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
/ / es6方法
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
}
其他回答
类似于Ian Grainger的解决方案(但在打字稿中):
function findDiffs(arrayOne: string[], arrayTwo: string[]) {
let onlyInArrayOne = []
let onlyInArrayTwo = []
let share = []
let [arrayOneCopy, arrayTwoCopy] = [[...arrayOne], [...arrayTwo]]
arrayOneCopy.sort(); arrayTwoCopy.sort()
while (arrayOneCopy.length !== 0 && arrayTwoCopy.length !== 0) {
if (arrayOneCopy[0] == arrayTwoCopy[0]) {
share.push(arrayOneCopy[0])
arrayOneCopy.splice(0, 1)
arrayTwoCopy.splice(0, 1)
}
if (arrayOneCopy[0] < arrayTwoCopy[0]) {
onlyInArrayOne.push(arrayOneCopy[0])
arrayOneCopy.splice(0, 1)
}
if (arrayOneCopy[0] > arrayTwoCopy[0]) {
onlyInArrayTwo.push(arrayTwoCopy[0])
arrayTwoCopy.splice(0, 1)
}
}
onlyInArrayTwo = onlyInArrayTwo.concat(arrayTwoCopy)
onlyInArrayOne = onlyInArrayOne.concat(arrayOneCopy)
return {
onlyInArrayOne,
onlyInArrayTwo,
share,
diff: onlyInArrayOne.concat(onlyInArrayTwo)
}
}
// arrayOne: [ 'a', 'b', 'c', 'm', 'y' ]
// arrayTwo: [ 'c', 'b', 'f', 'h' ]
//
// Results:
// {
// onlyInArrayOne: [ 'a', 'm', 'y' ],
// onlyInArrayTwo: [ 'f', 'h' ],
// share: [ 'b', 'c' ],
// diff: [ 'a', 'm', 'y', 'f', 'h' ]
// }
对于想要从一个数组中减去另一个数组的人的回应…
如果不超过1000个元素试试这个…
设置一个新变量来复制Array01，并将其命名为Array03。
现在，使用冒泡排序算法比较Array01和Array02的元素，并在找到匹配时进行比较 对Array03执行以下操作…
if (Array01[x]==Array02[y]) {Array03.splice(x,1);}
注意:我们正在修改Array03而不是Array01，这样就不会搞砸冒泡排序的嵌套循环!
最后，用一个简单的赋值操作将Array03的内容复制到Array01，就完成了。
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。
纯JavaScript解决方案(没有库) 与旧浏览器兼容(不使用过滤器) O (n ^ 2) 可选的fn回调参数，用于指定如何比较数组项
function diff(a, b, fn){ var max = Math.max(a.length, b.length); d = []; fn = typeof fn === 'function' ? fn : false for(var i=0; i < max; i++){ var ac = i < a.length ? a[i] : undefined bc = i < b.length ? b[i] : undefined; for(var k=0; k < max; k++){ ac = ac === undefined || (k < b.length && (fn ? fn(ac, b[k]) : ac == b[k])) ? undefined : ac; bc = bc === undefined || (k < a.length && (fn ? fn(bc, a[k]) : bc == a[k])) ? undefined : bc; if(ac == undefined && bc == undefined) break; } ac !== undefined && d.push(ac); bc !== undefined && d.push(bc); } return d; } alert( "Test 1: " + diff( [1, 2, 3, 4], [1, 4, 5, 6, 7] ).join(', ') + "\nTest 2: " + diff( [{id:'a',toString:function(){return this.id}},{id:'b',toString:function(){return this.id}},{id:'c',toString:function(){return this.id}},{id:'d',toString:function(){return this.id}}], [{id:'a',toString:function(){return this.id}},{id:'e',toString:function(){return this.id}},{id:'f',toString:function(){return this.id}},{id:'d',toString:function(){return this.id}}], function(a, b){ return a.id == b.id; } ).join(', ') );
Array.prototype.difference = function(e) {
return this.filter(function(i) {return e.indexOf(i) < 0;});
};
eg:-
[1,2,3,4,5,6,7].difference( [3,4,5] );
=> [1, 2, 6 , 7]
