How to count duplicate value in an array in javascript
JavascriptArraysJavascript Problem Overview
Currently, I got an array like that:
var uniqueCount = Array();
After a few steps, my array looks like that:
uniqueCount = [a,b,c,d,d,e,a,b,c,f,g,h,h,h,e,a];
How can I count how many a,b,c are there in the array? I want to have a result like:
a = 3
b = 1
c = 2
d = 2
etc.
Javascript Solutions
Solution 1 - Javascript
const counts = {};
const sampleArray = ['a', 'a', 'b', 'c'];
sampleArray.forEach(function (x) { counts[x] = (counts[x] || 0) + 1; });
console.log(counts)
Solution 2 - Javascript
Something like this:
uniqueCount = ["a","b","c","d","d","e","a","b","c","f","g","h","h","h","e","a"];
var count = {};
uniqueCount.forEach(function(i) { count[i] = (count[i]||0) + 1;});
console.log(count);
Use a simple for loop instead of forEach if you don't want this to break in older browsers.
Solution 3 - Javascript
I stumbled across this (very old) question. Interestingly the most obvious and elegant solution (imho) is missing: Array.prototype.reduce(...). All major browsers support this feature since about 2011 (IE) or even earlier (all others):
var arr = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
var map = arr.reduce(function(prev, cur) {
prev[cur] = (prev[cur] || 0) + 1;
return prev;
}, {});
// map is an associative array mapping the elements to their frequency:
console.log(map);
// prints {"a": 3, "b": 2, "c": 2, "d": 2, "e": 2, "f": 1, "g": 1, "h": 3}
EDIT:
By using the comma operator in an arrow function, we can write it in one single line of code:
var arr = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
var map = arr.reduce((cnt, cur) => (cnt[cur] = cnt[cur] + 1 || 1, cnt), {});
// map is an associative array mapping the elements to their frequency:
console.log(map);
// prints {"a": 3, "b": 2, "c": 2, "d": 2, "e": 2, "f": 1, "g": 1, "h": 3}
However, as this may be harder to read/understand, one should probably stick to the first version.
Solution 4 - Javascript
function count() {
array_elements = ["a", "b", "c", "d", "e", "a", "b", "c", "f", "g", "h", "h", "h", "e", "a"];
array_elements.sort();
var current = null;
var cnt = 0;
for (var i = 0; i < array_elements.length; i++) {
if (array_elements[i] != current) {
if (cnt > 0) {
document.write(current + ' comes --> ' + cnt + ' times<br>');
}
current = array_elements[i];
cnt = 1;
} else {
cnt++;
}
}
if (cnt > 0) {
document.write(current + ' comes --> ' + cnt + ' times');
}
}
count();
http://jsfiddle.net/aQsuP/9/"> Demo Fiddle
You can use higher-order functions too to do the operation. See this answer
Solution 5 - Javascript
Simple is better, one variable, one function :)
const arr = ["a", "b", "c", "d", "d", "e", "a", "b", "c", "f", "g", "h", "h", "h", "e", "a"];
const counts = arr.reduce((acc, value) => ({
...acc,
[value]: (acc[value] || 0) + 1
}), {});
console.log(counts);
Solution 6 - Javascript
Single line based on reduce array function
const uniqueCount = ["a", "b", "c", "d", "d", "e", "a", "b", "c", "f", "g", "h", "h", "h", "e", "a"];
const distribution = uniqueCount.reduce((acum,cur) => Object.assign(acum,{[cur]: (acum[cur] || 0)+1}),{});
console.log(JSON.stringify(distribution,null,2));
Solution 7 - Javascript
// Initial array
let array = ['a', 'b', 'c', 'd', 'd', 'e', 'a', 'b', 'c', 'f', 'g', 'h', 'h', 'h', 'e', 'a'];
// Unique array without duplicates ['a', 'b', ... , 'h']
let unique = [...new Set(array)];
// This array counts duplicates [['a', 3], ['b', 2], ... , ['h', 3]]
let duplicates = unique.map(value => [value, array.filter(str => str === value).length]);
Solution 8 - Javascript
Nobody responding seems to be using the Map() built-in for this, which tends to be my go-to combined with Array.prototype.reduce():
const data = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a']; const result = data.reduce((a, c) => a.set(c, (a.get(c) || 0) + 1), new Map()); console.log(...result);
N.b., you'll have to polyfill Map() if wanting to use it in older browsers.
