replace items which appear in both arrays in javascript - javascript

I have two arrays and want to remove from one all elements which exist in the other as well.
Can this be done with native JS?
Is there a jQuery function to do it?
What are best practices to do so (the
faster the better)
p.s.: just post code in other languages too, maybe I can port it to Javascript
Update, after accepting answer to help the JS dudes ;-)
// Array Remove - By John Resig (MIT Licensed)
Array.prototype.remove = function(from, to) {
var rest = this.slice((to || from) + 1 || this.length);
this.length = from < 0 ? this.length + from : from;
return this.push.apply(this, rest);
};
// Array Contains - By Helmuth Lammer (TU Vienna)
Array.prototype.contains = function(key){
for(var i = 0; i<this.length; i++){
if(this[i] == key) return i;
}
return false;
}
(There is a native JS method named contains too, but it should work)

Given two sets a and b, to remove all elements present in a from b. Translation to array formatting and javascript left as an exercise to the reader. Optimizations exist for sorted arrays.
for(element e : b) {
if(a.contains(e)) {
b.remove(e);
}
}

var foo = [1, 2, 3, 4, 5];
function removeAll(a, b) {
var i = 0,
j;
while (i < b.length) {
for (j = 0; j < a.length; j++) {
if (a[j] === b[i]) {
b.splice(i, 1);
i--;
break;
}
}
i++;
}
}
removeAll([2, 4], foo);
// foo === [1, 3, 5]

[See it in action]
// two arrays
var arr1 = [1,2,3,4,5];
var arr2 = [4,5,6,7,8];
// returns the element's index in an array
// or -1 if there isn't such an element
function indexOf( arr, el ) {
for ( var i = arr.length; i--; ) {
if ( arr[i] === el ) return i;
}
return -1;
}
// go through array1 and remove all
// elements which is also in array2
​for ( var i = arr1.length; i--; )​ {
if ( indexOf( arr2, arr1[i] ) !== -1 ) {
arr1.splice(i, 1);
}
}
// result should be: arr1 = [1,2,3]
alert( arr1 );
​

I had to write a code once where an Ajax call would return an array containing a list of element which must be deleted from a local (client-side) array.
I ended implementing it this way:
// The filter method creates a new array with all elements that pass the
// test implemented by the provided function.
local_array = local_array.filter( function (element) {
// Now I check if this element is present in the "delete" array. In order to do
// that I use the "some" method which runs a callback function for each of the
// array elements and returns true or false.
return !items_to_delete.some( function(item) {
return item.pk === element.pk;
});
});
Edit:
In order to make it cross-browser (I am talking about IE here). You will have to define the some and filter functions. Just put this somewhere in your code:
if (!Array.prototype.some)
{
Array.prototype.some = function(fun /*, thisp*/)
{
var i = 0,
len = this.length >>> 0;
if (typeof fun != "function")
throw new TypeError();
var thisp = arguments[1];
for (; i < len; i++)
{
if (i in this &&
fun.call(thisp, this[i], i, this))
return true;
}
return false;
};
}
if (!Array.prototype.filter)
{
Array.prototype.filter = function(fun /*, thisp*/)
{
var len = this.length >>> 0;
if (typeof fun != "function")
throw new TypeError();
var res = [];
var thisp = arguments[1];
for (var i = 0; i < len; i++)
{
if (i in this)
{
var val = this[i]; // in case fun mutates this
if (fun.call(thisp, val, i, this))
res.push(val);
}
}
return res;
};
}

Related

How can I use a Function to determine if a Value exists in a Javascript Array?

