Develop Reference

Check to see if an array contains all elements of another array, including whether duplicates appear twice

I need to check whether one array contains all of the elements of another array, including the same duplicates. The second array can have extra elements. I'm using every...includes, but it's not catching that the second array doesn't have the right duplicates.
For example:
const arr1 = [1, 2, 2, 3, 5, 5, 6, 6]
const arr2 = [1, 2, 3, 5, 6, 7]
if(arr1.every(elem => arr2.includes(elem))){
return true // should return false because arr2 does not have the same duplicates
}
Thanks!
Edit: arr1 is one of many arrays that I am looping through which are coming out of a graph traversal algorithm, so I'd like to avoid restructuring them into an object to create a dictionary data structure if possible.

Try creating this function:
function containsAll (target, toTest) {
const dictionary = {}
target.forEach(element => {
if (dictionary[element] === undefined) {
dictionary[element] = 1;
return;
}
dictionary[element]++;
});
toTest.forEach(element => {
if (dictionary[element] !== undefined)
dictionary[element]--;
})
for (let key in dictionary) {
if (dictionary[key] > 0) return false;
}
return true;
}
Then invoke it like this:
const arr1 = [1, 2, 2, 3, 5, 5, 6, 6]
const arr2 = [1, 2, 3, 5, 6, 7]
console.log(containsAll(arr1, arr2)) // returns false

const arr1 = [1, 2, 2, 3, 5, 5, 6, 6];
//const arr2 = [1, 2, 3, 5, 6, 7];
const arr2 = [1, 2, 2, 3, 5, 5];
let includesAll1 = true;
let includesAll2 = true;
const checkObj1 = {
};
const checkObj2 = {
};
arr1.forEach((el)=> {
if(checkObj1[el] === undefined) {
checkObj1[el] = 1;
} else {
checkObj1[el]++;
}
});
arr2.forEach((el)=> {
if(checkObj2[el] === undefined) {
checkObj2[el] = 1;
} else {
checkObj2[el]++;
}
});
const check1Keys = Object.keys(checkObj1);
const check2Keys = Object.keys(checkObj2);
if(check1Keys.length > check2Keys.length) {
includesAll2 = false;
check2Keys.forEach((key)=> {
const value1 = checkObj1[key];
const value2 = checkObj2[key];
if(!arr1.includes(parseInt(key)) || value1 != value2) {
includesAll1 = false;
}
});
} else {
includesAll1 = false;
check1Keys.forEach((key)=> {
const value1 = checkObj1[key];
const value2 = checkObj2[key];
console.log(value1, value2, key);
if(!arr2.includes(parseInt(key)) || value1 != value2) {
includesAll2 = false;
}
});
}
console.log(includesAll1);
console.log(includesAll2);

Does this solve your problem?
const arr = [1, 2, 3, 5, 6, 7, 2, 10, 2, 3, 2];
const subArr = [1, 2, 2, 3, 2]
const contains = subArr.every(num => subArr.filter(n => n == num).length <= arr.filter(n => n== num).length);
console.log(contains);

You indicate order does not matter in your comments. That makes this very simple.
Sort both arrays
Check if corresponding elements are equal
consider errors associated with sparse or short arrays
Use .reduce() to boil it down to a single result
So this really comes down to a single statement once the arrays are sorted:
matcher.reduce((acc, value , idx)=>matcher[idx] === test[idx], false);
You also mentioned testing this against many arrays. So the full example below does that for demo purposes.
let toMatch = [1, 2, 2, 3, 5, 5, 6, 6]
let arrayOfArrays = [[1,2],[1, 2, 3, 5, 6, 7, 3, 9, 8, 2, 7],[1, 2, 3, 3, 6, 7],[1, 3, 3, 5, 6, 7],[1, 2, 3, 5, 6, 6], [3,5,2,1,6,2,5,6]];
let toMatchSorted = toMatch.slice().sort();
arrayOfArrays.forEach(arr=>{
let toTestSorted = arr.slice().sort();
let out = toMatchSorted.reduce((acc, value , idx)=>toMatchSorted[idx] === toTestSorted[idx], false);
console.log(`Input: ${arr}, Result: ${out}`);
});

Problem with using for-in/of when returning an object in a function

I just got the result "[object Object]'s score is 0" printed on the terminal.
The result 27 was all fine until I separated the function into a return object.
How do I get 27 if I have to return an object?
How do I get "alex" printed on the console.log instead of [object Object]?
const alex = {
first: [1, 2, 9, 8],
second: [3],
third: [0, 0, 0, 1, 3]
};
const gordon = {
first: [3],
second: [2, 2, 4, 5, 6, 6, 7, 8]
}
function createPlayer(object) {
let score = 0;
return {
add: function() {
for (const key in object) {
for (const item in object[key]) {
score += object[key][item]
}
}
},
result: function() {
return `${object}\'s score is ${score}`
}
}
}
createPlayer(alex).add()
console.log(createPlayer(alex).result())

const alex = {
first: [1, 2, 9, 8],
second: [3],
third: [0, 0, 0, 1, 3]
};
const gordon = {
first: [3],
second: [2, 2, 4, 5, 6, 6, 7, 8]
}
function createPlayer(object, name) {
let score = 0;
return {
add: function() {
for (const key in object) {
for (const item in object[key]) {
score += object[key][item]
}
}
return this; // <<<<<
},
result: function() {
return `${name}\'s score is ${score}`
}
}
}
console.log(createPlayer(alex, 'Alex').add().result())

