在 javascript 中实现数组交集的最简单、无库的代码是什么？我想写
intersection([1,2,3], [2,3,4,5])
并得到
[2, 3]
问问题
601804 次
39 回答
1595
使用Array.prototype.filter和的组合Array.prototype.includes：
const filteredArray = array1.filter(value => array2.includes(value));
对于较旧的浏览器，带有Array.prototype.indexOf和不带有箭头功能：
var filteredArray = array1.filter(function(n) {
return array2.indexOf(n) !== -1;
});
注意！两者.includes和.indexOf内部都使用 比较数组中的元素===，因此如果数组包含对象，它将仅比较对象引用（而不是它们的内容）。如果要指定自己的比较逻辑，请Array.prototype.some改用。
于 2009-12-11T03:08:37.137 回答
160
Destructive 似乎最简单，尤其是如果我们可以假设输入已排序：
/* destructively finds the intersection of
* two arrays in a simple fashion.
*
* PARAMS
* a - first array, must already be sorted
* b - second array, must already be sorted
*
* NOTES
* State of input arrays is undefined when
* the function returns. They should be
* (prolly) be dumped.
*
* Should have O(n) operations, where n is
* n = MIN(a.length, b.length)
*/
function intersection_destructive(a, b)
{
var result = [];
while( a.length > 0 && b.length > 0 )
{
if (a[0] < b[0] ){ a.shift(); }
else if (a[0] > b[0] ){ b.shift(); }
else /* they're equal */
{
result.push(a.shift());
b.shift();
}
}
return result;
}
非破坏性必须是更复杂的头发，因为我们必须跟踪索引：
/* finds the intersection of
* two arrays in a simple fashion.
*
* PARAMS
* a - first array, must already be sorted
* b - second array, must already be sorted
*
* NOTES
*
* Should have O(n) operations, where n is
* n = MIN(a.length(), b.length())
*/
function intersect_safe(a, b)
{
var ai=0, bi=0;
var result = [];
while( ai < a.length && bi < b.length )
{
if (a[ai] < b[bi] ){ ai++; }
else if (a[ai] > b[bi] ){ bi++; }
else /* they're equal */
{
result.push(a[ai]);
ai++;
bi++;
}
}
return result;
}
于 2009-12-11T03:45:00.573 回答
103
如果您的环境支持ECMAScript 6 Set，一种简单且据称有效（参见规范链接）的方式：
function intersect(a, b) {
var setA = new Set(a);
var setB = new Set(b);
var intersection = new Set([...setA].filter(x => setB.has(x)));
return Array.from(intersection);
}
更短，但可读性更低（也没有创建额外的交集Set）：
function intersect(a, b) {
var setB = new Set(b);
return [...new Set(a)].filter(x => setB.has(x));
}
请注意，使用集合时，您只会得到不同的值，因此new Set([1, 2, 3, 3]).size计算结果为3.
于 2016-05-05T03:21:26.750 回答
71
使用 Underscore.js或lodash.js
_.intersection( [0,345,324] , [1,0,324] ) // gives [0,324]
于 2016-09-29T07:51:36.140 回答
40
// Return elements of array a that are also in b in linear time:
function intersect(a, b) {
return a.filter(Set.prototype.has, new Set(b));
}
// Example:
console.log(intersect([1,2,3], [2,3,4,5]));我推荐上面简洁的解决方案，它在大输入上优于其他实现。如果小输入的性能很重要，请检查以下替代方案。
替代方案和性能比较：
有关替代实现，请参阅以下代码段，并检查https://jsperf.com/array-intersection-comparison以进行性能比较。
function intersect_for(a, b) {
const result = [];
const alen = a.length;
const blen = b.length;
for (let i = 0; i < alen; ++i) {
const ai = a[i];
for (let j = 0; j < blen; ++j) {
if (ai === b[j]) {
result.push(ai);
break;
}
}
}
return result;
}
function intersect_filter_indexOf(a, b) {
return a.filter(el => b.indexOf(el) !== -1);
}
function intersect_filter_in(a, b) {
const map = b.reduce((map, el) => {map[el] = true; return map}, {});
return a.filter(el => el in map);
}
function intersect_for_in(a, b) {
const result = [];
const map = {};
for (let i = 0, length = b.length; i < length; ++i) {
map[b[i]] = true;
}
for (let i = 0, length = a.length; i < length; ++i) {
