CODEWITHSUNDEEP
javascriptjqueryarrays
TK.
TK.

Reputation: 28153

Extracting the most duplicate value from an array in JavaScript (with jQuery)

I have several array to deal with. I need to extract the most duplicate value from each array.

From [3, 7, 7, 7], I need to find the value 7. Each array size is 4. For now, I don't have to think about when the most duplicate values are more than one such as [3, 7, 7, 7]. All the values are a number.

I looked around the web. I found several ways to make an array to become uniq(). But I haven't found a way to get the duplicate value. I am using jQuery, but raw JavaScript is fine for this task.

Upvotes: 4

Views: 7631

Answers (5)

gaetanoM
gaetanoM

Reputation: 42044

Another solution can be based on Array.reduce():

var arr = [1,1,2,5,4,2,10,10,1,1,1,3,10,10,3,4,3,10,3,5,6,2,3,1,1,10,10,2,4,3,6,10,6,6];

var result = arr.reduce(function(acc, e) {
    acc[e] = (acc[e] || 0) + 1;
    if (acc[e] > acc.mostFreq.freq) {
        acc.mostFreq.value = e;
        acc.mostFreq.freq = acc[e];
    }
    return acc;
}, {"mostFreq": {"value": 0, "freq": 0}}).mostFreq;

console.log('The most duplicated elements is: ' + JSON.stringify(result));

Upvotes: 0

Xavi
Xavi

Reputation: 20439

Here's a quick example using javascript:

function mostFrequent(arr) {
    var uniqs = {};

    for(var i = 0; i < arr.length; i++) {
        uniqs[arr[i]] = (uniqs[arr[i]] || 0) + 1;
    }

    var max = { val: arr[0], count: 1 };
    for(var u in uniqs) {
        if(max.count < uniqs[u]) { max = { val: u, count: uniqs[u] }; }
    }

    return max.val;
}

A quick note on algorithmic complexity -- because you have to look at each value in the array at least once, you cannot do better than O(n). This is assuming that you have no knowledge of the contents of the array. If you do have prior knowledge (e.g. the array is sorted and only contains 1s and 0s), then you can devise an algorithm with a run time that is some fraction of n; though technically speaking, it's complexity is still O(n).

Upvotes: 3

Nick Craver
Nick Craver

Reputation: 630379

Here's a simpler and faster version using only JavaScript:

var arr = [3, 7, 7, 7, 10, 10, 8, 5, 5, 5, 5, 20, 20, 1];
var counts = {}, max = 0, res;
for (var v in arr) {
  counts[arr[v]] = (counts[arr[v]] || 0) + 1;
  if (counts[arr[v]] > max) { 
    max = counts[arr[v]];
    res = arr[v];
  }
}
alert(res + " occurs " + counts[res] + " times");

Note that this is a much more efficient since you're looping over the data once, if you're sorting very large arrays this will start to matter.

Upvotes: 6

kennebec
kennebec

Reputation: 104760

Array.prototype.mostFreq=function(){
 var what, a= this.concat(), ax, freq,
 count, max=0, limit= a.length/2;
 while(a.length){
  what= a.shift();
  count=1; 
  while((ax= a.indexOf(what))!= -1){
   a.splice(ax,1); // remove counted items  
   ++count;
  }
  // if any item has more than half the array, quit counting
  if(count> limit) return what; 
  if(count> max){
   freq= what;
   max= count;
  }
 }
 return freq;
}
var a=[1,1,2,5,4,2,7,7,1,1,1,3,7,7,3,4,3,7,3,5,6,2,3,1,1,7,7,2,4,3,6,7,6,6]
alert(a.mostFreq())

Upvotes: 1

Max Shawabkeh
Max Shawabkeh

Reputation: 38603

Not perfect in terms of efficiency, but does the job:

var nums = [3, 7, 7, 7];
var freqs = {};
var max_index;
var max_value = -1/0; // Negative infinity.

$.each(nums, function(i, v) {
  if (freqs[v] != undefined) {
    freqs[v]++;
  } else {
    freqs[v] = 1;
  }
});
$.each(freqs, function(num, freq) {
  if (freq > max_value) {
    max_value = freq;
    max_index = num;
  }
});

if (max_index != undefined) {
  alert("Most common element is " + max_index + " with " + max_value + " repetition(s).");
}
​

Upvotes: 8

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