how to get index of duplicates in a javascript array?

Question

I have a sorted array like this:

var a= [
    {id:"1", name:"A", address:"A"},
    {id:"2", name:"A", address:"B"},
    {id:"3", name:"A", address:"C"},
    {id:"4", name:"B", address:"A"},
    {id:"5", name:"B", address:"B"},
    {id:"6", name:"C", address:"A"},
    {id:"7", name:"C", address:"B"},
    {id:"8", name:"C", address:"C"},..

What I want to do is get the index of those duplicated 'names' like

index 0-2 => "A";
index 3-4 => "B";
index 5-7 => "C";

Or even counting those duplicates will do such as 3 for A, 2 for B, 3 for C.

        var counter = 0;
        var prev = "";
        var next = "";
        var prevLink = a[0].name;

        for (var i = 0; i < a.length; i++) { 

         if(i >= ssRow.length){
             //do nothing
        }else{        
          next  = a[i+1].name;  
        } 

        if(prev== next){
          counter++;
        }else{        
          counter = 0;
          prevLink = a[i].name;      
        }
          //play with counter variable
        }

But it's not working as expected. Is there a much better way or efficient way to do it?

did you checked my answer ? I hope it will work as per your expectation. — Rohìt Jíndal
– Rohìt Jíndal, Commented May 23, 2017 at 17:16

pranay kumar · Accepted Answer · 2017-05-23 11:12:17Z

4

 let dupes = {};
 const array= [
        {id:"1", name:"A", address:"A"},
        {id:"2", name:"A", address:"B"},
        {id:"3", name:"A", address:"C"},
        {id:"4", name:"B", address:"A"},
        {id:"5", name:"B", address:"B"},
        {id:"6", name:"C", address:"A"},
        {id:"7", name:"C", address:"B"},
        {id:"8", name:"C", address:"C"}];
 array.forEach((item,index) => {
   dupes[item.name] = dupes[item.name] || [];
   dupes[item.name].push(index);
 });       
 for(let name in dupes) console.log(name+'->indexes->'+dupes[name]+'->count->'+dupes[name].length)

edited May 23, 2017 at 11:12

answered May 23, 2017 at 10:38

pranay kumar

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6 Comments

silent_coder14 Over a year ago

array.forEach((item,index) => { dupes[item.name] = dupes[item.name] || []; dupes[item.name].push(index); }); can this be done in a normal for loop?

Mihai Alexandru-Ionut Over a year ago

@silent_coder14, yes, you should use a for statement

silent_coder14 Over a year ago

thanx.. also, what this line means: dupes[item.name] = dupes[item.name] || [];

Mihai Alexandru-Ionut Over a year ago

This line mean following: if dupes[item.name] is undefined then initialize it with new array, and if not remains same.

pranay kumar Over a year ago

@silent_coder14 yes a normal for loop can be used with the same block of code even for the conventional for loop. either for(let index in array){const item = array[index]; //rest same code block from original answer } or for(let index=0; index<array.length; index++){const item=array[index]; //rest same code block from original answer} can be used!

|

Nenad Vracar · Accepted Answer · 2017-05-23 10:17:27Z

You can use reduce() to return object as result with both count and start-end index for each name.

var arr = [{"id":"1","name":"A","address":"A"},{"id":"2","name":"A","address":"B"},{"id":"3","name":"A","address":"C"},{"id":"4","name":"B","address":"A"},{"id":"5","name":"B","address":"B"},{"id":"6","name":"C","address":"A"},{"id":"7","name":"C","address":"B"},{"id":"8","name":"C","address":"C"}]

var l = null;

var result = arr.reduce(function(r, e, i) {
  if (!r[e.name]) {
    l = e.name;
    r[e.name] = {
      count: 1,
      index: i.toString()
    }
  } else {
    if (arr[i + 1] && arr[i + 1].name != l || !arr[i + 1]) r[e.name].index += '-' + i;
    r[e.name].count++
  }
  return r;
}, {})

console.log(result)

Rohìt Jíndal · Accepted Answer · 2017-05-23 17:15:36Z

You can use ES6 Array.from() method.

