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In a Given Array, I am trying to get the Value that are having a even number of occurences,

I am trying to get my hand dirty with Javascript
I want to solve the following exercise (found on Codewars):
Given an array of integers, find the one that appears an odd number of times.
There will always be only one integer that appears an odd number of
times.
[7] should return 7, because it occurs 1 time (which is odd). [0]
should return 0, because it occurs 1 time (which is odd). [1,1,2]
should return 2, because it occurs 1 time (which is odd). [0,1,0,1,0]
should return 0, because it occurs 3 times (which is odd).
[1,2,2,3,3,3,4,3,3,3,2,2,1] should return 4, because it appears 1 time
(which is odd).
this is my code:
function findOdd(A) {
const isOdd = (x) => {
return x%2;
}
const counts ={};
for ( i of A)
{
counts[i]= counts[i] ? counts[i] +1:1;
}
const numOccurences = (Object.values(counts));
const occurences = Object.getOwnPropertyNames(counts);
let evenOccurence=0;
let oddOccurence;
for (let y=0;y<numOccurences.length;y++)
{
if(isOdd(numOccurences[y]))
{
console.log("numOccurences[y] is equal to: ",numOccurences[y])
evenOccurence = numOccurences[y];
console.log(`value ${occurences[y]} is appearing ${numOccurences} times which is even`)
}
// }
// console.log("evenOccurence",evenOccurence);
// return evenOccurence;
}
return evenOccurence;
}
// console.log(findOdd([7]));
console.log(findOdd([7,7,8,8,7,5,5,5,5,5,5,5]));
my issue is the value I am putting in evenOccurrence, I would expect an integer, but it comes as 7,3,2 in the array given as an example.
could anyone help me please?
Thanks in advance

You could take a bitwise XOR ^ which wipes out even count values.
const
odd = array => array.reduce((a, b) => a ^ b, 0);
console.log(odd([1, 2, 2, 3, 3, 3, 4, 3, 3, 3, 2, 2, 1]));
console.log(odd([7]));
console.log(odd([1, 1, 2]));
Another approach. With an object for keeping the state. The value is either true or false depending on the odd appearance of the value of the array.
Finally take all keys and find the one with the true value.
const
odd = array => {
const states = {};
for (const value of array) states[value] = !states[value];
return +Object.keys(states).find(k => states[k]);
};
console.log(odd([1, 2, 2, 3, 3, 3, 4, 3, 3, 3, 2, 2, 1]));
console.log(odd([7]));
console.log(odd([1, 1, 2]));

Updated your answer with comment.
function findOdd(A) {
const isOdd = (x) => {
return x % 2;
}
const counts = {}; // key with will be unique set of A and value will be part will be occurence count
for (i of A) {
counts[i] = counts[i] ? counts[i] + 1 : 1;
}
// const numOccurences = (Object.values(counts)); // her you are just getting list of how many time they occured, without any reference to A or their keys
// const occurences = Object.getOwnPropertyNames(counts);
let evenOccurence = []; // if you want to return a list, create a list and push values in it
for (let key of Object.keys(counts)) { // key is unique value of A
let occurenceCount = counts[key];
if (isOdd(occurenceCount)) {
console.log("numOccurences of key [", key,"] is equal to: ", occurenceCount, ' odd')
} else {
console.log("numOccurences of key [", key,"] is equal to: ", occurenceCount, ' event')
evenOccurence.push(key); }
}
return evenOccurence;
}
console.log(findOdd([7, 7, 8, 8, 7, 5, 5, 5, 5, 5, 5, 5]));

#Nina Scholz's first algorithm is a masterpiece. Basically it is very simple if you are ok with bitwise operations. I just would like to simplify the same thing down to daily math.
var odd = array => array.sort().reduce((a,b,i) => i%2 ? a-b : a+b);
console.log(odd([1, 2, 2, 3, 3, 3, 4, 3, 3, 3, 2, 2, 1]));
console.log(odd([7]));
console.log(odd([1, 1, 2]));

Returning the first two odd numbers in an array

I have to run a code that picks out the odd numbers in an array, and returns the first two odd numbers in that array. I've figured out how to return all of the odd numbers from the array, just not the first two.
code so far:
function oddCouple(_numberList) {
let numberList = [5, 2, 3, 1];
let oddNumbers = numberList.filter((number) => number % 2 !== 0);
console.log(oddNumbers);
}
oddCouple();
Any help would be appreciated!

