Return index of greatest value in an array

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?

Answers:

Answer

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

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;
}
Answer

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

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

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

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

Simple

maxarr: function(){
let maxval = 0;
let maxindex = null;
for (i = 0; i < arr.length; i++){
if (arr[i] > maxval) {
maxval = arr[i];
maxindex = i;
}
}
}
Answer

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

Here is another solution, If you are using ES6 using spread operator:

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

const indexOfMaxValue = arr.indexOf(Math.max(...arr));
Answer

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
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

 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]

Tags

Recent Questions

Top Questions

Home Tags Terms of Service Privacy Policy DMCA Contact Us

©2020 All rights reserved.