Solution 9 - Javascript
You can have an object that contains counts. Walk over the list and increment the count for each element:
var counts = {};
uniqueCount.forEach(function(element) {
counts[element] = (counts[element] || 0) + 1;
});
for (var element in counts) {
console.log(element + ' = ' + counts[element]);
}
Solution 10 - Javascript
I think this is the simplest way how to count occurrences with same value in array.
var a = [true, false, false, false];
a.filter(function(value){
return value === false;
}).length
Solution 11 - Javascript
You can solve it without using any for/while loops ou forEach.
function myCounter(inputWords) {
return inputWords.reduce( (countWords, word) => {
countWords[word] = ++countWords[word] || 1;
return countWords;
}, {});
}
Hope it helps you!
Solution 12 - Javascript
// new example.
var str= [20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5];
function findOdd(para) {
var count = {};
para.forEach(function(para) {
count[para] = (count[para] || 0) + 1;
});
return count;
}
console.log(findOdd(str));
Solution 13 - Javascript
const obj = {};
const uniqueCount = [ 'a', 'b', 'c', 'd', 'e', 'a', 'b', 'c', 'f', 'g', 'h', 'h', 'h', 'e', 'a' ];
for (let i of uniqueCount) obj[i] ? obj[i]++ : (obj[i] = 1);
console.log(obj);
Solution 14 - Javascript
You can do something like that:
uniqueCount = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
var map = new Object();
for(var i = 0; i < uniqueCount.length; i++) {
if(map[uniqueCount[i]] != null) {
map[uniqueCount[i]] += 1;
} else {
map[uniqueCount[i]] = 1;
}
}
now you have a map with all characters count
Solution 15 - Javascript
It is simple in javascript using array reduce method:
const arr = ['a','d','r','a','a','f','d'];
const result = arr.reduce((json,val)=>({...json, [val]:(json[val] | 0) + 1}),{});
console.log(result)
//{ a:3,d:2,r:1,f:1 }
Solution 16 - Javascript
Duplicates in an array containing alphabets:
var arr = ["a", "b", "a", "z", "e", "a", "b", "f", "d", "f"],
sortedArr = [],
count = 1;
sortedArr = arr.sort();
for (var i = 0; i < sortedArr.length; i = i + count) {
count = 1;
for (var j = i + 1; j < sortedArr.length; j++) {
if (sortedArr[i] === sortedArr[j])
count++;
}
document.write(sortedArr[i] + " = " + count + "<br>");
}
Duplicates in an array containing numbers:
var arr = [2, 1, 3, 2, 8, 9, 1, 3, 1, 1, 1, 2, 24, 25, 67, 10, 54, 2, 1, 9, 8, 1],
sortedArr = [],
count = 1;
sortedArr = arr.sort(function(a, b) {
return a - b
});
for (var i = 0; i < sortedArr.length; i = i + count) {
count = 1;
for (var j = i + 1; j < sortedArr.length; j++) {
if (sortedArr[i] === sortedArr[j])
count++;
}
document.write(sortedArr[i] + " = " + count + "<br>");
}
Solution 17 - Javascript
uniqueCount = ["a","b","a","c","b","a","d","b","c","f","g","h","h","h","e","a"];
var count = {};
uniqueCount.forEach((i) => { count[i] = ++count[i]|| 1});
console.log(count);
Solution 18 - Javascript
CODE:
function getUniqueDataCount(objArr, propName) {
var data = [];
if (Array.isArray(propName)) {
propName.forEach(prop => {
objArr.forEach(function(d, index) {
if (d[prop]) {
data.push(d[prop]);
}
});
});
} else {
objArr.forEach(function(d, index) {
if (d[propName]) {
data.push(d[propName]);
}
});
}
var uniqueList = [...new Set(data)];
var dataSet = {};
for (var i = 0; i < uniqueList.length; i++) {
dataSet[uniqueList[i]] = data.filter(x => x == uniqueList[i]).length;
}
return dataSet;
}
Snippet
var data= [
{day:'Friday' , name: 'John' },
{day:'Friday' , name: 'John' },
{day:'Friday' , name: 'Marium' },
{day:'Wednesday', name: 'Stephanie' },
{day:'Monday' , name: 'Chris' },
{day:'Monday' , name: 'Marium' },
];
console.log(getUniqueDataCount(data, ['day','name']));
function getUniqueDataCount(objArr, propName) {
var data = [];
if (Array.isArray(propName)) {
propName.forEach(prop => {
objArr.forEach(function(d, index) {
if (d[prop]) {
data.push(d[prop]);
}
});
});
} else {
objArr.forEach(function(d, index) {
if (d[propName]) {
data.push(d[propName]);
}
});
}
var uniqueList = [...new Set(data)];
var dataSet = {};
for (var i = 0; i < uniqueList.length; i++) {
dataSet[uniqueList[i]] = data.filter(x => x == uniqueList[i]).length;
}
return dataSet;
}
Solution 19 - Javascript
var uniqueCount = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
// here we will collect only unique items from the array
var uniqueChars = [];
// iterate through each item of uniqueCount
for (i of uniqueCount) {
// if this is an item that was not earlier in uniqueCount,
// put it into the uniqueChars array
if (uniqueChars.indexOf(i) == -1) {
uniqueChars.push(i);
}
}
// after iterating through all uniqueCount take each item in uniqueChars
// and compare it with each item in uniqueCount. If this uniqueChars item
// corresponds to an item in uniqueCount, increase letterAccumulator by one.