You would not show alex for an object named alex
You might mean this
const alex = {
Name: "Alex",
first: [1, 2, 9, 8],
second: [3],
third: [0, 0, 0, 1, 3]
};
const gordon = {
Name: "Gordon",
first: [3],
second: [2, 2, 4, 5, 6, 6, 7, 8]
}
function createPlayer(object) {
let score = 0;
return {
add: function() {
for (const key in object) {
if (key!=="Name") {
for (const item in object[key]) {
score += object[key][item]
}
}
}
},
result: function() {
return `${object.Name}\'s score is ${score}`
}
}
}
const player1 = createPlayer(alex)
player1.add()
console.log(player1.result())

Why can't I use an array as the 'specified condition' in a while loop?

So I have this code:
var myArray = [];
var value = 5;
while (myArray != [5, 4, 3, 2, 1, 0]) {
myArray.push(value)
value--;
console.log(myArray);
}
when I look at the console, it goes on for an infinite loop like so..
[ 5 ]
[ 5, 4 ]
[ 5, 4, 3 ]
[ 5, 4, 3, 2 ]
[ 5, 4, 3, 2, 1 ]
[ 5, 4, 3, 2, 1, 0 ]
[ 5, 4, 3, 2, 1, 0, -1 ]
[ 5, 4, 3, 2, 1, 0, -1, -2 ]
[ 5, 4, 3, 2, 1, 0, -1, -2, -3 ]
..........
Why doesn't it stop at [5,4,3,2,1,0] ? myArray = that at a point and the for loop should stop no?
Sorry for the noob question.

JavaScript does not provide built-in support for structural-equality of Arrays, but it's straightforward to implement a comparator:
function arraysEqual(a, b, orderSensitive = true) {
// Function from https://stackoverflow.com/a/16436975/159145
// But modified to add the `orderSensitive` option.
if (a === b) return true;
if (a == null || b == null) return false;
if (a.length != b.length) return false;
if (!orderSensitive) {
a = Array.from(a).sort();
b = Array.from(b).sort();
}
for (var i = 0; i < a.length; ++i) {
if (a[i] !== b[i]) return false;
}
return true;
}
function yourCode() {
var myArray = [];
var value = 5;
const finalArray = [5, 4, 3, 2, 1, 0];
while (!arraysEqual(myArray,finalArray)) {
myArray.push(value)
value--;
console.log(myArray);
}
}

you can use the following way to fix the issue if index and values of both arrays are different
while(sortArray(myArray).toString() !== sortArray([5, 4, 3, 2, 1, 0]).toString()) {
// your code
}
sortArray(array) {
return array.filter((a, b) => a-b)
}
If you are sure the array index and values for both arrays are same you can use
while(myArray.toString() !== [5, 4, 3, 2, 1, 0].toString()) {
// your code
}

#Christopher Barreto and welcome to StackOverflow. Good luck with your interesting question.
although #Dai right with his full answer there is much simpler built in a way to convert array to string and then compare them.
That will work for you:
var myArray = [];
var value = 5;
while (myArray.toString() != [5, 4, 3, 2, 1, 0].toString()) {
myArray.push(value)
value--;
console.log(myArray);
}
Or this if you prefer:
while ((''+myArray) != ('' + [5, 4, 3, 2, 1, 0]))

Removing duplicates in an ordered array

What I'm trying to do is find how many times an array elements repeats itself in array, push the element along with the number of repeats it has in an object and after that delete the element and all its duplicates.
At the moment I have this function :
function getDuplicates(arr) {
let lastIndex = null;
let obj = {};
for ( let i = 0; i < arr.length; i++ ) {
lastIndex = arr.lastIndexOf(arr[i]);
obj[arr[i]] = lastIndex + 1;
arr.splice(0, lastIndex + 1 );
}
console.log(obj);
}
getDuplicates([ 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 6 ]);
which logs : { '1': 4, '2': 2, '3': 4, '5': 5 }
It works great for the first 3 numbers ( 1,2 and 3 ) but 4 doesnt show up, 5 is messed up and 6 doesnt show due to lastIndex +1. Am I missing something or is there a better way to do this ?
Thank you.