I will be using an input field where a person types in a value.
I create the variable that they will type in: 'Ben'.
I want the function to loop thru the nameArray and return true or false.
The script is not working and I know it is rather simple.
function validateNames() {
var name = "Ben";
var nameArray = ["Bill", "Barry", "Zack", "Will"];
var arrayLength = nameArray.length;
for (var i = 0; i < arrayLength; i++) {
//Do something
if (name != nameArray[i]) {
alert((nameArray[i]) + "Name is not valid.");
console.log("Name is not found in array.");
} else {
return true;
console.log(nameArray[i]);
}
}
}
The only way for your loop logic to know that the value is not in the array is to go through the whole array first. Your loop will be alerting on every iteration until it finds a match. It also doesn't make sense to put a console.log after a return because the former will never execute:
function validateNames() {
var name = "Ben";
var nameArray = ["Bill", "Barry", "Zack", "Will"];
var arrayLength = nameArray.length;
for (var i = 0; i < arrayLength; i++) {
if (name === nameArray[i]) {
console.log(nameArray[i]);
return true;
}
}
console.log("Name is not found in array.");
return false;
}
validateNames();
Javascript arrays also provide a handy method for checking whether they contain a certain value. it's called .indexOf(), and it returns -1 when there's no match:
function validateNames() {
var name = "Ben";
var nameArray = ["Bill","Barry","Zack","Will"];
return nameArray.indexOf(name) !== -1;
}
You could use .indexOf():
var nameArray = ["Bill","Barry","Zack","Will"];
nameArray.indexOf("Bill"); // Returns 0, Bill is found
nameArray.indexOf("Hillary"); // Returns -1, Hillary not found
So your function could look like:
function validateName(name) {
return nameArray.indexOf(name) !== -1;
}
Please bear in mind it does not work on IE8 or below.
easy fix :)
var nameArray = ["Bill", "Barry", "Zack", "Will"];
console.log(validateName("Ben", nameArrray)); // False
console.log(validateName("Will", nameArrray)); // True
function validateName(name, arr){
var found = 0;
for(var i = 0; i < arr.length; i++){
found += (name === arr[i])? 1 : 0;
}
var result = (found>0)? true: false;
return result;
}
This will be false, because Ben does not exist in the array
if (nameArray.indexOf(name) > -1)
You can add a contains() method to the Array class so that you do not have to type this out each time.
// Static method
if (Array.contains === undefined) {
Array.contains = function(arr, val) {
return arr.indexOf(val) > -1;
}
}
// Instance method
if (Array.prototype.contains === undefined) {
Array.prototype.contains = function(val) {
return this.indexOf(val) > -1;
}
}
var nameArray = ["Bill", "Barry", "Zack", "Will"];
var name = "Ben";
Array.contains(nameArray, name); // false
nameArray.contains(name); // false
You could also use some Array.prototype.some.
if (Array.prototype.contains === undefined) {
Array.prototype.contains = function(val) {
return this.some(function(item) {
return item === name;
});
}
}
A better approach is to polyfill Array.prototype.includes(). This is an upcoming method in ECMAScript 7.
Polyfill
if (![].includes) {
Array.prototype.includes = function(searchElement /*, fromIndex*/ ) {'use strict';
var O = Object(this);
var len = parseInt(O.length) || 0;
if (len === 0) {
return false;
}
var n = parseInt(arguments[1]) || 0;
var k;
if (n >= 0) {
k = n;
} else {
k = len + n;
if (k < 0) {k = 0;}
}
var currentElement;
while (k < len) {
currentElement = O[k];
if (searchElement === currentElement ||
(searchElement !== searchElement && currentElement !== currentElement)) {
return true;
}
k++;
}
return false;
};
}
Usage
[1, 2, 3].includes(2); // true
[1, 2, 3].includes(4); // false
[1, 2, 3].includes(3, 3); // false
[1, 2, 3].includes(3, -1); // true
[1, 2, NaN].includes(NaN); // true