Array.slice returns the selected number of items from the array starting from a specified index.
function oddCouple(_numberList) {
let oddNumbers = _numberList.filter((number) => number % 2 !== 0).slice(0, 2);
console.log(oddNumbers);
}
oddCouple([1, 2, 3, 4, 5, 6, 7, 8, 9]);

The built in method would be Array.slice. This code would be suitable for a small arrays.
For larger arrays it would be more efficient to manually loop through the source array so you are only evaluating the modulus operation for the first N odd numbers.
Additionally, division is expensive. You can check if a number is odd using the binary & operator; e.g.: (number & 1) == 1
function oddCouple(_numberList) {
let numberList = [5, 2, 3, 1];
let oddNumbers = numberList.filter((number) => (number & 1) == 1).slice(0,2);
console.log(oddNumbers);
}
oddCouple();

You can use for..of loop, and break when the array contains 2 items like so:
function oddCouple(_numberList) {
let numberList = [5, 2, 3, 1];
let arr = [];
for(let number of numberList) {
(number % 2) !== 0 && arr.push(number);
if(arr.length === 2) break;
}
console.log(arr);
}
oddCouple();

function oddCouple(_numberList) {
let numberList = [5, 2, 3, 1];
let oddNumbers = numberList.filter((number) => number % 2 !== 0).slice(0,2);
console.log(oddNumbers);
}

check if values in array have common difference

I have a sample array as
const arr=[0,2,4,6]
I want to check if they have a common difference in them, which is 2 and yes in above array.
There could be more efficient way to do this. I tried using a simple for loop as:
const arr = [0,2,4,6];
for(let i=0;i<arr.length-1;i++){
if(arr[i]+2==arr[i+1]){
console.log(true);
}
else{
console.log(false);
}
}
This could give me result as true,true,true and push all values to a new array then check if all are true and finally I would get true
It would be helpful if there is a more efficient way to do this; any help appreciated.

Assuming the array is sorted, you can do it pretty simply with .every().
const arr = [0,2,4,6];
const itvl = arr[1] - arr[0];
const result = arr.slice(0, -1).every((n, i) =>
itvl === arr[i + 1] - n
)
console.log(result);
This starts off by calculating the first interval, then it iterates the array (except the last index) using .every() to test that the next index minus the current one is equal to that pre-calculated index.

You can do this. It's better to stop as soon as you find a false rather than testing all the values.
const arr=[0,2,4,6]
var r = true
for (var i=1; i<arr.length-1 && r; i++) {
r = ((arr[i+1] - arr[i]) == (arr[i] - arr[i-1]))
}
console.log(r)

I assume what you mean by difference is the step value. For that, first find the difference between the first two elements ; then check whether the same difference holds for all consecutive values.
const arr = [0, 2, 4, 6];
let flag = true;
diff = arr[1] - arr[0];
for(let i = 1; i < arr.length-1; i++){
if(arr[i+1] - arr[i] != diff ){
flag = false;
break;
}
}
console.log(flag)
The code assumes the array size is atleast 2. Add an if guard if this is not always guaranteed.

Checks if numbers sum up to desired sum and if rest of numbers have same difference as desired in comments. (This should be a separate question.)
add up until s >= sum. return false if s isn't equal to set sum
if numbers left <= 2 return true
get difference of first 2 numbers
check difference of numbers, if not same difference return false
differences all same, return true
let arr = [0, 1, 3, 4, 6, 8, 10, 12];
const sum = 4
check = arr => {
let s = 0,
idx = arr.findIndex(x => (s+=x) >= sum ) + 1
if(s !== sum) return false
if(arr.length - idx < 3) return true
const diff = arr[idx+1] - arr[idx]
while(++idx<arr.length-1)
if(arr[idx+1] - arr[idx] !== diff) return false
return true
}
arr = [0, 1, 3, 4, 6, 8, 10, 12]
console.log(
check(arr)
)
arr = [0, 1, 3]
console.log(
check(arr)
)
arr = [0, 1, 3, 4, 6, 8, 10, 12, 11]
console.log(
check(arr)
)

