How to count duplicate value in an array in javascript

Question

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.

Answer 1

const counts = {};
const sampleArray = ['a', 'a', 'b', 'c'];
sampleArray.forEach(function (x) { counts[x] = (counts[x] || 0) + 1; });
console.log(counts)

Answer 2

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.

Answer 3

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.

Answer 4

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();

Demo Fiddle

You can use higher-order functions too to do the operation. See this answer

Answer 5

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);

Answer 6

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));

Answer 7

// 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]);

Answer 8

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.

Answer 9

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                                      

Answer 10

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!

Answer 11

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 }

Answer 12

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);

Answer 13

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]);
} 

Answer 14

// 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));

Answer 15

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

Answer 16

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);

Answer 17

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>");
}

Answer 18

simplified sheet.js answare

var counts = {};
var aarr=['a','b','a'];
aarr.forEach(x=>counts[x]=(counts[x] || 0)+1 );
console.log(counts)

Answer 19

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;
}

Answer 20

Using this solution you can now get map of repeated items:

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);

Answer 21

Steps : first check if in accumulator has the current value or not if not ,than for that particular value set the count as 1 and in else condition ,if value alreadt exist in accumulator the simple increment the count.

const testarr = [1,2,1,3,1,2,4];

var count = testarr.reduce((acc,currentval)=>{

if(acc[currentval]){ acc[currentval] = ++acc[currentval]; }else{ acc[currentval] = 1; } return acc; },{})

console.log(count);

Answer 22

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}`);
}

Answer 23

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);

Answer 24

use forEach() method for convinience

var uniqueCount="a","b","c","d","d","e","a","b","c","f","g","h","h","h","e","a"];
var count=0;
var obj={};      
uniqueCount.forEach((i,j)=>{
count=0;
var now=i;
uniqueCount.forEach((i,j)=>{
if(now==uniqueCount[j]){
count++;
obj[i]=count;
}
});
});

console.log(obj);

Answer 25

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.

Answer 26

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);

Answer 27

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 
*/

Answer 28

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 ] ]

Answer 29

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']);

Answer 30

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;
}

Answer 31

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

Answer 32

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,.....}

Answer 33

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()

Answer 34

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'))

Answer 35

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");

}   

}