# data-structures js数组交集 - 在javascript中最简单的数组交集

## js数组补集 es6数组比较 (25)

``````intersection([1,2,3], [2,3,4,5])
``````

``````[2, 3]
``````

``````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;
}
``````

``````// 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;
}
``````

``````function intersect(a, b) {
var aa = {};
a.forEach(function(v) { aa[v]=1; });
return b.filter(function(v) { return v in aa; });
}
``````

``````/* 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;
}
``````

``````/* 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;
}
``````

## 采用ES2015的功能性方法

``````// 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) );``````

### 避免重复

``````// 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) );``````

### 计算任意数量的`Array`的交集

``````// 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

// 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) );``````

``````_.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;
};
``````

``````var a = [1,2,3];
var b = [2,3,4,5];
var c = \$(b).not(\$(b).not(a));
``````

``````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;
}
``````

``````var listA = [1,2,3,4,5,6,7];
var listB = [2,4,6,8];

var result = listA.filter(itemA=> {
return listB.some(itemB => itemB === itemA);
});
``````

``````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;
};
}
``````

``````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;
}
``````

``````// 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]));``````

``````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：

• 大阵列上的Ops /秒（10,000个元素）：

``````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
``````
• 小阵列上的Ops /秒（100个元素）：

``````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
``````

“indexOf”适用于IE 9.0，Chrome，Firefox，Opera，

``````    function intersection(a,b){
var rs = [], x = a.length;
while (x--) b.indexOf(a[x])!=-1 && rs.push(a[x]);
return rs.sort();
}

intersection([1,2,3], [2,3,4,5]);
//Result:  [2,3]
``````

``````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 ]
``````

``````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"];
``````

``````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;
}
``````

``````array1.filter(function(n) {
return array2.indexOf(n) !== -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()
``````

``````'use strict'

// Example 1
function intersection(a1, a2) {
return a1.filter(x => a2.indexOf(x) > -1)
}

// Example 2 (prototype function)
Array.prototype.intersection = function(arr) {
return this.filter(x => arr.indexOf(x) > -1)
}

const a1 = [1, 2, 3]
const a2 = [2, 3, 4, 5]

console.log(intersection(a1, a2))
console.log(a1.intersection(a2))``````

``````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);
}
``````

``````function intersect(a, b) {
return [...new Set(a)].filter(x => new Set(b).has(x));
}
``````

``````// 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]
``````

• 污垢简单
• 数据中心
• 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
• 保留形状（任何阵列中的第一次出现次序）
• 如有可能，尽早退出
• 内存安全，不会篡改函数/数组原型

• 内存使用量更高
• 更高的CPU使用率
• 需要了解减少
• 需要了解数据流

1. 把它分类
2. 从索引0逐个检查，从中创建新的数组。

``````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;
}

``````

PS：该算法仅用于Numbers和Normal Strings，任意对象数组的交集可能不起作用。

``````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;
});
});
}
``````

``````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"]``````

``````let intersection = function(a, b) {
a = new Set(a), b = new Set(b);
return [...a].filter(v => b.has(v));
};

console.log(intersection([1,2,1,2,3], [2,3,5,4,5,3]));

``````_.without([1, 2, 1, 0, 3, 1, 4], 0, 1); // => [2, 3, 4]