How might I find the largest number contained in a JavaScript array?

How might I find the largest number contained in a JavaScript array?

I have a simple JavaScript Array object containing a few numbers.
[267, 306, 108]

Is there a function that would find the largest number in this array?

Solutions/Answers:

Solution 1:

Resig to the rescue:

Array.max = function( array ){
    return Math.max.apply( Math, array );
};

Warning: since the maximum number of arguments is as low as 65535 on some VMs, use a for loop if you’re not certain the array is that small.

Solution 2:

You can use the apply function, to call Math.max:

var array = [267, 306, 108];
var largest = Math.max.apply(Math, array); // 306

How it works?

The apply function is used to call another function, with a given context and arguments, provided as an array. The min and max functions can take an arbitrary number of input arguments: Math.max(val1, val2, …, valN)

So if we call:

Math.min.apply(Math, [1,2,3,4]);

The apply function will execute:

Math.min(1,2,3,4);

Note that the first parameter, the context, is not important for these functions since they are static, they will work regardless of what is passed as the context.

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Solution 3:

The easiest syntax, with the new spread operator:

var arr = [1, 2, 3];
var max = Math.max(...arr);

Source : Mozilla MDN

Solution 4:

I’m no JS expert, but I wanted to see how these methods stack up, so this was good practice for me. I don’t know if this is technically the right way to performance test these, but I just ran them one right after another, as you can see in my code.

Sorting and getting the 0th value is by far the worst method (and it modifies the order of your array, which may not be desirable). For the others, the difference is negligible unless you’re talking millions of indices.

Average results of five runs with a 100,000-index array of random numbers:

  • reduce took 4.0392ms to run
  • Math.max.apply took 3.3742ms to run
  • sorting and getting the 0th value took 67.4724ms to run
  • Math.max within reduce() took 6.5804ms to run
  • custom findmax function took 1.6102ms to run
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var performance = window.performance

function findmax(array)
{
  var max = 0,
      a = array.length,
      counter

  for (counter=0;counter<a;counter++)
  {
      if (array[counter] > max)
      {
          max = array[counter]
      }
  }
  return max
}

function findBiggestNumber(num) {
  var counts = []
  var i
  for (i = 0; i < num; i++) {
    counts.push(Math.random())
  }

  var a, b

  a = performance.now()
  var biggest = counts.reduce(function(highest, count){
        return highest > count ? highest : count
      }, 0)
  b = performance.now()
  console.log('reduce took ' + (b - a) + ' ms to run')

  a = performance.now()
  var biggest2 = Math.max.apply(Math, counts)
  b = performance.now()
  console.log('Math.max.apply took ' + (b - a) + ' ms to run')

  a = performance.now()
  var biggest3 = counts.sort(function(a,b){return b-a;})[0]
  b = performance.now()
  console.log('sorting and getting the 0th value took ' + (b - a) + ' ms to run')

  a = performance.now()
  var biggest4 = counts.reduce(function(highest, count){
        return Math.max(highest,count)
      }, 0)
  b = performance.now()
  console.log('Math.max within reduce() took ' + (b - a) + ' ms to run')

  a = performance.now()
  var biggest5 = findmax(counts)
  b = performance.now()
  console.log('custom findmax function took ' + (b - a) + ' ms to run')
  console.log(biggest + '-' + biggest2 + '-' + biggest3 + '-' + biggest4 + '-' + biggest5)

}

findBiggestNumber(1E5)

Solution 5:

I’ve found that for bigger arrays (~100k elements), it actually pays to simply iterate the array with a humble for loop, performing ~30% better than Math.max.apply():

function mymax(a)
{
    var m = -Infinity, i = 0, n = a.length;

    for (; i != n; ++i) {
        if (a[i] > m) {
            m = a[i];
        }
    }

    return m;
}

Benchmark results

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Solution 6:

You could sort the array in descending order and get the first item:

[267, 306, 108].sort(function(a,b){return b-a;})[0]