if (a[i] in map) result.push(a[i]);
}
return result;
}
function intersect_filter_includes(a, b) {
return a.filter(el => b.includes(el));
}
function intersect_filter_has_this(a, b) {
return a.filter(Set.prototype.has, new Set(b));
}
function intersect_filter_has_arrow(a, b) {
const set = new Set(b);
return a.filter(el => set.has(el));
}
function intersect_for_has(a, b) {
const result = [];
const set = new Set(b);
for (let i = 0, length = a.length; i < length; ++i) {
if (set.has(a[i])) result.push(a[i]);
}
return result;
}Firefox 53 中的结果：
大型数组（10,000 个元素）上的 Ops/sec：
filter + has (this) 523 (this answer) for + has 482 for-loop + in 279 filter + in 242 for-loops 24 filter + includes 14 filter + indexOf 10小数组（100 个元素）上的 Ops/sec：
for-loop + in 384,426 filter + in 192,066 for-loops 159,137 filter + includes 104,068 filter + indexOf 71,598 filter + has (this) 43,531 (this answer) filter + has (arrow function) 35,588
于 2017-05-06T12:34:30.563 回答
18
我在 ES6 方面的贡献。一般来说，它会找到一个数组与作为参数提供的不定数量数组的交集。
Array.prototype.intersect = function(...a) {
return [this,...a].reduce((p,c) => p.filter(e => c.includes(e)));
}
var arrs = [[0,2,4,6,8],[4,5,6,7],[4,6]],
arr = [0,1,2,3,4,5,6,7,8,9];
document.write("<pre>" + JSON.stringify(arr.intersect(...arrs)) + "</pre>");于 2016-05-15T12:23:13.920 回答
12
只使用关联数组怎么样？
function intersect(a, b) {
var d1 = {};
var d2 = {};
var results = [];
for (var i = 0; i < a.length; i++) {
d1[a[i]] = true;
}
for (var j = 0; j < b.length; j++) {
d2[b[j]] = true;
}
for (var k in d1) {
if (d2[k])
results.push(k);
}
return results;
}
编辑：
// new version
function intersect(a, b) {
var d = {};
var results = [];
for (var i = 0; i < b.length; i++) {
d[b[i]] = true;
}
for (var j = 0; j < a.length; j++) {
if (d[a[j]])
results.push(a[j]);
}
return results;
}
于 2009-12-11T04:22:33.570 回答
10
通过使用 .pop 而不是 .shift，可以提高 @atk 对已排序基元数组的实现的性能。
function intersect(array1, array2) {
var result = [];
// Don't destroy the original arrays
var a = array1.slice(0);
var b = array2.slice(0);
var aLast = a.length - 1;
var bLast = b.length - 1;
while (aLast >= 0 && bLast >= 0) {
if (a[aLast] > b[bLast] ) {
a.pop();
aLast--;
} else if (a[aLast] < b[bLast] ){
b.pop();
bLast--;
} else /* they're equal */ {
result.push(a.pop());
b.pop();
aLast--;
bLast--;
}
}
return result;
}
我使用 jsPerf 创建了一个基准测试：http: //bit.ly/P9FrZK。使用 .pop 大约快三倍。
于 2012-07-19T19:33:22.907 回答
9
- 把它分类
- 从索引 0 开始一一检查，从中创建新数组。
像这样的东西，虽然没有很好地测试。
function intersection(x,y){
x.sort();y.sort();
var i=j=0;ret=[];
while(i<x.length && j<y.length){
if(x[i]<y[j])i++;
else if(y[j]<x[i])j++;
else {
ret.push(x[i]);
i++,j++;
}
}
return ret;
}
alert(intersection([1,2,3], [2,3,4,5]));
PS：该算法仅适用于数字和普通字符串，任意对象数组的交集可能不起作用。
于 2009-12-11T03:15:42.363 回答
9
使用jQuery：
var a = [1,2,3];
var b = [2,3,4,5];
var c = $(b).not($(b).not(a));
alert(c);
于 2013-12-25T07:20:49.147 回答
9
如果您需要让它处理相交的多个数组：
const intersect = (a1, a2, ...rest) => {
const a12 = a1.filter(value => a2.includes(value))
if (rest.length === 0) { return a12; }
return intersect(a12, ...rest);
};
console.log(intersect([1,2,3,4,5], [1,2], [1, 2, 3,4,5], [2, 10, 1])) 于 2018-03-09T04:41:38.137 回答
6
对于仅包含字符串或数字的数组，您可以根据其他一些答案进行排序。对于任意对象数组的一般情况，我认为您无法避免长期这样做。以下将为您提供作为参数提供的任意数量的数组的交集arrayIntersection：
var arrayContains = Array.prototype.indexOf ?