Javascript filter function polyfill

I am unable to understand why check corresponding to line if (i in t) - Line no.18 is placed in filter function polyfill :
if (!Array.prototype.filter) {
Array.prototype.filter = function(fun/*, thisArg*/) {
'use strict';
if (this === void 0 || this === null) {
throw new TypeError();
}
var t = Object(this);
var len = t.length >>> 0;
if (typeof fun !== 'function') {
throw new TypeError();
}
var res = [];
var thisArg = arguments.length >= 2 ? arguments[1] : void 0;
for (var i = 0; i < len; i++) {
if (i in t) {
var val = t[i];
// NOTE: Technically this should Object.defineProperty at
// the next index, as push can be affected by
// properties on Object.prototype and Array.prototype.
// But that method's new, and collisions should be
// rare, so use the more-compatible alternative.
if (fun.call(thisArg, val, i, t)) {
res.push(val);
}
}
}
return res;
};
}
It is to avoid the elements which are not defined yet, in the sparse arrays. See the following example,
var array = [];
array[3] = 10;
console.log(array.length);
// 4
So, the length of the array is 4, but only the element at index 3 is defined, all others are not defined yet. So, if you do
for (var i = 0; i < array.length; i += 1) {
console.log(i, array[i]);
}
you will get
0 undefined
1 undefined
2 undefined
3 10
Arrays are special JavaScript objects. The indices are just properties in the array object. Whenever you extend the array object with an index which is not in the array, the length property will be adjusted internally. In this case, there is only one property in the array object, with the name 3 defined. But we are trying to access elements from 0 to 3. So it returns undefined for all the indices which are not present in the array object yet.
To avoid that, we are checking if the current index is really defined in the array object, with that if statement.
for (var i = 0; i < array.length; i += 1) {
if (i in array) {
console.log(i, array[i]);
}
}
would print
3 10
This is because it's possible for JavaScript arrays to have gaps.
For example, consider the following:
var a = ["hello", "howdy", "welcome"];
delete greetings[1];
"0" in a; // true
"1" in a; // false
"2" in a; // true
Array.prototype.myFilter = function(callBack) {
let newArray = [];
for (let i = 0; i < this.length; i++) {
let result = callBack(this[i], i, this);
if (result) {
newArray.push(this[i]);
}
}
return newArray;
IN is a reserved keyword which can be used in for and if statement.
18th line they are doing exist check
Refer this dot vs in.
Create own filter() method
Array.prototype.newFilter = function(func){
let filtered = [], n = this.length;
for(let i = 0; i<n; i++) {
if(func(this[i],i,this))
filtered.push(this[i]);
}
return filtered;
}
let result = [1,2,3,4,5,6,7].newFilter(item => item > 4);
console.log(result);
Array.prototype.myFilter = function(callBackFn) {
let res = [];
if (!Array.isArray(this)) {
throw new Error(`${this} is not a function`);
}
for (let i = 0; i<this.length; i++) {
if (callBackFn(this[i], i, this)) {
res.push(this[i])
}
}
return res
}

Check if an array contains duplicate values [duplicate]