You can simplify a lot the logic. Just an object to count and an if statement to increment values or define as 1 if it wasn't defined.
function countDuplicates(arr) {
// Contains a pair of values an instances.
var counting = {};
// Iterate array: check if already counted. If yes, increment, if not define as 1.
for (el of arr) (counting[el]) ? counting[el]++ : counting[el] = 1;
console.log(counting);
return counting;
}
countDuplicates([ 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 6 ]);
Adding, if you also want to get the unique elements, you can just use E6 set:
var set = new Set([ 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 6 ]);

You can count and print as you would want like this:
function getDuplicates(arr) {
var counts = {};
arr.forEach(function(x) { counts[x] = (counts[x] || 0)+1; });
console.log(counts);
}

function getDuplicates(arr) {
let lastNum = null;
let obj = {};
for ( let i = 0; i < arr.length; i++ ) {
if (arr[i] != lastNum){
lastNum = arr[i];
obj[arr[i]] = 1;
}else{
obj[arr[i]]++;
}
}
console.log(obj);
}

You can simply use Array#reduce() to count the occurrences and Array#filter() to remove the duplicates
getDuplicates([1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 6]);
function getDuplicates(arr) {
var obj = arr.reduce((map, item) => (map[item] = ++map[item] || 1, map),{} );
var withoutDup = arr.filter((item, pos) => arr.indexOf(item) == pos);
console.log(JSON.stringify(obj));
console.log(JSON.stringify(withoutDup));
}

Here's one method how to solve it.
Firstly I've removed all duplicated elements from the given array, using new Set() and then iterated over it using Array#forEach and checked with Array#filter how many times given element appears in the passed array.
function getDuplicates(arr){
var filtered = [...new Set(arr)],
result = {};
filtered.forEach(function(v){
result[v] = arr.filter(c => c == v).length;
})
console.log(result);
}
getDuplicates([ 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 6 ]);
Array#reduce solution.
function getDuplicates(arr) {
var res = arr.reduce(function(s, a) {
s[a] = arr.filter(c => c == a).length;
return s;
}, {});
console.log(res);
}
getDuplicates([1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 6]);

It looks as if you want to COUNT duplicates, but if all you want to do is remove duplicates (As headline states), as per #ChantryCargill s suggestion:
function removeDuplicates (arr) {
var results = [];
for(var i = 0; i < arr.length; i++) {
var item = arr[i];
if(results.indexOf(item) === -1) {
results.push(item);
}
}
return results;
}
console.log(removeDuplicates([ 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 6 ]));
//[1, 2, 3, 4, 5, 6]
If you want to COUNT duplicates:
function getDuplicates(arr) {
var results = {};
for(var item of arr) {
if(!results[item]) {
results[item] = 0;
}
results[item]++;
}
return results;
}
console.log(getDuplicates([ 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 6 ]));
//{"1":4,"2":2,"3":4,"4":2,"5":3,"6":1}

Try this:
function getDuplicates(){
var numbers=Array.prototype.slice.call(arguments);
var duplicates={};
for(var index in numbers){
if(numbers.indexOf(numbers[index])==index)
continue;
duplicates[numbers[index]]= (duplicates[numbers[index]] || 0) + 1;
}
return duplicates;
}
console.log(getDuplicates(1,2,3,1,1,3,4,5,6,7,8,6));
/*
prints {
1: 2,
3: 1,
6: 1
}
*/

How to remove duplicate from object list in javasccript

$(".search").keyup(function(){
var val = this.value;
$.ajax({
type: "POST",
url: "search?entered=" + val,
beforeSend: function(){
$(".search").css("background","#FFF");
},
success: function(data){
for (var i = 0; i < data.length; i++) {
$(".suggesstions").append("<ul>"+data[i]["category"]+" "+data[i]["productTitle"]+"</ul>");
}
}
});
});
Here is my code,
I want to remove duplicate entries from list and append to.

I modified the code with remove duplicate value based on productTitle.
success: function(data){
var array = [],
Finalresult = [];
$.each(data, function (index, value) {
if ($.inArray(value.productTitle, array) == -1) {
array.push(value.productTitle);
Finalresult.push(value);
}
});
for (var i = 0; i < Finalresult.length; i++) {
$(".suggesstions").append("<ul>"+Finalresult[i]["category"]+" "+Finalresult[i]["productTitle"]+"</ul>");
}
console.log(Finalresult);
}

First, you have to remove all duplicate entries from array , like this
var myArr = [2, 5, 5, 7, 7, 8, 9, 9, 9, 1, 3, 3, 4, 5];
var newArr = $.unique(myArr.sort()).sort();
for (var i = 0; i < newArr.length; i++) {
//your code
}

One Liner
var arrOutput = Array.from(new Set([1,2,3,4,5,5,6,7,8,9,6,7]));
alert(arrOutput);

let myArray = [2, 5, 5, 7, 7, 8, 9, 9, 9, 1, 3, 3, 4, 5];
let arrayOutput = [];
myArray?.map((c) => {
if (!arrayOutput.includes(c)) {
arrayOutput.push(c);
}
});
console.log(arrayOutput)

let myArray = [2, 5, 5, 7, 7, 8, 9, 9, 9, 1, 3, 3, 4, 5];
let arrayOutput = [];
myArray?.map((c) => {
(!(arrayOutput.indexOf(c) > -1)) ? arrayOutput.push(c) : null;
});
console.log(arrayOutput)