Javascript have a function return no value

I've been trying to make a function that will not return any value if two values are the same between two arrays. I've searched my question several ways on here and on google, and everything I find says that a function HAS to return a value, and there's no way to avoid doing it. So I'm directly asking if there is a way, and if not how can I get around what I'm trying to do?
var bannedValues = [2, 4, 6, 8]; //Even numbers 1-8
var inputValues = [1, 2, 3, 4, 5, 6, 7, 8]; //All numbers 1-8
var filterValues = function(value)
{
for (var number of bannedValues)
{
if (value == number)
{
return; //I want this to return absolutely nothing, not even an empty value
}
}
return value;
}
var filteredValues = inputValues.map(filterValues);
console.log(filteredValues);
/* The array just gains "undefined" values in the places I want there
to be no value, so the array is a length of 8 instead of the 4 I want.
Can I somehow take those out or can I just not have them there in the first place? */

If you are using map, you are actually iterating through your array and manipulate it if necessary, but this will not remove item from array. Instead of map try filter
var bannedValues = [2, 4, 6, 8]; //Even numbers 1-8
var inputValues = [1, 2, 3, 4, 5, 6, 7, 8]; //All numbers 1-8
var filterValues = function(value) {
return bannedValues.indexOf(value)===-1;
}
var filteredValues = inputValues.filter(filterValues);
console.log(filteredValues);
result will be
(4) [1, 3, 5, 7]

Use the Array.filter() method:
var bannedValues = [2, 4, 6, 8]; //Even numbers 1-8
var inputValues = [1, 2, 3, 4, 5, 6, 7, 8]; //All numbers 1-8
filteredValues = inputValues.filter( function( el ) {
return bannedValues.indexOf( el ) < 0;
} );
/*This is another way to do it, but not supported in IE*/
filteredValues = inputValues.filter( function( el ) {
return !bannedValues.includes( el );
} );
console.log(filteredValues);
Lodash has an utility function for this as well: see here

JavaScript functions always have a return value. There's nothing you can do about that.
It's impossible I think to tell the difference between a function that explicitly returns undefined and one that just returns with no value expression.
If you want to do something like .map() but have the option of skipping values, use .reduce() instead:
var filterValues = function(accumulator, value)
{
for (var number of bannedValues)
{
if (value == number)
{
return accumulator;
}
}
accumulator.push(value);
return accumulator;
}
var filteredValues = inputValues.reduce(filterValues, []);

How to count certain elements in array?

I have an array:
[1, 2, 3, 5, 2, 8, 9, 2]
I would like to know how many 2s are in the array.
What is the most elegant way to do it in JavaScript without looping with for loop?