This question already has answers here:
In Javascript, how do I check if an array has duplicate values?
(9 answers)
Closed 10 months ago.
I wanted to write a javascript function which checks if array contains duplicate values or not.
I have written the following code but its giving answer as "true" always.
Can anybody please tell me what am I missing.
function checkIfArrayIsUnique(myArray)
{
for (var i = 0; i < myArray.length; i++)
{
for (var j = 0; j < myArray.length; j++)
{
if (i != j)
{
if (myArray[i] == myArray[j])
{
return true; // means there are duplicate values
}
}
}
}
return false; // means there are no duplicate values.
}
An easy solution, if you've got ES6, uses Set:
function checkIfArrayIsUnique(myArray) {
return myArray.length === new Set(myArray).size;
}
let uniqueArray = [1, 2, 3, 4, 5];
console.log(`${uniqueArray} is unique : ${checkIfArrayIsUnique(uniqueArray)}`);
let nonUniqueArray = [1, 1, 2, 3, 4, 5];
console.log(`${nonUniqueArray} is unique : ${checkIfArrayIsUnique(nonUniqueArray)}`);
let arr = [11,22,11,22];
let hasDuplicate = arr.some((val, i) => arr.indexOf(val) !== i);
// hasDuplicate = true
True -> array has duplicates
False -> uniqe array
This should work with only one loop:
function checkIfArrayIsUnique(arr) {
var map = {}, i, size;
for (i = 0, size = arr.length; i < size; i++){
if (map[arr[i]]){
return false;
}
map[arr[i]] = true;
}
return true;
}
You got the return values the wrong way round:
As soon as you find two values that are equal, you can conclude that the array is not unique and return false.
At the very end, after you've checked all the pairs, you can return true.
If you do this a lot, and the arrays are large, you might want to investigate the possibility of sorting the array and then only comparing adjacent elements. This will have better asymptotic complexity than your current method.
Assuming you're targeting browsers that aren't IE8,
this would work as well:
function checkIfArrayIsUnique(myArray)
{
for (var i = 0; i < myArray.length; i++)
{
if (myArray.indexOf(myArray[i]) !== myArray.lastIndexOf(myArray[i])) {
return false;
}
}
return true; // this means not unique
}
Here's an O(n) solution:
function hasDupes(arr) {
/* temporary object */
var uniqOb = {};
/* create object attribute with name=value in array, this will not keep dupes*/
for (var i in arr)
uniqOb[arr[i]] = "";
/* if object's attributes match array, then no dupes! */
if (arr.length == Object.keys(uniqOb).length)
alert('NO dupes');
else
alert('HAS dupes');
}
var arr = ["1/1/2016", "1/1/2016", "2/1/2016"];
hasDupes(arr);
https://jsfiddle.net/7kkgy1j3/
Another solution:
Array.prototype.checkIfArrayIsUnique = function() {
this.sort();
for ( var i = 1; i < this.length; i++ ){
if(this[i-1] == this[i])
return false;
}
return true;
}
function hasNoDuplicates(arr) {
return arr.every(num => arr.indexOf(num) === arr.lastIndexOf(num));
}
hasNoDuplicates accepts an array and returns true if there are no duplicate values. If there are any duplicates, the function returns false.
Without a for loop, only using Map().
You can also return the duplicates.
(function(a){
let map = new Map();
a.forEach(e => {
if(map.has(e)) {
let count = map.get(e);
console.log(count)
map.set(e, count + 1);
} else {
map.set(e, 1);
}
});
let hasDup = false;
let dups = [];
map.forEach((value, key) => {
if(value > 1) {
hasDup = true;
dups.push(key);
}
});
console.log(dups);
return hasDup;
})([2,4,6,2,1,4]);
Late answer but can be helpful
function areThereDuplicates(args) {
let count = {};
for(let i = 0; i < args.length; i++){
count[args[i]] = 1 + (count[args[i]] || 0);
}
let found = Object.keys(count).filter(function(key) {
return count[key] > 1;
});
return found.length ? true : false;
}
areThereDuplicates([1,2,5]);
The code given in the question can be better written as follows
function checkIfArrayIsUnique(myArray)
{
for (var i = 0; i < myArray.length; i++)
{
for (var j = i+1; j < myArray.length; j++)
{
if (myArray[i] == myArray[j])
{
return true; // means there are duplicate values
}
}
}
return false; // means there are no duplicate values.
}
Returns the duplicate item in array and creates a new array with no duplicates:
var a = ["hello", "hi", "hi", "juice", "juice", "test"];
var b = ["ding", "dong", "hi", "juice", "juice", "test"];
var c = a.concat(b);
var dupClearArr = [];
function dupArray(arr) {
for (i = 0; i < arr.length; i++) {
if (arr.indexOf(arr[i]) != i && arr.indexOf(arr[i]) != -1) {
console.log('duplicate item ' + arr[i]);
} else {
dupClearArr.push(arr[i])
}
}