Try this it will work as per your expectation :

var a = [
    {id:"1", name:"A", address:"A"},
    {id:"2", name:"A", address:"B"},
    {id:"3", name:"A", address:"C"},
    {id:"4", name:"B", address:"A"},
    {id:"5", name:"B", address:"B"},
    {id:"6", name:"C", address:"A"},
    {id:"7", name:"C", address:"B"},
    {id:"8", name:"C", address:"C"}];

// fetch all the name property value into an array.
var nameArr = Array.from(a, x => x.name);

var obj = {};

for (var i = 0; i < nameArr.length; i++) {
    var elem = nameArr[i];

    // if we haven't seen the element yet, 
    // we have to create a new entry in the map
    if (!obj[elem]) {
        obj[elem] = [i];
    }
    else {
        if(obj[elem].indexOf(nameArr.lastIndexOf(elem)) == -1) {
          // otherwise append to the existing array
          obj[elem].push(nameArr.lastIndexOf(elem));
        }
    }
}
 
for (var i in obj) {
  console.log('index ' +obj[i].join('-')+ ' => "'+i+'"');
}

Simple_Programmer · Accepted Answer · 2017-05-23 10:23:34Z

For counting the duplicates, you can use Underscore.js which has a method called _.groupBy()(see this). It gives you an object of key-value pairs where keys are the value you want to check for duplication and value is the array of objects what have that key. Its like this -

  var groupedData = _.groupBy(array, function(elem){
       return elem.name; //here you can specify whichever key you want to check duplicates for
  });

  console.log(groupedData['A'].length);

Nina Scholz · Accepted Answer · 2017-05-23 11:02:02Z

You could use Array#reduce and check the predecessor if the name is not the same or at the first index, then create a new array with the index information.

var data = [{ id: "1", name: "A", address: "A" }, { id: "2", name: "A", address: "B" }, { id: "3", name: "A", address: "C" }, { id: "4", name: "B", address: "A" }, { id: "5", name: "B", address: "B" }, { id: "6", name: "C", address: "A" }, { id: "7", name: "C", address: "B" }, { id: "8", name: "C", address: "C" }],
    groups = data.reduce(function (r, a, i, aa) {
        (!i || aa[i - 1].name !== a.name) && r.push([i, i, a.name]);
        r[r.length - 1][1] = i;
        return r;
    }, []);

console.log(groups);

.as-console-wrapper { max-height: 100% !important; top: 0; }

AncientSwordRage · Accepted Answer · 2017-05-23 11:27:49Z

You could try to use the groupBy method in underscore. It doesn't do the full job of formatting it, but gets you to a good place to carry on with it.

var data = [{ id: "1", name: "A", address: "A" }, { id: "2", name: "A", address: "B" }, { id: "3", name: "A", address: "C" }, { id: "4", name: "B", address: "A" }, { id: "5", name: "B", address: "B" }, { id: "6", name: "C", address: "A" }, { id: "7", name: "C", address: "B" }, { id: "8", name: "C", address: "C" }],

groups = _.groupBy(data, 'name');
console.log(groups);

.as-console-wrapper { max-height: 100% !important; top: 0; }

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>

@silent_coder14 that's a shame, the above is very handy and readable. I hope it's useful for future reader's with the same problem but not the same restrictions

Mihai Alexandru-Ionut · Accepted Answer · 2017-05-23 12:16:38Z

You can do more simply using filter method in combination with forEach.

The first step is to find out the unique names from your array. This can be achieved using Set method from ES6.

var names=[...new Set(array.map(a=>a.name))];

or

UPDATE

 names=array.map(a=>a.name).filter(function (x, i, a) { 
      return a.indexOf(x) == i; 
 });

Then you should use filter method, which accepts a callback provided function , in order to get the items which have name equals to one element from names array.

var array= [
    {id:"1", name:"A", address:"A"},
    {id:"2", name:"A", address:"B"},
    {id:"3", name:"A", address:"C"},
    {id:"4", name:"B", address:"A"},
    {id:"5", name:"B", address:"B"},
    {id:"6", name:"C", address:"A"},
    {id:"7", name:"C", address:"B"},
    {id:"8", name:"C", address:"C"}];
    
var names=[...new Set(array.map(a=>a.name))];
names.forEach(function(item){
  indexes=array.map(function(item,i){
     item["index"]=i;
     return item;
  })
  .filter(a=>a.name==item)
  .map(a=>a.index);
  console.log(indexes[0]+'-'+indexes[indexes.length-1]+'->'+item+', count= '+indexes.length);
});

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