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
试一试。
var first = [ 1, 2, 3, 4, 5 ];
var second = [ 4, 5, 6 ];
var difference = first.filter(x => second.indexOf(x) === -1);
console.log(difference);
Output: [ 1, 2, 3]
Var first = [1,2,3,4,5]; Var second = [4,5,6]; Var差=第一。filter(x => second.indexOf(x) === -1); console.log(差异);
其他回答
下面是一个稍微修改的版本,它使用Object来存储哈希,可以处理数组中的数字和字符串。
function arrDiff(a, b) {
const hash = {};
a.forEach(n => { hash[n] = n; });
b.forEach(n => {
if (hash[n]) {
delete hash[n];
} else {
hash[n] = n;
}
});
return Object.values(hash);
}
您可以使用一个公共对象并计算第一个数组中每个值的频率。对于第二个数组,减少公共对象中的值。然后遍历所有键并添加所有值大于1的键。
常量差值= (a1, a2) => { Var obj = {}; a1。forEach(obj[v] = (obj[v] || 0) + 1); a2。forEach(v => obj[v] = (obj[v] || 0) - 1); 返回对象 . keys (obj) .reduce((r,k) => { If (obj[k] > 0) r = r.concat (Array.from({长度:obj [k]}) .fill (k)); 返回r; }, []); }; const =结果不同([' a ', ' ', ' b ', ' c ', ' d '], [a, b]); console.log(结果);
下面是另一个可以返回差异的解决方案,就像git diff一样:(它已经用typescript编写,如果你不使用typescript版本,只需删除类型)
/**
* util function to calculate the difference between two arrays (pay attention to 'from' and 'to'),
* it would return the mutations from 'from' to 'to'
* @param { T[] } from
* @param { T[] } to
* @returns { { [x in string]: boolean } } it would return the stringified version of array element, true means added,
* false means removed
*/
export function arrDiff<T>(from: T[], to: T[]): { [x in string]: boolean } {
var diff: { [x in string]: boolean } = {};
var newItems: T[] = []
diff = from.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
for (var i = 0; i < to.length; i++) {
if (diff[JSON.stringify(to[i])]) {
delete diff[JSON.stringify(to[i])]
} else {
newItems.push(to[i])
}
}
return {
...Object.keys(diff).reduce((a, e) => ({ ...a, [e]: false }), {}),
...newItems.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
}
}
下面是一个用法示例:
arrDiff(['a', 'b', 'c'], ['a', 'd', 'c', 'f']) //{"b": false, "d": true, "f": true}
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(', ') );
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)))