How i can find and count duplicate values in javascript object

Question

I have the below array. Did console.log for the array and it is as below

(3) [Array(4), Array(4), Array(4)]
0 ["jackpot", "cherry", "bar", "pear"]
1 ["raspberry", "raspberry", "raspberry", "lemon"]
2 ["plum", "crown", "lemon", "jackpot"]

I tried the below and in console i just see my array and 1 next to it

var counts = {};
currentResults.forEach(function(obj) {
var key = JSON.stringify(obj)
counts[key] = (counts[key] || 0) + 1
})
console.log(counts);

How can i search and find the duplicated values in the above array/object and count how many duplicated items I have with javascript/jquery?

Duplicates per child array, or duplicates within all child arrays? Also, please show what you've tried. At the moment this is a 'write my code for me' question, which is likely to be downvoted and/or closed — Rory McCrossan
– Rory McCrossan, Commented Sep 20, 2017 at 11:11
possible duplicate of:- stackoverflow.com/questions/10541068/… — Death-is-the-real-truth
– Death-is-the-real-truth, Commented Sep 20, 2017 at 11:13
Duplicates within all child arrays @RoryMcCrossan. I tried so many ways that's why not sure which one to add — Anahit Ghazaryan
– Anahit Ghazaryan, Commented Sep 20, 2017 at 11:13
An off topic question for everyone: What does (counts[key] || 0) + 1mean? — Reporter
– Reporter, Commented Sep 20, 2017 at 11:21
Just so you know, your answer you had was pretty close. The reason you got 1 next to each array is because obj is an array and you were trying to compare arrays to arrays, so counts[obj] would be 0 + 1 giving you 1 after each line. You just needed obj.forEach loop after currentResults.forEach function and making your key equal to that function loop. Also no need to JSON.stringify. — Matt
– Matt, Commented Sep 20, 2017 at 13:54

Shiladitya · Accepted Answer · 2017-09-20 13:58:30Z

1

Here you go with a solution

var data = [["jackpot", "cherry", "bar", "pear"], ["raspberry", "raspberry", "raspberry", "lemon"],["plum", "crown", "lemon", "jackpot"]];

var key = {};
for(var i=0; i<data.length; i++){
  for(var j=0; j<data[i].length; j++){
    if(typeof key[data[i][j]] === 'undefined'){
      key[data[i][j]] = 1;
    } else {
      key[data[i][j]] = parseInt(key[data[i][j]]) + 1;
    }
  }
}

console.log(key);

I assumed your data as an array of array. Loop through both the array & for every item, kept an count in the form of a JSON (key: value).

Key as item & value is the count.

Updated Solution to get the duplicate items

var data = [["jackpot", "cherry", "bar", "pear"], ["raspberry", "raspberry", "raspberry", "lemon"],["plum", "crown", "lemon", "jackpot"]];

var key = {};
var duplicate = [];
for(var i=0; i<data.length; i++){
	for(var j=0; j<data[i].length; j++){
  	if(typeof key[data[i][j]] === 'undefined'){
    	key[data[i][j]] = 1;
    } else {
    	key[data[i][j]] = parseInt(key[data[i][j]]) + 1;
      if(parseInt(key[data[i][j]]) >= 3 && duplicate.indexOf(data[i][j]) == -1){
      	duplicate.push(data[i][j]);
      }
    }
  }
}

console.log(duplicate);

Updated Solution for finding duplicates within child array

var data = [["jackpot", "cherry", "bar", "pear"], ["raspberry", "raspberry", "raspberry", "lemon"],["plum","jackpot", "crown", "lemon", "jackpot", "jackpot"]];

var key = {};
var duplicate = [];
for(var i=0; i<data.length; i++){
	key = {};
	for(var j=0; j<data[i].length; j++){
  	if(typeof key[data[i][j]] === 'undefined'){
    	key[data[i][j]] = 1;
    } else {
    	key[data[i][j]] = parseInt(key[data[i][j]]) + 1;
      if(parseInt(key[data[i][j]]) >= 3 && duplicate.indexOf(data[i][j]) == -1){
      	duplicate.push(data[i][j]);
      }
    }
  }
}

console.log(duplicate);

Hope this will help you.

edited Sep 20, 2017 at 13:58

answered Sep 20, 2017 at 11:17

Shiladitya

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9 Comments

Anahit Ghazaryan Over a year ago

Hey @Shiladitya yes is what i needed. I just need to get only the element value which is duplicated 3 times not all but is very useful and exact snippet i need. How i can retrive the key value only when number is 3 or more?