function(arr, val) {
return arr.indexOf(val) > -1;
} :
function(arr, val) {
var i = arr.length;
while (i--) {
if (arr[i] === val) {
return true;
}
}
return false;
};
function arrayIntersection() {
var val, arrayCount, firstArray, i, j, intersection = [], missing;
var arrays = Array.prototype.slice.call(arguments); // Convert arguments into a real array
// Search for common values
firstArray = arrays.pop();
if (firstArray) {
j = firstArray.length;
arrayCount = arrays.length;
while (j--) {
val = firstArray[j];
missing = false;
// Check val is present in each remaining array
i = arrayCount;
while (!missing && i--) {
if ( !arrayContains(arrays[i], val) ) {
missing = true;
}
}
if (!missing) {
intersection.push(val);
}
}
}
return intersection;
}
arrayIntersection( [1, 2, 3, "a"], [1, "a", 2], ["a", 1] ); // Gives [1, "a"];
于 2009-12-11T11:37:28.023 回答
5
另一种能够一次处理任意数量的数组的索引方法：
// Calculate intersection of multiple array or object values.
function intersect (arrList) {
var arrLength = Object.keys(arrList).length;
// (Also accepts regular objects as input)
var index = {};
for (var i in arrList) {
for (var j in arrList[i]) {
var v = arrList[i][j];
if (index[v] === undefined) index[v] = 0;
index[v]++;
};
};
var retv = [];
for (var i in index) {
if (index[i] == arrLength) retv.push(i);
};
return retv;
};
它仅适用于可以作为字符串评估的值，您应该将它们作为数组传递，例如：
intersect ([arr1, arr2, arr3...]);
...但它透明地接受对象作为参数或任何要相交的元素（总是返回公共值的数组）。例子：
intersect ({foo: [1, 2, 3, 4], bar: {a: 2, j:4}}); // [2, 4]
intersect ([{x: "hello", y: "world"}, ["hello", "user"]]); // ["hello"]
编辑：我只是注意到这在某种程度上有点错误。
那就是：我编码它认为输入数组本身不能包含重复（因为提供的示例没有）。
但是如果输入数组碰巧包含重复，那会产生错误的结果。示例（使用以下实现）：
intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]);
// Expected: [ '1' ]
// Actual: [ '1', '3' ]
幸运的是，这很容易通过简单地添加二级索引来解决。那是：
改变：
if (index[v] === undefined) index[v] = 0;
index[v]++;
经过：
if (index[v] === undefined) index[v] = {};
index[v][i] = true; // Mark as present in i input.
...和：
if (index[i] == arrLength) retv.push(i);
经过：
if (Object.keys(index[i]).length == arrLength) retv.push(i);
完整示例：
// Calculate intersection of multiple array or object values.
function intersect (arrList) {
var arrLength = Object.keys(arrList).length;
// (Also accepts regular objects as input)
var index = {};
for (var i in arrList) {
for (var j in arrList[i]) {
var v = arrList[i][j];
if (index[v] === undefined) index[v] = {};
index[v][i] = true; // Mark as present in i input.