for (x of uniqueChars) {
let letterAccumulator = 0;
for (i of uniqueCount) {
if (i == x) {letterAccumulator++;}
}
console.log(`${x} = ${letterAccumulator}`);
}
Solution 20 - Javascript
var testArray = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
var newArr = [];
testArray.forEach((item) => {
newArr[item] = testArray.filter((el) => {
return el === item;
}).length;
})
console.log(newArr);
Solution 21 - Javascript
simplified sheet.js answare
var counts = {};
var aarr=['a','b','a'];
aarr.forEach(x=>counts[x]=(counts[x] || 0)+1 );
console.log(counts)
Solution 22 - Javascript
A combination of good answers:
var count = {};
var arr = ['a', 'b', 'c', 'd', 'd', 'e', 'a', 'b', 'c', 'f', 'g', 'h', 'h', 'h', 'e', 'a'];
var iterator = function (element) {
count[element] = (count[element] || 0) + 1;
}
if (arr.forEach) {
arr.forEach(function (element) {
iterator(element);
});
} else {
for (var i = 0; i < arr.length; i++) {
iterator(arr[i]);
}
}
Hope it's helpful.
Solution 23 - Javascript
By using array.map we can reduce the loop, see this on jsfiddle
function Check(){
var arr = Array.prototype.slice.call(arguments);
var result = [];
for(i=0; i< arr.length; i++){
var duplicate = 0;
var val = arr[i];
arr.map(function(x){
if(val === x) duplicate++;
})
result.push(duplicate>= 2);
}
return result;
}
To Test:
var test = new Check(1,2,1,4,1);
console.log(test);
Solution 24 - Javascript
var string = ['a','a','b','c','c','c','c','c','a','a','a'];
function stringCompress(string){
var obj = {},str = "";
string.forEach(function(i) {
obj[i] = (obj[i]||0) + 1;
});
for(var key in obj){
str += (key+obj[key]);
}
console.log(obj);
console.log(str);
}stringCompress(string)
/*
Always open to improvement ,please share
*/
Solution 25 - Javascript
Create a file for example demo.js and run it in console with node demo.js and you will get occurrence of elements in the form of matrix.
var multipleDuplicateArr = Array(10).fill(0).map(()=>{return Math.floor(Math.random() * Math.floor(9))});
console.log(multipleDuplicateArr);
var resultArr = Array(Array('KEYS','OCCURRENCE'));
for (var i = 0; i < multipleDuplicateArr.length; i++) {
var flag = true;
for (var j = 0; j < resultArr.length; j++) {
if(resultArr[j][0] == multipleDuplicateArr[i]){
resultArr[j][1] = resultArr[j][1] + 1;
flag = false;
}
}
if(flag){
resultArr.push(Array(multipleDuplicateArr[i],1));
}
}
console.log(resultArr);
You will get result in console as below:
[ 1, 4, 5, 2, 6, 8, 7, 5, 0, 5 ] . // multipleDuplicateArr
[ [ 'KEYS', 'OCCURENCE' ], // resultArr
[ 1, 1 ],
[ 4, 1 ],
[ 5, 3 ],
[ 2, 1 ],
[ 6, 1 ],
[ 8, 1 ],
[ 7, 1 ],
[ 0, 1 ] ]
Solution 26 - Javascript
Quickest way:
Сomputational complexity is O(n).