[this answer is a bit dated: read the edits, in the notion of 'equal' in javascript is ambiguous]
Say hello to your friends: map and filter and reduce and forEach and every etc.
(I only occasionally write for-loops in javascript, because of block-level scoping is missing, so you have to use a function as the body of the loop anyway if you need to capture or clone your iteration index or value. For-loops are more efficient generally, but sometimes you need a closure.)
The most readable way:
[....].filter(x => x==2).length
(We could have written .filter(function(x){return x==2}).length instead)
The following is more space-efficient (O(1) rather than O(N)), but I'm not sure how much of a benefit/penalty you might pay in terms of time (not more than a constant factor since you visit each element exactly once):
[....].reduce((total,x) => (x==2 ? total+1 : total), 0)
or as a commenter kindly pointed out:
[....].reduce((total,x) => total+(x==2), 0)
(If you need to optimize this particular piece of code, a for loop might be faster on some browsers... you can test things on jsperf.com.)
You can then be elegant and turn it into a prototype function:
[1, 2, 3, 5, 2, 8, 9, 2].count(2)
Like this:
Object.defineProperties(Array.prototype, {
count: {
value: function(value) {
return this.filter(x => x==value).length;
}
}
});
You can also stick the regular old for-loop technique (see other answers) inside the above property definition (again, that would likely be much faster).
2017 edit:
Whoops, this answer has gotten more popular than the correct answer. Actually, just use the accepted answer. While this answer may be cute, the js compilers probably don't (or can't due to spec) optimize such cases. So you should really write a simple for loop:
Object.defineProperties(Array.prototype, {
count: {
value: function(query) {
/*
Counts number of occurrences of query in array, an integer >= 0
Uses the javascript == notion of equality.
*/
var count = 0;
for(let i=0; i<this.length; i++)
if (this[i]==query)
count++;
return count;
}
}
});
You could define a version .countStrictEq(...) which used the === notion of equality. The notion of equality may be important to what you're doing! (for example [1,10,3,'10'].count(10)==2, because numbers like '4'==4 in javascript... hence calling it .countEq or .countNonstrict stresses it uses the == operator.)
Caveat:
Defining a common name on the prototype should be done with care. It is fine if you control your code, but bad if everyone wants to declare their own [].count function, especially if they behave differently. You may ask yourself "but .count(query) surely sounds quite perfect and canonical"... but consider perhaps you could do something like [].count(x=> someExpr of x). In that case you define functions like countIn(query, container) (under myModuleName.countIn), or something, or [].myModuleName_count().
Also consider using your own multiset data structure (e.g. like python's 'collections.Counter') to avoid having to do the counting in the first place. This works for exact matches of the form [].filter(x=> x==???).length (worst case O(N) down to O(1)), and modified will speed up queries of the form [].filter(filterFunction).length (roughly by a factor of #total/#duplicates).
class Multiset extends Map {
constructor(...args) {
super(...args);
}
add(elem) {
if (!this.has(elem))
this.set(elem, 1);
else
this.set(elem, this.get(elem)+1);
}
remove(elem) {
var count = this.has(elem) ? this.get(elem) : 0;
if (count>1) {
this.set(elem, count-1);
} else if (count==1) {
this.delete(elem);
} else if (count==0)
throw `tried to remove element ${elem} of type ${typeof elem} from Multiset, but does not exist in Multiset (count is 0 and cannot go negative)`;
// alternatively do nothing {}
}
}
Demo:
> counts = new Multiset([['a',1],['b',3]])
Map(2) {"a" => 1, "b" => 3}
> counts.add('c')
> counts
Map(3) {"a" => 1, "b" => 3, "c" => 1}
> counts.remove('a')
> counts
Map(2) {"b" => 3, "c" => 1}
> counts.remove('a')
Uncaught tried to remove element a of type string from Multiset, but does not exist in Multiset (count is 0 and cannot go negative)
sidenote: Though, if you still wanted the functional-programming way (or a throwaway one-liner without overriding Array.prototype), you could write it more tersely nowadays as [...].filter(x => x==2).length. If you care about performance, note that while this is asymptotically the same performance as the for-loop (O(N) time), it may require O(N) extra memory (instead of O(1) memory) because it will almost certainly generate an intermediate array and then count the elements of that intermediate array.

Modern JavaScript:
Note that you should always use triple equals === when doing comparison in JavaScript (JS). The triple equals make sure, that JS comparison behaves like double equals == in other languages (there is one exception, see below). The following solution shows how to solve this the functional way, which will ensure that you will never have out of bounds error:
// Let has local scope
let array = [1, 2, 3, 5, 2, 8, 9, 2]
// Functional filter with an Arrow function
// Filter all elements equal to 2 and return the length (count)
array.filter(x => x === 2).length // -> 3
The following anonymous Arrow function (lambda function) in JavaScript:
(x) => {
const k = 2
return k * x
}
may be simplified to this concise form for a single input:
x => 2 * x
where the return is implied.
Always use triple equals: === for comparison in JS, with the exception of when checking for nullability: if (something == null) {} as it includes a check for undefined, if you only use double equals as in this case.