Shiladitya Over a year ago

@AnahitGhazaryan I've updated the answer, have a look.

Anahit Ghazaryan Over a year ago

Yes @Shiladitya thank you for the good snippet. can you pls explain what this means duplicate.indexOf(data[i][j]) == -1 ?

Shiladitya Over a year ago

duplicate.indexOf(data[i][j]) == -1 this part will check if the item is already exists in duplicate array or not. -1 means the item is not present in duplicate array.

Anahit Ghazaryan Over a year ago

Aha ok. Thank you very much @Shiladitya for the good idea and useful snippet

|

Rohit Agrawal · Accepted Answer · 2017-09-20 11:24:51Z

0

You can iterate over each array element inside your main array and push unique values in a new array and if the value already exists than you got a duplicate and store it in result array.

Here I am not writing the whole code as I can not see what you have tried.

You can use this method to check if an element already exist in new array

arr.indexOf()

Edit: A better approach would be to iterate over each array element inside your main array and store them as a property of an object and their count as the value of the property. At the end your object would be something like this:

{
  "raspberry": 3,
  "lemon" : 2
}

edited Sep 20, 2017 at 11:24

answered Sep 20, 2017 at 11:13

Rohit Agrawal

1,52111 silver badges21 bronze badges

1 Comment

Anahit Ghazaryan Over a year ago

Yes right. @Shiladitya answer was doign that which is really good idea.

Aymen · Accepted Answer · 2017-09-20 11:24:51Z

0

use this :)

var test=[["jackpot", "cherry", "bar", "pear"], ["raspberry", "raspberry", "raspberry", "lemon"],["plum", "crown", "lemon", "jackpot"]];
var counts = {};
for(var i=0;i<test.length;i++)
test[i].forEach((x)=> { counts[x] = (counts[x] || 0)+1; });
console.log(counts);

edited Sep 20, 2017 at 11:24

answered Sep 20, 2017 at 11:10

Aymen

1,50619 silver badges29 bronze badges

4 Comments

Rajesh Dixit Over a year ago

Why? Does it solves problem? I don't know, may be explaining your changes might help

Aymen Over a year ago

the question is to count the duplicates; with this method can solve it

Anahit Ghazaryan Over a year ago

but how this counts? i dont see it @DevAymen

Aymen Over a year ago

@AnahitGhazaryan he expression counts[x] || 0 returns the value of counts[x] if it is set, otherwise 0. Then just add one and set it again in the object and the count is done

Murat Acarsoy · Accepted Answer · 2017-09-20 11:44:41Z

You can check duplications for each arrays or you can merge arrays in one array for checking all arrays.

        var array1 = ["jackpot", "cherry", "bar", "pear"];
        var array2 = ["raspberry", "raspberry", "raspberry", "lemon"];
        var array3 = ["plum", "crown", "lemon", "jackpot"];

        var parentArray = [array1, array2, array3];
        var joinedArray = [];

        // ForEach Array
        var duplicatedElementsForEachArray = [];
        parentArray.forEach(function (childArray) {

            // Merge Arrays in joinedArray
            joinedArray = joinedArray.concat(childArray);

            var duplicatedElements = 0;
            var sortedChildArray = childArray.slice().sort(); // Sorting each Arrays
            var i = 0, l = sortedChildArray.length - 1;
            while (i < l) { if (sortedChildArray[i] == sortedChildArray[i + 1]) duplicatedElements++; i++; } // If i. and (i+1). elements are same, this will be duplication
            duplicatedElementsForEachArray.push(duplicatedElements);
            console.log(duplicatedElements);
        });

        // Joined Array, you can check duplications for merged array with same algorithm
        var sortedJoinedArray = joinedArray.slice().sort();
        var duplicatedElementsForJoinedArray = 0;
        var i = 0, l = sortedJoinedArray.length - 1;
        while (i < l) { if (sortedJoinedArray[i] == sortedJoinedArray[i + 1]) duplicatedElementsForJoinedArray++; i++; }
        console.log(duplicatedElementsForJoinedArray);

Thanks for the answer but i have one array not 3 separate but i will use the above snippet for rest cases i would need :)

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