){
newArr.push(f);
}
}
return newArr;
}
还有一个答案,但似乎没有人提到jsperf,它比较了几种算法和技术支持:https://jsperf.com/array-difference-javascript 似乎使用filter得到最佳结果。谢谢。
如果你想找到两个对象数组之间的差异,可以按照以下方式进行:
let arrObj = [{id: 1},{id: 2},{id: 3}]
let arrObj2 = [{id: 1},{id: 3}]
let result = arrObj.filter(x => arrObj2.every(x2 => x2.id !== x.id))
console.log(result)
您可以使用underscore.js: http://underscorejs.org/#intersection
您需要的数组方法有:
_.difference([1, 2, 3, 4, 5], [5, 2, 10]);
=> [1, 3, 4]
_.intersection([1, 2, 3], [101, 2, 1, 10], [2, 1]);
=> [1, 2]
//ES6方式
function diff(a, b) {
var u = a.slice(); //dup the array
b.map(e => {
if (u.indexOf(e) > -1) delete u[u.indexOf(e)]
else u.push(e) //add non existing item to temp array
})
return u.filter((x) => {return (x != null)}) //flatten result
}
使用额外的内存来做这件事。这样你可以以更少的时间复杂度解决它,O(n)而不是o(n*n)。
function getDiff(arr1,arr2){
let k = {};
let diff = []
arr1.map(i=>{
if (!k.hasOwnProperty(i)) {
k[i] = 1
}
}
)
arr2.map(j=>{
if (!k.hasOwnProperty(j)) {
k[j] = 1;
} else {
k[j] = 2;
}
}
)
for (var i in k) {
if (k[i] === 1)
diff.push(+i)
}
return diff
}
getDiff([4, 3, 52, 3, 5, 67, 9, 3],[4, 5, 6, 75, 3, 334, 5, 5, 6])
function diff(arr1, arr2) {
var filteredArr1 = arr1.filter(function(ele) {
return arr2.indexOf(ele) == -1;
});
var filteredArr2 = arr2.filter(function(ele) {
return arr1.indexOf(ele) == -1;
});
return filteredArr1.concat(filteredArr2);
}
diff([1, "calf", 3, "piglet"], [1, "calf", 3, 4]); // Log ["piglet",4]
该函数会比较所有值,并返回一个仅包含不重复数值的数组。
var main = [9, '$', 'x', 'r', 3, 'A', '#', 0, 1];
var arr0 = ['Z', 9, 'e', '$', 'r'];
var arr1 = ['x', 'r', 3, 'A', '#'];
var arr2 = ['m', '#', 'a', 0, 'r'];
var arr3 = ['$', 1, 'n', '!', 'A'];
Array.prototype.diff = function(arrays) {
var items = [].concat.apply(this, arguments);
var diff = [].slice.call(items), i, l, x, pos;
// go through all items
for (x = 0, i = 0, l = items.length; i < l; x = 0, i++) {
// find all positions
while ((pos = diff.indexOf(items[i])) > -1) {
// remove item + increase found count
diff.splice(pos, 1) && x++;
}
// if item was found just once, put it back
if (x === 1) diff.push(items[i]);
}
// get all not duplicated items
return diff;
};
main.diff(arr0, arr1, arr2, arr3).join(''); // returns "Zeman!"
[].diff(main, arr0, arr1, arr2, arr3).join(''); // returns "Zeman!"
diff
作为函数,并且该函数具有与您的函数签名不同的函数签名,则会破坏您的代码或使用此函数的外部库。 - t.niese我在寻找一个简单的答案,不涉及使用不同的库,然后我自己想出了一个方法,我认为这里没有提到过。我不知道它有多高效,但它可以工作;
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,我也需要去除重复项,但我想这并不总是被优先考虑的。
我想主要的缺点就是可能会比较已经被拒绝的许多选项。
O(a1.length x log(a2.length))
- 这种性能在JavaScript中是否可能? - Raul