How to get the item that appear the most time in an array?

Question

var store = ['1','2','2','3','4'];

I want to find out that 2 appear the most in the array. How do I go about doing that?

Answer 1

Make a histogram, find the key for the maximum number in the histogram.

var hist = [];
for (var i = 0; i < store.length; i++) {
  var n = store[i];
  if (hist[n] === undefined) hist[n] = 0;
  else hist[n]++;
}

var best_count = hist[store[0]];
var best = store[0];
for (var i = 0; i < store.length; i++) {
  if (hist[store[i]] > best_count) {
    best_count = hist[store[i]];
    best = store[i];
  }
}

alert(best + ' occurs the most at ' + best_count + ' occurrences');

This assumes either there are no ties, or you don't care which is selected.

Answer 2

I would do something like:

var store = ['1','2','2','3','4'];
var frequency = {};  // array of frequency.
var max = 0;  // holds the max frequency.
var result;   // holds the max frequency element.
for(var v in store) {
        frequency[store[v]]=(frequency[store[v]] || 0)+1; // increment frequency.
        if(frequency[store[v]] > max) { // is this frequency > max so far ?
                max = frequency[store[v]];  // update max.
                result = store[v];          // update result.
        }
}

Answer 3

If the array is sorted this should work:

function popular(array) { 
   if (array.length == 0) return [null, 0];
   var n = max = 1, maxNum = array[0], pv, cv;

   for(var i = 0; i < array.length; i++, pv = array[i-1], cv = array[i]) {
      if (pv == cv) { 
        if (++n >= max) {
           max = n; maxNum = cv;
        }
      } else n = 1;
   }

   return [maxNum, max];
};

popular([1,2,2,3,4,9,9,9,9,1,1])
[9, 4]

popular([1,2,2,3,4,9,9,9,9,1,1,10,10,10,10,10])
[10, 5]

Answer 4

This version will quit looking when the count exceeds the number of items not yet counted.

It works without sorting the array.

Array.prototype.most= function(){
    var L= this.length, freq= [], unique= [], 
    tem, max= 1, index, count;
    while(L>= max){
        tem= this[--L];
        if(unique.indexOf(tem)== -1){
            unique.push(tem);
            index= -1, count= 0;
            while((index= this.indexOf(tem, index+1))!= -1){
                ++count;
            }
            if(count> max){
                freq= [tem];
                max= count;
            }
            else if(count== max) freq.push(tem);
        }
    }
    return [freq, max];
}

    //test
    var A= ["apples","oranges","oranges","oranges","bananas",
   "bananas","oranges","bananas"];
    alert(A.most()) // [oranges,4]

    A.push("bananas");
    alert(A.most()) // [bananas,oranges,4]

Answer 5

<?php
$res=mysql_query("select * FROM `check`");
while($row=mysql_fetch_array($res)){
$input[]=$row['value'];
}
# returns most common value in input array
function array_most_common($input)
{
  $counted = array_count_values($input);
  arsort($counted);
  return(key($counted));    
} 
?>

Answer 6

If you have really large data, you may need something like the count-min sketch.