console.log('actual array \n' + arr + ' \nno duplicate items array \n' + dupClearArr)
}
dupArray(c);
const containsMatches = (a1, a2) => a1.some((v) => a2.includes(v));
If your array nests other arrays/objects, using the Set approach may not be what you want since comparing two objects compares their references. If you want to check that their contained values are equal, something else is needed. Here are a couple different approaches.
Approach 1: Map using JSON.stringify for keys
If you want to consider objects with the same contained values as equal, here's one simple way to do it using a Map object. It uses JSON.stringify to make a unique id for each element in the array.
I believe the runtime of this would be O(n * m) on arrays, assuming JSON.stringify serializes in linear time. n is the length of the outer array, m is size of the arrays. If the objects get very large, however, this may slow down since the keys will be very long. Not a very space-efficient implementation, but it is simple and works for many data types.
function checkArrayDupeFree(myArray, idFunc) {
const dupeMap = new Map();
for (const el of myArray) {
const id = idFunc(el);
if (dupeMap.has(id))
return false;
dupeMap.set(id, el);
}
return true;
}
const notUnique = [ [1, 2], [1, 3], [1, 2] ];
console.log(`${JSON.stringify(notUnique)} has no duplicates? ${checkArrayDupeFree(notUnique, JSON.stringify)}`);
const unique = [ [2, 1], [1, 3], [1, 2] ];
console.log(`${JSON.stringify(unique)} has no duplicates? ${checkArrayDupeFree(unique, JSON.stringify)}`);
Of course, you could also write your own id-generator function, though I'm not sure you can do much better than JSON.stringify.
Approach 2: Custom HashMap, Hashcode, and Equality implementations
If you have a lot of big arrays, it may be better performance-wise to implement your own hash/equality functions and use a Map as a HashMap.
In the following implementation, we hash the array. If there is a collision, map a key to an array of collided values, and check to see if any of the array values match according to the equality function.
The downside of this approach is that you may have to consider a wide range of types for which to make hashcode/equality functions, depending on what's in the array.
function checkArrayDupeFreeWHashes(myArray, hashFunc, eqFunc) {
const hashMap = new Map();
for (const el of myArray) {
const hash = hashFunc(el);
const hit = hashMap.get(hash);
if (hit == null)
hashMap.set(hash, [el]);
else if (hit.some(v => eqFunc(v, el)))
return false;
else
hit.push(el);
}
return true;
}
Here's a demo of the custom HashMap in action. I implemented a hashing function and an equality function for arrays of arrays.
function checkArrayDupeFreeWHashes(myArray, hashFunc, eqFunc) {
const hashMap = new Map();
for (const el of myArray) {
const hash = hashFunc(el);
const hit = hashMap.get(hash);
if (hit == null)
hashMap.set(hash, [el]);
else if (hit.some(v => eqFunc(v, el)))
return false;
else
hit.push(el);
}
return true;
}
function arrayHasher(arr) {
let hash = 19;
for (let i = 0; i < arr.length; i++) {
const el = arr[i];
const toHash = Array.isArray(el)
? arrayHasher(el)
: el * 23;
hash = hash * 31 + toHash;
}
return hash;
}
function arrayEq(a, b) {
if (a.length != b.length)
return false;
for (let i = 0; i < a.length; i++) {
if ((Array.isArray(a) || Array.isArray(b)) && !arrayEq(a[i], b[i]))
return false;
else if (a[i] !== b[i])
return false;
}
return true;
}
const notUnique = [ [1, 2], [1, 3], [1, 2] ];
const unique = [ [2, 1], [1, 3], [1, 2] ];
console.log(`${JSON.stringify(notUnique)} has no duplicates? ${checkArrayDupeFreeWHashes(notUnique, arrayHasher, arrayEq)}`);
console.log(`${JSON.stringify(unique)} has no duplicates? ${checkArrayDupeFreeWHashes(unique, arrayHasher, arrayEq)}`);
function checkIfArrayIsUnique(myArray)
{
isUnique=true
for (var i = 0; i < myArray.length; i++)
{
for (var j = 0; j < myArray.length; j++)
{
if (i != j)
{
if (myArray[i] == myArray[j])
{
isUnique=false
}
}
}
}
return isUnique;
}
This assume that the array is unique at the start.
If find two equals values, then change to false
i think this is the simple way
$(document).ready(function() {
var arr = [1,2,3,9,6,5,6];
console.log( "result =>"+ if_duplicate_value (arr));
});
function if_duplicate_value (arr){
for(i=0;i<arr.length-1;i++){
for(j=i+1;j<arr.length;j++){
if(arr[i]==arr[j]){
return true;
}
}
}
return false;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
var c=[2,2,3,3,5,5,4,4,8,8];
for(var i=0; i<b.length; i++){
for(var j=i+1; j<b.length; j++){
if(c[i]==c[j]){
console.log(c[j]);
}
}
}

