Return index of greatest value in an array

Question

I have this:

var arr = [0, 21, 22, 7];

What's the best way to return the index of the highest value into another variable?

Answer 1

To find the index of the greatest value in an array, copy the original array into the new array and then sort the original array in decreasing order to get the output [22, 21, 7, 0]; now find the value 22 index in the copyNumbers array using this code copyNumbers.indexOf(numbers[0]);

const numbers = [0, 21, 22, 7];
const copyNumbers = [];
copyNumbers.push(...numbers);
numbers.sort(function(a, b) {
  return b - a
});
const index = copyNumbers.indexOf(numbers[0]);
console.log(index);

Answer 2

Make this

const max = arr.reduce((m, n) => Math.max(m, n))

then the indexes of the max

get index with findIndex

var index = arr.findIndex(i => i === max)

Answer 3

Simple, clear solution using for

const arr = [1, 2, 60, 2, 10, 20];

for (var i = 0, biggerIndex = 0; i < arr.length; i++) {
    if (arr[i] > arr[biggerIndex]) {
        biggerIndex = i;
    }
}

console.log(biggerIndex);

Answer 4

A minor modification revised from the "reduce" version of @traxium 's solution taking the empty array into consideration:

function indexOfMaxElement(array) {
    return array.reduce((iMax, x, i, arr) => 
        arr[iMax] === undefined ? i :
        x > arr[iMax]           ? i : iMax
        , -1            // return -1 if empty
    );
}

Answer 5

To complete the work of @VFDan, I benchmarked the 3 methods: the accepted one (custom loop), reduce, and find(max(arr)) on an array of 10000 floats.

Results on chromimum 85 linux (higher is better):

• custom loop: 100%
• reduce: 94.36%
• indexOf(max): 70%

Results on firefox 80 linux (higher is better):

• custom loop: 100%
• reduce: 96.39%
• indexOf(max): 31.16%

Conclusion:

If you need your code to run fast, don't use indexOf(max). reduce is ok but use the custom loop if you need the best performances.

You can run this benchmark on other browser using this link: https://jsben.ch/wkd4c

Answer 6

If you create a copy of the array and sort it descending, the first element of the copy will be the largest. Than you can find its index in the original array.

var sorted = [...arr].sort((a,b) => b - a)
arr.indexOf(sorted[0])

Time complexity is O(n) for the copy, O(n*log(n)) for sorting and O(n) for the indexOf.

If you need to do it faster, Ry's answer is O(n).

Answer 7

function findIndicesOf(haystack, needle)
{
    var indices = [];

    var j = 0;
    for (var i = 0; i < haystack.length; ++i) {
        if (haystack[i] == needle)
            indices[j++] = i;
    }
    return indices;
}

pass array to haystack and Math.max(...array) to needle. This will give all max elements of the array, and it is more extensible (for example, you also need to find min values)

Answer 8

This is probably the best way, since it’s reliable and works on old browsers:

function indexOfMax(arr) {
    if (arr.length === 0) {
        return -1;
    }

    var max = arr[0];
    var maxIndex = 0;

    for (var i = 1; i < arr.length; i++) {
        if (arr[i] > max) {
            maxIndex = i;
            max = arr[i];
        }
    }

    return maxIndex;
}

There’s also this one-liner:

let i = arr.indexOf(Math.max(...arr));

It performs twice as many comparisons as necessary and will throw a RangeError on large arrays, though. I’d stick to the function.

Answer 9

Another solution of max using reduce:

[1,2,5,0,4].reduce((a,b,i) => a[0] < b ? [b,i] : a, [Number.MIN_VALUE,-1])
//[5,2]

This returns [5e-324, -1] if the array is empty. If you want just the index, put [1] after.

Min via (Change to > and MAX_VALUE):

[1,2,5,0,4].reduce((a,b,i) => a[0] > b ? [b,i] : a, [Number.MAX_VALUE,-1])
//[0, 3]

Answer 10

EDIT: Years ago I gave an answer to this that was gross, too specific, and too complicated. So I'm editing it. I favor the functional answers above for their neat factor but not their readability; but if I were more familiar with javascript then I might like them for that, too.

Pseudo code:

Track index that contains largest value. Assume index 0 is largest initially. Compare against current index. Update index with largest value if necessary.

Code:

var mountains = [3, 1, 5, 9, 4];

function largestIndex(array){
  var counter = 1;
  var max = 0;

  for(counter; counter < array.length; counter++){
    if(array[max] < array[counter]){
        max = counter;
    }
  }
  return max;
}

console.log("index with largest value is: " +largestIndex(mountains));
// index with largest value is: 3

Answer 11

If you are utilizing underscore, you can use this nice short one-liner:

_.indexOf(arr, _.max(arr))

It will first find the value of the largest item in the array, in this case 22. Then it will return the index of where 22 is within the array, in this case 2.

Answer 12

In one line and probably faster then arr.indexOf(Math.max.apply(Math, arr)):

var a = [0, 21, 22, 7];
var indexOfMaxValue = a.reduce((iMax, x, i, arr) => x > arr[iMax] ? i : iMax, 0);

document.write("indexOfMaxValue = " + indexOfMaxValue); // prints "indexOfMaxValue = 2"

Where:

• iMax - the best index so far (the index of the max element so far, on the first iteration iMax = 0 because the second argument to reduce() is 0, we can't omit the second argument to reduce() in our case)
• x - the currently tested element from the array
• i - the currently tested index
• arr - our array ([0, 21, 22, 7])

About the reduce() method (from "JavaScript: The Definitive Guide" by David Flanagan):

reduce() takes two arguments. The first is the function that performs the reduction operation. The task of this reduction function is to somehow combine or reduce two values into a single value, and to return that reduced value.

Functions used with reduce() are different than the functions used with forEach() and map(). The familiar value, index, and array values are passed as the second, third, and fourth arguments. The first argument is the accumulated result of the reduction so far. On the first call to the function, this first argument is the initial value you passed as the second argument to reduce(). On subsequent calls, it is the value returned by the previous invocation of the function.

When you invoke reduce() with no initial value, it uses the first element of the array as the initial value. This means that the first call to the reduction function will have the first and second array elements as its first and second arguments.

Answer 13

 var arr=[0,6,7,7,7];
 var largest=[0];
 //find the largest num;
 for(var i=0;i<arr.length;i++){
   var comp=(arr[i]-largest[0])>0;
      if(comp){
	  largest =[];
	  largest.push(arr[i]);
	  }
 }
 alert(largest )//7
 
 //find the index of 'arr'
 var arrIndex=[];
 for(var i=0;i<arr.length;i++){
    var comp=arr[i]-largest[0]==0;
	if(comp){
	arrIndex.push(i);
	}
 }
 alert(arrIndex);//[2,3,4]

Answer 14

A stable version of this function looks like this:

// not defined for empty array
function max_index(elements) {
    var i = 1;
    var mi = 0;
    while (i < elements.length) {
        if (!(elements[i] < elements[mi]))
            mi = i;
        i += 1;
    }
    return mi;
}

Answer 15

Unless I'm mistaken, I'd say it's to write your own function.

function findIndexOfGreatest(array) {
  var greatest;
  var indexOfGreatest;
  for (var i = 0; i < array.length; i++) {
    if (!greatest || array[i] > greatest) {
      greatest = array[i];
      indexOfGreatest = i;
    }
  }
  return indexOfGreatest;
}