Get the item that appears the most times 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

This solution returns an array of the most appearing numbers in an array, in case multiple numbers appear at the "max" times.

    function mode(numbers) {
      var counterObj = {}; 
      var max = 0;
      var result = [];
      for(let num in numbers) {
        counterObj[numbers[num]] = (counterObj[numbers[num]] || 0) + 1; 
        if(counterObj[numbers[num]] >= max) { 
          max = counterObj[numbers[num]];
        }
      }
      for (let num in counterObj) {
        if(counterObj[num] == max) {
          result.push(parseInt(num));
        }
      }
      return result;
    }

Answer 2

A fairly short solution.

function mostCommon(list) {
  var keyCounts = {};
  var topCount = 0;
  var topKey = {};
  list.forEach(function(item, val) {
    keyCounts[item] = keyCounts[item] + 1 || 1;
    if (keyCounts[item] > topCount) {
      topKey = item;
      topCount = keyCounts[item];
    }
  });

  return topKey;
}

document.write(mostCommon(['AA', 'AA', 'AB', 'AC']))

Answer 3

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 4

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 5

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 6

This is my solution.

var max_frequent_elements = function(arr){
var a = [], b = [], prev;
arr.sort();
for ( var i = 0; i < arr.length; i++ ) {
    if ( arr[i] !== prev ) {
        a.push(arr[i]);
        b.push(1);
    } else {
        b[b.length-1]++;
    }
    prev = arr[i];
}


var max = b[0]
for(var p=1;p<b.length;p++){
       if(b[p]>max)max=b[p]
 }

var indices = []
for(var q=0;q<a.length;q++){
   if(b[q]==max){indices.push(a[q])}
}
return indices;

};