};
};
var retv = [];
for (var i in index) {
if (Object.keys(index[i]).length == arrLength) retv.push(i);
};
return retv;
};
intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]); // [ '1' ]
于 2015-03-04T12:51:24.087 回答
5
对此处最小的一个（filter/indexOf 解决方案）进行微小的调整，即使用 JavaScript 对象在其中一个数组中创建值的索引，将其从 O(N*M) 减少到“可能”的线性时间。来源1 来源 2
function intersect(a, b) {
var aa = {};
a.forEach(function(v) { aa[v]=1; });
return b.filter(function(v) { return v in aa; });
}
这不是最简单的解决方案（它比filter+indexOf更多的代码），也不是最快的（可能比intersect_safe()慢一个常数因子），但似乎是一个很好的平衡。它非常简单，同时提供了良好的性能，并且不需要预先排序的输入。
于 2015-11-26T18:46:08.377 回答
4
由于对数据有一些限制，您可以在线性时间内完成！
对于正整数：使用将值映射到“已见/未见”布尔值的数组。
function intersectIntegers(array1,array2) {
var seen=[],
result=[];
for (var i = 0; i < array1.length; i++) {
seen[array1[i]] = true;
}
for (var i = 0; i < array2.length; i++) {
if ( seen[array2[i]])
result.push(array2[i]);
}
return result;
}
对象也有类似的技术：获取一个虚拟键，为 array1 中的每个元素将其设置为“true”，然后在 array2 的元素中查找此键。完成后清理。
function intersectObjects(array1,array2) {
var result=[];
var key="tmpKey_intersect"
for (var i = 0; i < array1.length; i++) {
array1[i][key] = true;
}
for (var i = 0; i < array2.length; i++) {
if (array2[i][key])
result.push(array2[i]);
}
for (var i = 0; i < array1.length; i++) {
delete array1[i][key];
}
return result;
}
当然，您需要确保密钥之前没有出现，否则您将破坏您的数据......
于 2015-07-16T16:13:36.903 回答
3
function intersection(A,B){
var result = new Array();
for (i=0; i<A.length; i++) {
for (j=0; j<B.length; j++) {
if (A[i] == B[j] && $.inArray(A[i],result) == -1) {
result.push(A[i]);
}
}
}
return result;
}
于 2012-09-19T21:13:32.387 回答
3
为简单起见：
// Usage
const intersection = allLists
.reduce(intersect, allValues)
.reduce(removeDuplicates, []);
// Implementation
const intersect = (intersection, list) =>
intersection.filter(item =>
list.some(x => x === item));
const removeDuplicates = (uniques, item) =>
uniques.includes(item) ? uniques : uniques.concat(item);
// Example Data
const somePeople = [bob, doug, jill];
const otherPeople = [sarah, bob, jill];
const morePeople = [jack, jill];
const allPeople = [...somePeople, ...otherPeople, ...morePeople];
const allGroups = [somePeople, otherPeople, morePeople];
// Example Usage
const intersection = allGroups
.reduce(intersect, allPeople)
.reduce(removeDuplicates, []);
intersection; // [jill]
好处：
- 简单的污垢
- 以数据为中心
- 适用于任意数量的列表
- 适用于任意长度的列表
- 适用于任意类型的值
- 适用于任意排序顺序
- 保持形状（在任何数组中首次出现的顺序）
- 尽可能早退
- 内存安全，不会篡改函数/数组原型
缺点：
- 更高的内存使用率
- 更高的 CPU 使用率
- 需要了解reduce
- 需要了解数据流
您不希望将它用于 3D 引擎或内核工作，但如果您无法在基于事件的应用程序中运行它，那么您的设计就有更大的问题。
于 2017-02-02T20:22:16.980 回答
2
我将贡献对我最有效的东西：
if (!Array.prototype.intersect){
Array.prototype.intersect = function (arr1) {
var r = [], o = {}, l = this.length, i, v;
for (i = 0; i < l; i++) {
o[this[i]] = true;
}
l = arr1.length;
for (i = 0; i < l; i++) {
v = arr1[i];
if (v in o) {
r.push(v);
}
}
return r;
};
}
于 2013-05-18T13:36:47.067 回答
2
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，每个函数只涉及一个工作。
这些限制增强了所涉及功能的可组合性和可重用性。
// small, reusable auxiliary functions
const createSet = xs => new Set(xs);
const filter = f => xs => xs.filter(apply(f));
const apply = f => x => f(x);
// intersection
const intersect = xs => ys => {
const zs = createSet(ys);
return filter(x => zs.has(x)
? true
: false
) (xs);
};
// mock data
const xs = [1,2,2,3,4,5];
const ys = [0,1,2,3,3,3,6,7,8,9];
// run it
console.log( intersect(xs) (ys) );请注意，使用的Set是本机类型，它具有优越的查找性能。
避免重复
