How to count Matching values in Array of Javascript

Question

please tell me any good algorithm/code to get list of unique values from array and count of its occurrence in array. (i am using javascript).

Answer 1

Use an object as an associative array:

var histo = {}, val;
for (var i=0; i < arr.length; ++i) {
    val = arr[i];
    if (histo[val]) {
        ++histo[val];
    } else {
        histo[val] = 1;
    }
}

This should be at worst O(n*log(n)), depending on the timing for accessing object properties. If you want just the strings, loop over the object's properties:

for (val in histo) {...}

Answer 2

For a method that will strip the duplicates from an array and returns a new array with the unique values, you may want to check the following Array.unique implementation. With an O(n²) complexity, it is certainly not the quickest algorithm, but will do the job for small unsorted arrays.

It is licensed under GPLv3, so I should be allowed to paste the implementation here:

// **************************************************************************
// Copyright 2007 - 2009 Tavs Dokkedahl
// Contact: http://www.jslab.dk/contact.php
//
// This file is part of the JSLab Standard Library (JSL) Program.
//
// JSL is free software; you can redistribute it and/or modify
// it under the terms of the GNU General Public License as published by
// the Free Software Foundation; either version 3 of the License, or
// any later version.
//
// JSL is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.
//
// You should have received a copy of the GNU General Public License
// along with this program. If not, see <http://www.gnu.org/licenses/>.
// ***************************************************************************

Array.prototype.unique =
  function() {
    var a = [];
    var l = this.length;
    for(var i=0; i<l; i++) {
      for(var j=i+1; j<l; j++) { 
        if (this[i] === this[j])  // If this[i] is found later in the array
          j = ++i;
      }
      a.push(this[i]);
    }
    return a;
  };

You would be able to use it as follows:

var myArray = new Array("b", "c", "b", "a", "b", "g", "a", "b");
myArray.unique(); // returns: ["c", "g", "a", "b"]

You may want to tweak the above to somehow append the number of occurrences of each value.

Answer 3

a straightforward way is to loop over the array once and count values in a hash

a = [11, 22, 33, 22, 11];
count = {}
for(var i = 0; i < a.length; i++)
   count[a[i]] = (count[a[i]] || 0) + 1

the "count" would be like this { 11: 2, 22: 2, 33: 1 }

for sorted array the following would be faster

a = [11, 11, 11, 22, 33, 33, 33, 44];
a.sort()
uniq = [];
len = a.length
for(var i = 0; i < len;) {
   for(var k = i; k < len && a[k] == a[i]; k++);
   if(k == i + 1) uniq.push(a[i])
   i = k
}   
// here uniq contains elements that occur only once in a

Answer 4

This method works for arrays of primitives - strings, numbers, booleans,

and objects that can be compared(like dom elements)

Array.prototype.frequency= function(){
    var i= 0, ax, count, item, a1= this.slice(0);
    while(i<a1.length){
        count= 1;
        item= a1[i];
        ax= i+1;
        while(ax<a1.length && (ax= a1.indexOf(item, ax))!= -1){
            count+= 1;
            a1.splice(ax, 1);
        }
        a1[i]+= ':'+count;
        ++i;
    }
    return a1;
}

var arr= 'jgeeitpbedoowknnlfiaetgetatetiiayolnoaaxtek'.split('');
var arrfreq= arr.frequency();

The return value is in the order of the first instance of each unique element in the array.

You can sort it as you like- this sorts from highest to lowest frequency:

arrfreq.sort(function(a, b){
    a= a.split(':');
    b= b.split(':');
    if(a[1]== b[1]){
        if(a[0]== b[0]) return 0;
        return a[0]> b[0]? 1: -1;
    }
    return a[1]> b[1]? -1: 1;
});

arrfreq now returns(Array): ['e:7','t:6','a:5','i:4','o:4','n:3','g:2','k:2','l:2','b:1','d:1','f:1','j:1','p:1','w:1','x:1','y:1']

mustn't leave out IE:

Array.prototype.indexOf= Array.prototype.indexOf || 
function(what, index){
    index= index || 0;
    var L= this.length;
    while(index< L){
        if(this[index]=== what) return index;
        ++index;
    }
    return -1;
}

Answer 5

I know this is old post but I was lookimg for a simple solution so I figured I'd post what I worked out for anyone still looking into this

const test = [5, 3, 9, 5, 3, 5, 5]


//This function will return a new array of only the specified value
Array.prototype.unique = function(find) {
return this.filter(x => x == find)
}

//Usage
console.log(test.unique(5)) // returns [5,5,5,5]


//This Function will return the number of occurences in an array
Array.prototype.count = function(find) {
return this.filter(x => x == find).length
}

//Usage
console.log(test.count(5)) // returns 4