Very simple:
var count = 0;
for(var i = 0; i < array.length; ++i){
if(array[i] == 2)
count++;
}

2017:
If someone is still interested in the question, my solution is the following:
const arrayToCount = [1, 2, 3, 5, 2, 8, 9, 2];
const result = arrayToCount.filter(i => i === 2).length;
console.log('number of the found elements: ' + result);

Here is an ES2017+ way to get the counts for all array items in O(N):
const arr = [1, 2, 3, 5, 2, 8, 9, 2];
const counts = {};
arr.forEach((el) => {
counts[el] = counts[el] ? (counts[el] + 1) : 1;
});
You can also optionally sort the output:
const countsSorted = Object.entries(counts).sort(([_, a], [__, b]) => a - b);
console.log(countsSorted) for your example array:
[
[ '2', 3 ],
[ '1', 1 ],
[ '3', 1 ],
[ '5', 1 ],
[ '8', 1 ],
[ '9', 1 ]
]

If you are using lodash or underscore the _.countBy method will provide an object of aggregate totals keyed by each value in the array. You can turn this into a one-liner if you only need to count one value:
_.countBy(['foo', 'foo', 'bar'])['foo']; // 2
This also works fine on arrays of numbers. The one-liner for your example would be:
_.countBy([1, 2, 3, 5, 2, 8, 9, 2])[2]; // 3

Weirdest way I can think of doing this is:
(a.length-(' '+a.join(' ')+' ').split(' '+n+' ').join(' ').match(/ /g).length)+1
Where:
a is the array
n is the number to count in the array
My suggestion, use a while or for loop ;-)

Not using a loop usually means handing the process over to some method that does use a loop.
Here is a way our loop hating coder can satisfy his loathing, at a price:
var a=[1, 2, 3, 5, 2, 8, 9, 2];
alert(String(a).replace(/[^2]+/g,'').length);
/* returned value: (Number)
3
*/
You can also repeatedly call indexOf, if it is available as an array method, and move the search pointer each time.
This does not create a new array, and the loop is faster than a forEach or filter.
It could make a difference if you have a million members to look at.
function countItems(arr, what){
var count= 0, i;
while((i= arr.indexOf(what, i))!= -1){
++count;
++i;
}
return count
}
countItems(a,2)
/* returned value: (Number)
3
*/

I'm a begin fan of js array's reduce function.
const myArray =[1, 2, 3, 5, 2, 8, 9, 2];
const count = myArray.reduce((count, num) => num === 2 ? count + 1 : count, 0)
In fact if you really want to get fancy you can create a count function on the Array prototype. Then you can reuse it.
Array.prototype.count = function(filterMethod) {
return this.reduce((count, item) => filterMethod(item)? count + 1 : count, 0);
}
Then do
const myArray =[1, 2, 3, 5, 2, 8, 9, 2]
const count = myArray.count(x => x==2)

Most of the posted solutions using array functions such as filter are incomplete because they aren't parameterized.
Here goes a solution with which the element to count can be set at run time.
function elementsCount(elementToFind, total, number){
return total += number==elementToFind;
}
var ar = [1, 2, 3, 5, 2, 8, 9, 2];
var elementToFind=2;
var result = ar.reduce(elementsCount.bind(this, elementToFind), 0);
The advantage of this approach is that could easily change the function to count for instance the number of elements greater than X.
You may also declare the reduce function inline
var ar = [1, 2, 3, 5, 2, 8, 9, 2];
var elementToFind=2;
var result = ar.reduce(function (elementToFind, total, number){
return total += number==elementToFind;
}.bind(this, elementToFind), 0);

Really, why would you need map or filter for this?
reduce was "born" for these kind of operations:
[1, 2, 3, 5, 2, 8, 9, 2].reduce( (count,2)=>count+(item==val), 0);
that's it! (if item==val in each iteration, then 1 will be added to the accumulator count, as true will resolve to 1).
As a function:
function countInArray(arr, val) {
return arr.reduce((count,item)=>count+(item==val),0)
}
Or, go ahead and extend your arrays:
Array.prototype.count = function(val) {
return this.reduce((count,item)=>count+(item==val),0)
}

It is better to wrap it into function:
let countNumber = (array,specificNumber) => {
return array.filter(n => n == specificNumber).length
}
countNumber([1,2,3,4,5],3) // returns 1

I use this:
function countElement(array, element) {
let tot = 0;
for(var el of array) {
if(el == element) {
tot++;
}
}
return tot;
}
var arr = ["a", "b", "a", "c", "d", "a", "e", "f", "a"];
console.log(countElement(arr, "a")); // 4

var arrayCount = [1,2,3,2,5,6,2,8];
var co = 0;
function findElement(){
arrayCount.find(function(value, index) {
if(value == 2)
co++;
});
console.log( 'found' + ' ' + co + ' element with value 2');
}
I would do something like that:
var arrayCount = [1,2,3,4,5,6,7,8];
function countarr(){
var dd = 0;
arrayCount.forEach( function(s){
dd++;
});
console.log(dd);
}

I believe what you are looking for is functional approach
const arr = ['a', 'a', 'b', 'g', 'a', 'e'];
const count = arr.filter(elem => elem === 'a').length;
console.log(count); // Prints 3
elem === 'a' is the condition, replace it with your own.

