Count the number of unique occurrences in an array that contain a specific string with Javascript

Question

Here is my javascript array:

arr = ['blue-dots', 'blue', 'red-dots', 'orange-dots', 'blue-dots'];

With Javascript, how can I count the total number of all unique values in the array that contain the string “dots”. So, for the above array the answer would be 3 (blue-dots, orange-dots, and red-dots).

Possible duplicate of [Unique values in an array](http://stackoverflow.com/questions/1960473/unique-values-in-an-array) — Heretic Monkey, Jun 08 '16 at 19:50
This other question doesn't explain how to also only count elements that contain the string. — J. Lynx, Jun 10 '16 at 15:39

score 2 · Answer 1 · answered Jun 08 '16 at 17:49

var count = 0,
    arr1 = [];    
for (var i = 0; i < arr.length; i++) {
    if (arr[i].indexOf('dots') !== -1) {
        if (arr1.indexOf(arr[i]) === -1) {
            count++;
            arr1.push(arr[i]);
        }
    }
 }

you check if a certain element contains 'dots', and if it does, you check if it is already in arr1, if not increment count and add element to arr1.

score 1 · Answer 2 · answered Jun 08 '16 at 17:52

1

One way is to store element as key of an object, then get the count of the keys:

var arr = ["blue-dots", "blue", "red-dots", "orange-dots", "blue-dots"];
console.log(Object.keys(arr.reduce(function(o, x) {
  if (x.indexOf('dots') != -1) {
    o[x] = true;
  }
  return o
}, {})).length)

answered Jun 08 '16 at 17:52

Fabricator

12,722
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40

elegant, but could be improved, altho like it +1 – Medet Tleukabiluly Jun 08 '16 at 18:06

score 0 · Answer 3 · edited May 23 '17 at 10:34

From this question, I got the getUnique function.

Array.prototype.getUnique = function(){
   var u = {}, a = [];
   for(var i = 0, l = this.length; i < l; ++i){
      if(u.hasOwnProperty(this[i])) {
         continue;
      }
      a.push(this[i]);
      u[this[i]] = 1;
   }
   return a;
}

then you can add a function that counts ocurrences of a string inside an array of strings:

function getOcurrencesInStrings(targetString, arrayOfStrings){
  var ocurrencesCount = 0;
  for(var i = 0, arrayOfStrings.length; i++){
    if(arrayOfStrings[i].indexOf(targetString) > -1){
      ocurrencesCount++;
    }
  }
  return ocurrencesCount;
}

then you just:

getOcurrencesInStrings('dots', initialArray.getUnique())

This will return the number you want.

It's not the smallest piece of code, but It's highly reusable.

score 0 · Answer 4 · answered Jun 08 '16 at 17:53

Try this something like this:

// Create a custom function
function countDots(array) {
  var count = 0;

  // Get and store each value, so they are not repeated if present.
  var uniq_array = [];
  array.forEach(function(value) {
    if(uniq_array.indexOf(value) == -1) {
      uniq_array.push(value);

      // Add one to count if 'dots' word is present.
      if(value.indexOf('dots') != -1) {
        count += 1;
      }
    }
  });

  return count;
}

// This will print '3' on console
console.log( countDots(['blue-dots', 'blue', 'red-dots', 'orange-dots', 'blue-dots']) );

Medet Tleukabiluly · Answer 5 · 2016-06-08T18:03:50.057

0

var uniqueHolder = {};
var arr = ["blue-dots", "blue", "red-dots", "orange-dots", "blue-dots"];
arr.filter(function(item) {
    return item.indexOf('dots') > -1;
  })
  .forEach(function(item) {
    uniqueHolder[item] ? void(0) : uniqueHolder[item] = true;
  });
console.log('Count: ' + Object.keys(uniqueHolder).length);
console.log('Values: ' + Object.keys(uniqueHolder));

edited Jun 08 '16 at 18:03

answered Jun 08 '16 at 17:58

Medet Tleukabiluly

11,662
3
34
69

Brandon Anzaldi · Answer 6 · 2016-06-08T18:45:22.310

var arr = [ "blue-dots", "blue", "red-dots", "orange-dots", "blue-dots" ];
var fArr = []; // Empty array, which could replace arr after the filtering is done.
arr.forEach( function( v ) {
  v.indexOf( "dots" ) > -1 && fArr.indexOf( v ) === -1 ? fArr.push( v ) : null;
  // Filter if "dots" is in the string, and not already in the other array.
});

// Code for displaying result on page, not necessary to filter arr

document.querySelector( ".before" ).innerHTML = arr.join( ", " );
document.querySelector( ".after" ).innerHTML = fArr.join( ", " );

Before:
<pre class="before">
  </pre>
After:
<pre class="after">
  </pre>

To put this simply, it will loop through the array, and if dots is in the string, AND it doesn't already exist in fArr, it'll push it into fArr, otherwise it'll do nothing.

score 0 · Answer 7 · answered Jun 08 '16 at 18:33

Try this code,

arr = ["blue-dots", "blue", "red-dots", "orange-dots", "blue-dots"];

sample = [];    
for (var i = 0; i < arr.length; i++) {
    if ((arr[i].indexOf('dots') !== -1) && (sample.indexOf(arr[i]) === -1)){
       sample.push(arr[i]);

    }
 }

alert(sample.length);

1983 · Answer 8 · 2016-06-08T19:19:54.430

I'd separate the operations of string comparison and returning unique items, to make your code easier to test, read, and reuse.

var unique = function(a){
    return a.length === 0 ? [] : [a[0]].concat(unique(a.filter(function(x){
        return x !== a[0];
    })));
};

var has = function(x){
    return function(y){
        return y.indexOf(x) !== -1;
    };
};

var arr = ["blue-dots", "blue", "red-dots", "orange-dots", "blue-dots"];
var uniquedots = unique(arr.filter(has('dots')));
console.log(uniquedots);
console.log(uniquedots.length);

Count the number of unique occurrences in an array that contain a specific string with Javascript

8 Answers8