Check two arrays have elements with same values [duplicate]

What's the simplest, library-free code for implementing array intersections in javascript? I want to write
intersection([1,2,3], [2,3,4,5])
and get
[2, 3]
Use a combination of Array.prototype.filter and Array.prototype.includes:
const filteredArray = array1.filter(value => array2.includes(value));
For older browsers, with Array.prototype.indexOf and without an arrow function:
var filteredArray = array1.filter(function(n) {
return array2.indexOf(n) !== -1;
});
NB! Both .includes and .indexOf internally compares elements in the array by using ===, so if the array contains objects it will only compare object references (not their content). If you want to specify your own comparison logic, use Array.prototype.some instead.
Destructive seems simplest, especially if we can assume the input is sorted:
/* 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;
}
Non-destructive has to be a hair more complicated, since we’ve got to track indices:
/* 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;
}
If your environment supports ECMAScript 6 Set, one simple and supposedly efficient (see specification link) way:
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);
}
Shorter, but less readable (also without creating the additional intersection Set):
function intersect(a, b) {
var setB = new Set(b);
return [...new Set(a)].filter(x => setB.has(x));
}
Note that when using sets you will only get distinct values, thus new Set([1, 2, 3, 3]).size evaluates to 3.
Using Underscore.js or lodash.js
_.intersection( [0,345,324] , [1,0,324] ) // gives [0,324]
// 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]));
I recommend above succinct solution which outperforms other implementations on large inputs. If performance on small inputs matters, check the alternatives below.
Alternatives and performance comparison:
See the following snippet for alternative implementations and check https://jsperf.com/array-intersection-comparison for performance comparisons.
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;
}
Results in Firefox 53:
Ops/sec on large arrays (10,000 elements):
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
Ops/sec on small arrays (100 elements):
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
My contribution in ES6 terms. In general it finds the intersection of an array with indefinite number of arrays provided as arguments.
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>");
How about just using associative arrays?
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;
}
edit:
// 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;
}
The performance of #atk's implementation for sorted arrays of primitives can be improved by using .pop rather than .shift.
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;
}
I created a benchmark using jsPerf. It's about three times faster to use .pop.
If you need to have it handle intersecting multiple arrays:
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]))
Sort it
check one by one from the index 0, create new array from that.
Something like this, Not tested well though.
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:The algorithm only intended for Numbers and Normal Strings, intersection of arbitary object arrays may not work.
Using jQuery:
var a = [1,2,3];
var b = [2,3,4,5];
var c = $(b).not($(b).not(a));
alert(c);
A tiny tweak to the smallest one here (the filter/indexOf solution), namely creating an index of the values in one of the arrays using a JavaScript object, will reduce it from O(N*M) to "probably" linear time. source1 source2
function intersect(a, b) {
var aa = {};
a.forEach(function(v) { aa[v]=1; });
return b.filter(function(v) { return v in aa; });
}
This isn't the very simplest solution (it's more code than filter+indexOf), nor is it the very fastest (probably slower by a constant factor than intersect_safe()), but seems like a pretty good balance. It is on the very simple side, while providing good performance, and it doesn't require pre-sorted inputs.
For arrays containing only strings or numbers you can do something with sorting, as per some of the other answers. For the general case of arrays of arbitrary objects I don't think you can avoid doing it the long way. The following will give you the intersection of any number of arrays provided as parameters to 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"];
Simplest, fastest O(n) and shortest way:
function intersection (a, b) {
const setA = new Set(a);
return b.filter(value => setA.has(value));
}
console.log(intersection([1,2,3], [2,3,4,5]))
#nbarbosa has almost the same answer but he cast both arrays to Set and then back to array. There is no need for any extra casting.
Another indexed approach able to process any number of arrays at once:
// 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;
};
It works only for values that can be evaluated as strings and you should pass them as an array like:
intersect ([arr1, arr2, arr3...]);
...but it transparently accepts objects as parameter or as any of the elements to be intersected (always returning array of common values). Examples:
intersect ({foo: [1, 2, 3, 4], bar: {a: 2, j:4}}); // [2, 4]
intersect ([{x: "hello", y: "world"}, ["hello", "user"]]); // ["hello"]
EDIT: I just noticed that this is, in a way, slightly buggy.
That is: I coded it thinking that input arrays cannot itself contain repetitions (as provided example doesn't).
But if input arrays happen to contain repetitions, that would produce wrong results. Example (using below implementation):
intersect ([[1, 3, 4, 6, 3], [1, 8, 99]]);
// Expected: [ '1' ]
// Actual: [ '1', '3' ]