显然，第一个重复出现的项目Array被保留，而第二个Array被重复数据删除。这可能是也可能不是所需的行为。如果您需要一个独特的结果，只需应用于dedupe第一个参数：
// 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Arrays的交集
如果你想计算任意数量的Arrays 的交集，只需intersect用foldl. 这是一个便利功能：
// 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) );于 2016-10-11T09:07:18.443 回答
2
.reduce建立地图，并.filter找到交叉点。delete在.filter允许我们将第二个数组视为一个唯一的集合。
function intersection (a, b) {
var seen = a.reduce(function (h, k) {
h[k] = true;
return h;
}, {});
return b.filter(function (k) {
var exists = seen[k];
delete seen[k];
return exists;
});
}
我发现这种方法很容易推理。它在恒定时间内执行。
于 2017-03-21T09:19:51.680 回答
2
我编写了一个 intesection 函数，它甚至可以根据这些对象的特定属性检测对象数组的交集。
例如，
if arr1 = [{id: 10}, {id: 20}]
and arr2 = [{id: 20}, {id: 25}]
我们想要基于id属性的交集，那么输出应该是：
[{id: 20}]
因此，相同的功能（注意：ES6 代码）是：
const intersect = (arr1, arr2, accessors = [v => v, v => v]) => {
const [fn1, fn2] = accessors;
const set = new Set(arr2.map(v => fn2(v)));
return arr1.filter(value => set.has(fn1(value)));
};
您可以将函数调用为：
intersect(arr1, arr2, [elem => elem.id, elem => elem.id])
另请注意：考虑到第一个数组是主数组，此函数会找到交集，因此交集结果将是主数组的结果。
于 2019-05-22T09:17:54.610 回答
1
下面是underscore.js的实现：
_.intersection = function(array) {
if (array == null) return [];
var result = [];
var argsLength = arguments.length;
for (var i = 0, length = array.length; i < length; i++) {
var item = array[i];
if (_.contains(result, item)) continue;
for (var j = 1; j < argsLength; j++) {
if (!_.contains(arguments[j], item)) break;
}
if (j === argsLength) result.push(item);
}
return result;
};
于 2014-12-30T00:58:17.283 回答
1
这可能是最简单的，除了list1.filter(n => list2.includes(n))
var list1 = ['bread', 'ice cream', 'cereals', 'strawberry', 'chocolate']
var list2 = ['bread', 'cherry', 'ice cream', 'oats']
function check_common(list1, list2){
list3 = []
for (let i=0; i<list1.length; i++){
for (let j=0; j<list2.length; j++){
if (list1[i] === list2[j]){
list3.push(list1[i]);
}
}
}
return list3
}
check_common(list1, list2) // ["bread", "ice cream"]于 2018-02-03T04:55:20.873 回答
1
使用一个数组创建一个 Object 并遍历第二个数组以检查该值是否作为键存在。
function intersection(arr1, arr2) {
var myObj = {};
var myArr = [];
for (var i = 0, len = arr1.length; i < len; i += 1) {
if(myObj[arr1[i]]) {
myObj[arr1[i]] += 1;
} else {
myObj[arr1[i]] = 1;
}
}
for (var j = 0, len = arr2.length; j < len; j += 1) {
if(myObj[arr2[j]] && myArr.indexOf(arr2[j]) === -1) {
myArr.push(arr2[j]);
}
}
return myArr;
}
于 2018-11-18T09:31:03.570 回答
1
此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，第三个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，只要修改generateStrKey函数，肯定可以解决。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。如果任何数组为空，请避免循环。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;
}
于 2021-05-31T15:33:09.223 回答
1
我认为在内部使用对象可以帮助计算并且也可以是高性能的。
// 方法维护每个元素的计数，也适用于负元素
function intersect(a,b){
const A = {};
a.forEach((v)=>{A[v] ? ++A[v] : A[v] = 1});
const B = {};
b.forEach((v)=>{B[v] ? ++B[v] : B[v] = 1});
const C = {};
Object.entries(A).map((x)=>C[x[0]] = Math.min(x[1],B[x[0]]))
return Object.entries(C).map((x)=>Array(x[1]).fill(Number(x[0]))).flat();
}
const x = [1,1,-1,-1,0,0,2,2];
const y = [2,0,1,1,1,1,0,-1,-1,-1];
const result = intersect(x,y);
console.log(result); // (7) [0, 0, 1, 1, 2, -1, -1]
于 2021-08-14T16:58:20.373 回答
0
我扩展了 tarulen 的答案以使用任意数量的数组。它也应该适用于非整数值。
function intersect() {
const last = arguments.length - 1;
var seen={};
var result=[];
for (var i = 0; i < last; i++) {
