用javascript实现数组交叉的最简单、无库代码是什么?我想写
intersection([1,2,3], [2,3,4,5])
并获得
[2, 3]
当前回答
基于Anon的出色回答，这个函数返回两个或多个数组的交集。
function arrayIntersect(arrayOfArrays)
{
var arrayCopy = arrayOfArrays.slice(),
baseArray = arrayCopy.pop();
return baseArray.filter(function(item) {
return arrayCopy.every(function(itemList) {
return itemList.indexOf(item) !== -1;
});
});
}
其他回答
使用Array.prototype.filter和Array.prototype.includes的组合:
const filteredArray = array1.filter(value => array2.includes(value));
对于较旧的浏览器，使用Array.prototype.indexOf且不使用箭头函数:
var filteredArray = array1.filter(function(n) {
return array2.indexOf(n) !== -1;
});
NB !.includes和. indexof都在内部使用===来比较数组中的元素，所以如果数组包含对象，它只比较对象引用(而不是对象的内容)。如果你想指定自己的比较逻辑，请使用Array.prototype.some。
希望这有助于所有版本。
function diffArray(arr1, arr2) {
var newArr = [];
var large = arr1.length>=arr2.length?arr1:arr2;
var small = JSON.stringify(large) == JSON.stringify(arr1)?arr2:arr1;
for(var i=0;i<large.length;i++){
var copyExists = false;
for(var j =0;j<small.length;j++){
if(large[i]==small[j]){
copyExists= true;
break;
}
}
if(!copyExists)
{
newArr.push(large[i]);
}
}
for(var i=0;i<small.length;i++){
var copyExists = false;
for(var j =0;j<large.length;j++){
if(large[j]==small[i]){
copyExists= true;
break;
}
}
if(!copyExists)
{
newArr.push(small[i]);
}
}
return newArr;
}
下面是一个使用可选的比较函数处理多个数组的简单实现:
函数交叉(数组，compareFn = (val1, val2) => (val1 == val2)) { 如果数组。长度< 2)返回数组[0]?［］ Const array1 = arrays[0] const array2 =交集(arrays.slice(1)， compareFn) array1返回。过滤器(val1 =>数组2。if (val2 => compareFn(val1, val2))) ｝ console.log(十字路口([[1,2,3],[2、3、4、5]])) console.log(十字路口([[{id: 1}, {id: 2}], [{id: 1}, {id: 3}]], (val1, val2) => val1。Id === val2.id)
该函数利用字典的强大功能，避免了N^2问题。每个数组只循环一次，第三次更短的循环返回最终结果。 它还支持数字、字符串和对象。
function array_intersect(array1, array2)
{
var mergedElems = {},
result = [];
// Returns a unique reference string for the type and value of the element
function generateStrKey(elem) {
var typeOfElem = typeof elem;
if (typeOfElem === 'object') {
typeOfElem += Object.prototype.toString.call(elem);
}
return [typeOfElem, elem.toString(), JSON.stringify(elem)].join('__');
}
array1.forEach(function(elem) {
var key = generateStrKey(elem);
if (!(key in mergedElems)) {
mergedElems[key] = {elem: elem, inArray2: false};
}
});
array2.forEach(function(elem) {
var key = generateStrKey(elem);
if (key in mergedElems) {
mergedElems[key].inArray2 = true;
}
});
Object.values(mergedElems).forEach(function(elem) {
if (elem.inArray2) {
result.push(elem.elem);
}
});
return result;
}
如果有无法解决的特殊情况，仅通过修改generateStrKey函数就可以解决。这个函数的诀窍在于，它根据类型和值唯一地表示每个不同的数据。
这个变体有一些性能改进。在任何数组为空的情况下避免循环。它还首先遍历较短的数组，因此如果它在第二个数组中找到了第一个数组的所有值，则退出循环。
function array_intersect(array1, array2)
{
var mergedElems = {},
result = [],
firstArray, secondArray,
firstN = 0,
secondN = 0;
function generateStrKey(elem) {
var typeOfElem = typeof elem;
if (typeOfElem === 'object') {
typeOfElem += Object.prototype.toString.call(elem);
}
return [typeOfElem, elem.toString(), JSON.stringify(elem)].join('__');
}
// Executes the loops only if both arrays have values
if (array1.length && array2.length)
{
// Begins with the shortest array to optimize the algorithm
if (array1.length < array2.length) {
firstArray = array1;
secondArray = array2;
} else {
firstArray = array2;
secondArray = array1;
}
firstArray.forEach(function(elem) {
var key = generateStrKey(elem);
if (!(key in mergedElems)) {
mergedElems[key] = {elem: elem, inArray2: false};
// Increases the counter of unique values in the first array
firstN++;
}
});
secondArray.some(function(elem) {
var key = generateStrKey(elem);
if (key in mergedElems) {
if (!mergedElems[key].inArray2) {
mergedElems[key].inArray2 = true;
// Increases the counter of matches
secondN++;
// If all elements of first array have coincidence, then exits the loop
return (secondN === firstN);
}
}
});
Object.values(mergedElems).forEach(function(elem) {
if (elem.inArray2) {
result.push(elem.elem);
}
});
}
return result;
}
ES2015的函数式方法
函数式方法必须考虑只使用没有副作用的纯函数，每个函数只与单个作业有关。
这些限制增强了所涉及函数的可组合性和可重用性。
//小的，可重用的辅助函数 const createSet = xs => new Set(xs); Const filter = f => xs => xs.filter(apply(f)); Const apply = f => x => f(x); / /十字路口 Const相交= xs => ys => { const zs =创建集(ys); 返回过滤器(x => zs.has(x)) ？ 真正的 :假 ) (x); }; //模拟数据 Const xs = [1,2,2,3,4,5]; Const ys = [0,1,2,3,3,3,6,7,8,9]; //运行 Console.log (intersect(xs) (ys));
请注意，使用本机Set类型，这有一个优点 查找性能。
避免重复
显然，第一个数组中重复出现的项将被保留，而第二个数组将被去重。这可能是也可能不是理想的行为。如果你需要一个唯一的结果，只需对第一个参数应用重复数据删除:
// auxiliary functions const apply = f => x => f(x); const comp = f => g => x => f(g(x)); const afrom = apply(Array.from); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); // intersection const intersect = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? true : false ) (xs); }; // de-duplication const dedupe = comp(afrom) (createSet); // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; // unique result console.log( intersect(dedupe(xs)) (ys) );
计算任意数量数组的交集
如果你想计算任意数量的数组的交点，只需用compose intersect和foldl。这是一个方便函数:
// auxiliary functions const apply = f => x => f(x); const uncurry = f => (x, y) => f(x) (y); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); const foldl = f => acc => xs => xs.reduce(uncurry(f), acc); // intersection const intersect = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? true : false ) (xs); }; // intersection of an arbitrarily number of Arrays const intersectn = (head, ...tail) => foldl(intersect) (head) (tail); // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; const zs = [0,1,2,3,4,5,6]; // run console.log( intersectn(xs, ys, zs) );