Fortunately this is easy to fix by simply adding second level indexing. That is:
Change:
if (index[v] === undefined) index[v] = 0;
index[v]++;
by:
if (index[v] === undefined) index[v] = {};
index[v][i] = true; // Mark as present in i input.
...and:
if (index[i] == arrLength) retv.push(i);
by:
if (Object.keys(index[i]).length == arrLength) retv.push(i);
Complete example:
// 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' ]
With some restrictions on your data, you can do it in linear time!
For positive integers: use an array mapping the values to a "seen/not seen" boolean.
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;
}
There is a similar technique for objects: take a dummy key, set it to "true" for each element in array1, then look for this key in elements of array2. Clean up when you're done.
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;
}
Of course you need to be sure the key didn't appear before, otherwise you'll be destroying your data...
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;
}
For simplicity:
// 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]
Benefits:
dirt simple
data-centric
works for arbitrary number of lists
works for arbitrary lengths of lists
works for arbitrary types of values
works for arbitrary sort order
retains shape (order of first appearance in any array)
exits early where possible
memory safe, short of tampering with Function / Array prototypes
Drawbacks:
higher memory usage
higher CPU usage
requires an understanding of reduce
requires understanding of data flow
You wouldn't want to use this for 3D engine or kernel work, but if you have problems getting this to run in an event-based app, your design has bigger problems.
I'll contribute with what has been working out best for me:
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;
};
}
A functional approach with ES2015
A functional approach must consider using only pure functions without side effects, each of which is only concerned with a single job.
These restrictions enhance the composability and reusability of the functions involved.
// 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) );
Please note that the native Set type is used, which has an advantageous
lookup performance.
Avoid duplicates
Obviously repeatedly occurring items from the first Array are preserved, while the second Array is de-duplicated. This may be or may be not the desired behavior. If you need a unique result just apply dedupe to the first argument:
// 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) );
Compute the intersection of any number of Arrays
If you want to compute the intersection of an arbitrarily number of Arrays just compose intersect with foldl. Here is a convenience function:
// 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) );
.reduce to build a map, and .filter to find the intersection. delete within the .filter allows us to treat the second array as though it's a unique set.
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;
});
}
I find this approach pretty easy to reason about. It performs in constant time.
I have written an intesection function which can even detect intersection of array of objects based on particular property of those objects.
For instance,
if arr1 = [{id: 10}, {id: 20}]
and arr2 = [{id: 20}, {id: 25}]
and we want intersection based on the id property, then the output should be :
[{id: 20}]
As such, the function for the same (note: ES6 code) is :
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)));
};
and you can call the function as:
intersect(arr1, arr2, [elem => elem.id, elem => elem.id])
Also note: this function finds intersection considering the first array is the primary array and thus the intersection result will be that of the primary array.
This function avoids the N^2 problem, taking advantage of the power of dictionaries. Loops through each array only once, and a third and shorter loop to return the final result.
It also supports numbers, strings, and objects.
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;
}
If there is a special case that cannot be solved, just by modifying the generateStrKey function, it could surely be solved. The trick of this function is that it uniquely represents each different data according to type and value.
This variant has some performance improvements. Avoid loops in case any array is empty. It also starts by walking through the shorter array first, so if it finds all the values of the first array in the second array, exits the loop.
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;
}
Here is underscore.js implementation:
_.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;
};
Source: http://underscorejs.org/docs/underscore.html#section-62
Create an Object using one array and loop through the second array to check if the value exists as key.
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;
}
I think using an object internally can help with computations and could be performant too.
// Approach maintains a count of each element and works for negative elements too
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]
I am using map even object could be used.
//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;
}
This is a proposed standard: With the currently stage 2 proposal https://github.com/tc39/proposal-set-methods, you could use
mySet.intersection(mySet2);
Until then, you could use Immutable.js's Set, which inspired that proposal
Immutable.Set(mySet).intersect(mySet2)
I extended tarulen's answer to work with any number of arrays. It also should work with non-integer values.
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;
}
If your arrays are sorted, this should run in O(n), where n is 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;
}