for (var j = 0; j < arguments[i].length; j++) {
if (seen[arguments[i][j]]) {
seen[arguments[i][j]] += 1;
}
else if (!i) {
seen[arguments[i][j]] = 1;
}
}
}
for (var i = 0; i < arguments[last].length; i++) {
if ( seen[arguments[last][i]] === last)
result.push(arguments[last][i]);
}
return result;
}
于 2016-09-19T14:34:22.487 回答
0
function getIntersection(arr1, arr2){
var result = [];
arr1.forEach(function(elem){
arr2.forEach(function(elem2){
if(elem === elem2){
result.push(elem);
}
});
});
return result;
}
getIntersection([1,2,3], [2,3,4,5]); // [ 2, 3 ]
于 2017-10-20T23:36:17.820 回答
0
如果您的数组已排序，这应该在 O(n) 中运行，其中 n 是 min(a.length, b.length)
function intersect_1d( a, b ){
var out=[], ai=0, bi=0, acurr, bcurr, last=Number.MIN_SAFE_INTEGER;
while( ( acurr=a[ai] )!==undefined && ( bcurr=b[bi] )!==undefined ){
if( acurr < bcurr){
if( last===acurr ){
out.push( acurr );
}
last=acurr;
ai++;
}
else if( acurr > bcurr){
if( last===bcurr ){
out.push( bcurr );
}
last=bcurr;
bi++;
}
else {
out.push( acurr );
last=acurr;
ai++;
bi++;
}
}
return out;
}
于 2018-08-14T00:25:08.823 回答
0
var arrays = [
[1, 2, 3],
[2, 3, 4, 5]
]
function commonValue (...arr) {
let res = arr[0].filter(function (x) {
return arr.every((y) => y.includes(x))
})
return res;
}
commonValue(...arrays);于 2018-12-24T06:28:12.610 回答
0
function intersectionOfArrays(arr1, arr2) {
return arr1.filter((element) => arr2.indexOf(element) !== -1).filter((element, pos, self) => self.indexOf(element) == pos);
}
于 2019-03-04T02:03:29.850 回答
0
这是一种现代且简单的 ES6 方式，也非常灵活。它允许您将多个数组指定为要与主题数组进行比较的数组，并且可以在包含模式和独占模式下工作。
// =======================================
// The function
// =======================================
function assoc(subjectArray, otherArrays, { mustBeInAll = true } = {}) {
return subjectArray.filter((subjectItem) => {
if (mustBeInAll) {
return otherArrays.every((otherArray) =>
otherArray.includes(subjectItem)
);
} else {
return otherArrays.some((otherArray) => otherArray.includes(subjectItem));
}
});
}
// =======================================
// The usage
// =======================================
const cheeseList = ["stilton", "edam", "cheddar", "brie"];
const foodListCollection = [
["cakes", "ham", "stilton"],
["juice", "wine", "brie", "bread", "stilton"]
];
// Output will be: ['stilton', 'brie']
const inclusive = assoc(cheeseList, foodListCollection, { mustBeInAll: false }),
// Output will be: ['stilton']
const exclusive = assoc(cheeseList, foodListCollection, { mustBeInAll: true })
现场示例：https ://codesandbox.io/s/zealous-butterfly-h7dgf?fontsize=14&hidenavigation=1&theme=dark
于 2021-01-04T18:07:57.023 回答
0
我正在使用地图，甚至可以使用对象。
//find intersection of 2 arrs
const intersections = (arr1,arr2) => {
let arrf = arr1.concat(arr2)
let map = new Map();
let union = [];
for(let i=0; i<arrf.length; i++){
if(map.get(arrf[i])){
map.set(arrf[i],false);
}else{
map.set(arrf[i],true);
}
}
map.forEach((v,k)=>{if(!v){union.push(k);}})
return union;
}
于 2021-10-12T17:43:27.453 回答
-1
这是我正在使用的一个非常天真的实现。它是非破坏性的，并且确保不复制整体。
Array.prototype.contains = function(elem) {
return(this.indexOf(elem) > -1);
};
Array.prototype.intersect = function( array ) {
// this is naive--could use some optimization
var result = [];
for ( var i = 0; i < this.length; i++ ) {
if ( array.contains(this[i]) && !result.contains(this[i]) )
result.push( this[i] );
}
return result;
}
于 2011-12-16T16:17:04.470 回答
-1
Coffeescript中N个数组的交集
getIntersection: (arrays) ->