Array.prototype.count = function (v) {
var c = 0;
for (let i = 0; i < this.length; i++) {
if(this[i] === v){
c++;
}
}
return c;
}
var arr = [1, 2, 3, 5, 2, 8, 9, 2];
console.log(arr.count(2)); //3

Solution by recursion
function count(arr, value) {
if (arr.length === 1) {
return arr[0] === value ? 1 : 0;
} else {
return (arr.shift() === value ? 1 : 0) + count(arr, value);
}
}
count([1,2,2,3,4,5,2], 2); // 3

Create a new method for Array class in core level file and use it all over your project.
// say in app.js
Array.prototype.occurrence = function(val) {
return this.filter(e => e === val).length;
}
Use this anywhere in your project -
[1, 2, 4, 5, 2, 7, 2, 9].occurrence(2);
// above line returns 3

Here is a one liner in javascript.
Use map. Find the matching values (v === 2) in the array, returning an array of ones and zeros.
Use Reduce. Add all the values of the array for the total number found.
[1, 2, 3, 5, 2, 8, 9, 2]
.map(function(v) {
return v === 2 ? 1 : 0;
})
.reduce((a, b) => a + b, 0);
The result is 3.

Depending on how you want to run it:
const reduced = (array, val) => { // self explanatory
return array.filter((element) => element === val).length;
}
console.log(reduced([1, 2, 3, 5, 2, 8, 9, 2], 2));
// 3
const reducer = (array) => { // array to set > set.forEach > map.set
const count = new Map();
const values = new Set(array);
values.forEach((element)=> {
count.set(element, array.filter((arrayElement) => arrayElement === element).length);
});
return count;
}
console.log(reducer([1, 2, 3, 5, 2, 8, 9, 2]));
// Map(6) {1 => 1, 2 => 3, 3 => 1, 5 => 1, 8 => 1, …}

You can use built-in function Array.filter()
array.filter(x => x === element).length;
var arr = [1, 2, 3, 5, 2, 8, 9, 2];
// Count how many 2 there are in arr
var count = arr.filter(x => x === 2).length;
console.log(count);

One-liner function
const countBy = (a,f)=>a.reduce((p,v,i,x)=>p+!!f(v,i,x), 0)
countBy([1,2,3,4,5], v=>v%2===0) // 2

There are many ways to find out. I think the easiest way is to use the array filter method which is introduced in es6.
function itemCount(array, item) {
return array.filter(element => element === item).length
}
const myArray = [1,3,5,7,1,2,3,4,5,1,9,0,1]
const items = itemCount(myArray, 1)
console.log(items)

Something a little more generic and modern (in 2022):
import {pipe, count} from 'iter-ops';
const arr = [1, 2, 3, 5, 2, 8, 9, 2];
const n = pipe(arr, count(a => a === 2)).first; //=> 3
What's good about this:
It counts without creating a new array, so it is memory-efficient
It works the same for any Iterable and AsyncIterable

Another approach using RegExp
const list = [1, 2, 3, 5, 2, 8, 9, 2]
const d = 2;
const counter = (`${list.join()},`.match(new RegExp(`${d}\\,`, 'g')) || []).length
console.log(counter)
The Steps follows as below
Join the string using a comma Remember to append ',' after joining so as not to have incorrect values when value to be matched is at the end of the array
Match the number of occurrence of a combination between the digit and comma
Get length of matched items

I believe you can use the new Set array method of JavaScript to have unique values.
Example:
var arr = [1, 2, 3, 5, 2, 8, 9, 2]
var set = new Set(arr);
console.log(set);
// 1,2,3,5,8,9 . We get unique values as output.

You can use length property in JavaScript array:
var myarray = [];
var count = myarray.length;//return 0
myarray = [1,2];
count = myarray.length;//return 2