Create an array and check against it

I am not sure of how to do this, but what I want to do it create an array and be able to add new items to this array. Since the items are supposed to be a random number, when a new instance is created I would like it to be checked against the rest of the array and be sure that the number it has generated is not already in the array. How would I accomplish something like this?
I looked at Šime Vidas's answer and it seems to work, but I tried to shorten it to
var arr = [];
function add(a) {
var n = ~~(Math.random() * 100);
for (var i = 0; i < a.length; i++) {
if ( a[i] === n) { a.push(n) }
}
}
for (var i=0; i<5; i++){
add(arr)
}
document.getElementById('output').innerHTML += arr;
and I don't understand why this wouldn't work. It does pretty much the same thing, correct?
var arr = [];
function add(a) {
var n = ~~(Math.random() * 1000);
!is(a, n) && a.push(n);
}
function is(a, n) {
for (var i = 0; i < a.length; i++) {
if ( a[i] === n ) { return true; }
}
return false;
}
The add function creates a random integer number between 0 and 1000, and adds it to the array.
The is function checks whether the n number is somewhere inside the a array.
Demo: http://jsfiddle.net/kHhMp/2/
Demo 2: http://jsfiddle.net/kHhMp/3/
(Demo 2 shows that a number will only be added to the array if it's not already in it.)
btw
!is(a, n) && a.push(n);
is a short form of this:
if ( is(a, n) == false ) { a.push(n); }
The number is added to the array only if is(a, n) returns false.
UPDATE
var arr = [];
function add(a) {
var n = ~~(Math.random() * 1000),
ok = true;
for (var i = 0; i < a.length; i++) {
if ( a[i] === n ) { ok = false; }
}
ok && a.push(n);
}
If you enjoy fast code and you have many items in your array, you should use an Object rather than an Array.
Instead of doing var my_array=[]; my_array.push(my_number), use var my_object = {}; my_object[my_number] = true to add items in your structure.
With that approach, you can easily check if a new number is already in there with an if (my_object[my_number]) { /* already there */ } else { /* not there yet */ }
Once you're done, you can extract the list of numbers as an array by either using var keys = Object.keys(my_object), or if that's not available, var keys=[],i=0;for (keys[i++] in my_object);, more or less.
You may extend the built in Array object for your needs.
Array.prototype.pushUnique = function(value) {
var len = this.length;
for(var i = 0; i < len; i++) {
if(this[i]===value) return;
}
this.push(value);
}
var uniques = new Array();
uniques.pushUnique(1);
uniques.pushUnique(2);
uniques.pushUnique(1);
// array will contain only 1 and 2
The fastest, most cross-browser way is to iterate over the array using a loop:
var arr = [];
function addNum(num) {
for (i = 0, len = arr.length; i < len; i++) {
if ( arr[i] === num ) { return false; }
}
arr.push(num);
}
Be sure to get the length of the array before you run the loop so the length property isn't constantly checked.
var array = []
array[0] = 'Item'
array[0] === undefined
# returns false
array[1] === undefined
# returns true

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