if not arrays.length
return []
a1 = arrays[0]
for a2 in arrays.slice(1)
a = (val for val in a1 when val in a2)
a1 = a
return a1.unique()
于 2013-04-18T15:47:00.567 回答
-1
基于 Anon 的出色答案，这个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;
});
});
}
于 2017-01-26T08:54:01.983 回答
-1
希望这对所有版本都有帮助。
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;
}
于 2017-01-29T14:14:27.540 回答
-2
不是关于效率，而是易于理解，这里是集合的并集和交集的示例，它处理集合的数组和集合的集合。
http://jsfiddle.net/zhulien/NF68T/
// process array [element, element...], if allow abort ignore the result
function processArray(arr_a, cb_a, blnAllowAbort_a)
{
var arrResult = [];
var blnAborted = false;
var intI = 0;
while ((intI < arr_a.length) && (blnAborted === false))
{
if (blnAllowAbort_a)
{
blnAborted = cb_a(arr_a[intI]);
}
else
{
arrResult[intI] = cb_a(arr_a[intI]);
}
intI++;
}
return arrResult;
}
// process array of operations [operation,arguments...]
function processOperations(arrOperations_a)
{
var arrResult = [];
var fnOperationE;
for(var intI = 0, intR = 0; intI < arrOperations_a.length; intI+=2, intR++)
{
var fnOperation = arrOperations_a[intI+0];
var fnArgs = arrOperations_a[intI+1];
if (fnArgs === undefined)
{
arrResult[intR] = fnOperation();
}
else
{
arrResult[intR] = fnOperation(fnArgs);
}
}
return arrResult;
}
// return whether an element exists in an array
function find(arr_a, varElement_a)
{
var blnResult = false;
processArray(arr_a, function(varToMatch_a)
{
var blnAbort = false;
if (varToMatch_a === varElement_a)
{
blnResult = true;
blnAbort = true;
}
return blnAbort;
}, true);
return blnResult;
}
// return the union of all sets
function union(arr_a)
{
var arrResult = [];
var intI = 0;
processArray(arr_a, function(arrSet_a)
{
processArray(arrSet_a, function(varElement_a)
{
// if the element doesn't exist in our result
if (find(arrResult, varElement_a) === false)
{
// add it
arrResult[intI] = varElement_a;
intI++;
}
});
});
return arrResult;
}
// return the intersection of all sets
function intersection(arr_a)
{
var arrResult = [];
var intI = 0;
// for each set
processArray(arr_a, function(arrSet_a)
{
// every number is a candidate
processArray(arrSet_a, function(varCandidate_a)
{
var blnCandidate = true;
// for each set
processArray(arr_a, function(arrSet_a)
{
// check that the candidate exists
var blnFoundPart = find(arrSet_a, varCandidate_a);
// if the candidate does not exist
if (blnFoundPart === false)
{
// no longer a candidate
blnCandidate = false;
}
});
if (blnCandidate)
{
// if the candidate doesn't exist in our result
if (find(arrResult, varCandidate_a) === false)
{
// add it
arrResult[intI] = varCandidate_a;
intI++;
}
}
});
});
return arrResult;
}
var strOutput = ''
var arrSet1 = [1,2,3];
var arrSet2 = [2,5,6];
var arrSet3 = [7,8,9,2];
// return the union of the sets
strOutput = union([arrSet1, arrSet2, arrSet3]);
alert(strOutput);
// return the intersection of 3 sets
strOutput = intersection([arrSet1, arrSet2, arrSet3]);
alert(strOutput);
// of 3 sets of sets, which set is the intersecting set
strOutput = processOperations([intersection,[[arrSet1, arrSet2], [arrSet2], [arrSet2, arrSet3]]]);
alert(strOutput);
于 2014-04-03T07:42:13.120 回答
-5
IE 中的 Array 不支持“filter”和“indexOf”。这个怎么样：
var array1 = [1, 2, 3];
var array2 = [2, 3, 4, 5];
var intersection = [];
for (i in array1) {
for (j in array2) {
if (array1[i] == array2[j]) intersection.push(array1[i]);
}
}
于 2009-12-11T05:42:33.890 回答