function howMuchIsRepeated_es5(arr) {
const count = {};
for (let i = 0; i < arr.length; i++) {
const val = arr[i];
if (val in count) {
count[val] = count[val] + 1;
} else {
count[val] = 1;
}
}
for (let key in count) {
console.log("Value " + key + " is repeated " + count[key] + " times");
}
}
howMuchIsRepeated_es5(['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a']);
The shortest code:
Use ES6.
function howMuchIsRepeated_es6(arr) {
// count is [ [valX, count], [valY, count], [valZ, count]... ];
const count = [...new Set(arr)].map(val => [val, arr.join("").split(val).length - 1]);
for (let i = 0; i < count.length; i++) {
console.log(`Value ${count[i][0]} is repeated ${count[i][1]} times`);
}
}
howMuchIsRepeated_es6(['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a']);
Solution 27 - Javascript
var arr = ['a','d','r','a','a','f','d'];
//call function and pass your array, function will return an object with array values as keys and their count as the key values.
duplicatesArr(arr);
function duplicatesArr(arr){
var obj = {}
for(var i = 0; i < arr.length; i++){
obj[arr[i]] = [];
for(var x = 0; x < arr.length; x++){
(arr[i] == arr[x]) ? obj[arr[i]].push(x) : '';
}
obj[arr[i]] = obj[arr[i]].length;
}
console.log(obj);
return obj;
}
Solution 28 - Javascript
Declare an object arr to hold the unique set as keys. Populate arr by looping through the array once using map. If the key has not been previously found then add the key and assign a value of zero. On each iteration increment the key's value.
Given testArray:
var testArray = ['a','b','c','d','d','e','a','b','c','f','g','h','h','h','e','a'];
solution:
var arr = {};
testArray.map(x=>{ if(typeof(arr[x])=="undefined") arr[x]=0; arr[x]++;});
JSON.stringify(arr) will output
{"a":3,"b":2,"c":2,"d":2,"e":2,"f":1,"g":1,"h":3}
Object.keys(arr) will return ["a","b","c","d","e","f","g","h"]
To find the occurrences of any item e.g. b arr['b'] will output 2
Solution 29 - Javascript
let arr=[1,2,3,3,4,5,5,6,7,7]
let obj={}
for(var i=0;i<arr.length;i++){
obj[arr[i]]=obj[arr[i]]!=null ?obj[arr[i]]+1:1 //stores duplicate in an obj
}
console.log(obj)
//returns object {1:1,:1,3:2,.....}
Solution 30 - Javascript
Count the Letters provided in string
function countTheElements(){
var str = "ssdefrcgfrdesxfdrgs";
var arr = [];
var count = 0;
for(var i=0;i<str.length;i++){
arr.push(str[i]);
}
arr.sort();
for(var i=0;i<arr.length;i++){
if(arr[i] == arr[i-1]){
count++;
}else{
count = 1;
}
if(arr[i] != arr[i+1]){
console.log(arr[i] +": "+(count));
}
}
}
countTheElements()
Solution 31 - Javascript
Example how to return the character that is most commonly used in the string.
function maxChar(str) {
const charMap = {};
let maxCharacter = '';
let maxNumber = 0;
for (let item of str) {
charMap[item] = charMap[item] + 1 || 1;
}
for (let char in charMap) {
if (charMap[char] > maxNumber) {
maxNumber = charMap[char];
maxCharacter = char;
}
}
return maxCharacter;
}
console.log(maxChar('abcccccccd'))
Solution 32 - Javascript
Str= ['a','b','c','d','d','e','a','h','e','a'];
var obj= new Object();
for(var i = 0; i < Str.length; i++) {
if(obj[Str[i]] != null) {
obj[Str[i]] += 1;
} else {
obj[Str[i]] = 1;
}
}
console.log(obj);
Now you can get map of repeated items
Solution 33 - Javascript
public class CalculateCount {
public static void main(String[] args) {
int a[] = {1,2,1,1,5,4,3,2,2,1,4,4,5,3,4,5,4};
Arrays.sort(a);
int count=1;
int i;
for(i=0;i<a.length-1;i++){
if(a[i]!=a[i+1]){
System.out.println("The Number "+a[i]+" appears "+count+" times");
count=1;
}
else{
count++;
}
}
System.out.println("The Number "+a[i]+" appears "+count+" times");
}
}