'How to count duplicate value in an array in javascript
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.
Solution 1:[1]
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();You can use higher-order functions too to do the operation. See this answer
Solution 2:[2]
const counts = {};
const sampleArray = ['a', 'a', 'b', 'c'];
sampleArray.forEach(function (x) { counts[x] = (counts[x] || 0) + 1; });
console.log(counts)Solution 3:[3]
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 4:[4]
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 5:[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);Solution 6:[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));Solution 7:[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]);
Solution 8:[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.
Solution 9:[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
Solution 10:[10]
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 11:[11]
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:[12]
// 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:[13]
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:[14]
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:[15]
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:[16]
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:[17]
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:[18]
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:[19]
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:[20]
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:[21]
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:[22]
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:[23]
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:[24]
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:[25]
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:[26]
Quickest way:
?omputational 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:[27]
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:[28]
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:[29]
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:[30]
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()
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow