JavaScript finding the largest integer in an array of arrays

Question

function largestOfFour(arr) {
    var newArray = [];
    for(var i =0; i <=arr.length-1; i++){
        console.log(arr[i]);
        newArray[i] = Math.max(arr[i]);
    }
    console.log(newArray);
              // You can do this!
    return newArray;
}

largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);

So I do not understand why I am getting NaN even though i have provided an argument within my math.max function. In the console.log within my for loop, i get each array within the main array to display. Meaning if I use the same arr[i] within the max function, it should give me the max of that sub Array.

yes So what i am trying to accomplish is go through each subarray and find the max integer within it, and pass the max integer into a seperate array. So the function return an array with 4 integers each integer being the max integer of the 4 sub array within the main array. — Ali Aslam
– Ali Aslam, Commented Aug 6, 2015 at 23:16
Possible duplicate of why math.max() returning NaN with array of integers? — user202729
– user202729, Commented May 14, 2018 at 16:03

Oriol · Accepted Answer · 2015-08-06 23:17:07Z

3

You are passing an array to Math.max and expect it to return the maximum in that array.

However, Math.max returns the maximum among its arguments. So use

var newArray = [];
for(var i =0; i < arr.length; ++i)
  newArray[i] = Math.max.apply(void 0, arr[i]);

In ES6, you can use arrow functions and the spread operator to simplify:

arr.map(a => Math.max(...a));

answered Aug 6, 2015 at 23:17

Oriol

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3 Comments

Ali Aslam Over a year ago

Cool! Your solution works however before I use it, i want to know what does the void 0 mean within math.max.apply()?

Oriol Over a year ago

@AliAslam It's the value used as the this value inside the function. Since Math.max doesn't use it, it doesn't matter.

Pointy Over a year ago

Or just newArray = arr.map(Math.max.apply.bind(Math.max, Math)) :)

Kirill Slatin · Accepted Answer · 2015-08-06 23:16:38Z

2

It won't work like that. Math.max expect separate numbers. You can get desired output using apply

function largestOfFour(arr) {
    return arr.map(function(subArr){
        return Math.max.apply(Math, subArr);
    });
}

answered Aug 6, 2015 at 23:16

Kirill Slatin

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Comments

Daniel A. White · Accepted Answer · 2015-08-06 23:21:47Z

0

If you want the ultimate max do this.

var original = [[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]];
Math.max.apply(null, original.map(function(arr) { return Math.max.apply(null, arr); }));

or if you want a the max of each array just do

original.map(function(arr) { return Math.max.apply(null, arr); });

edited Aug 6, 2015 at 23:21

answered Aug 6, 2015 at 23:17

Daniel A. White

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1 Comment

Kirill Slatin Over a year ago

So this actually returns a single max among all arrays which is not the OP's intent. But can be a separate task

Jerome Anthony · Accepted Answer · 2015-08-06 23:35:18Z

0

Using lodash

var x = ([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
_.flatten(x).max();

answered Aug 6, 2015 at 23:35

Jerome Anthony

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Comments

Pastel Belem · Accepted Answer · 2015-08-06 23:34:59Z

-1

Edited: In order to add values to the array, I'd suggest using the push methodn. As in: NewArray.push(Math.max(...sub[i])); or newarray.push(math.max.apply(null, sub[i]). An alternative is alocating, when declarating the array, the size it will take: Var newArray = new Array(arr.length); There's this new spread operator (...) in javascript !

edited Aug 6, 2015 at 23:34

answered Aug 6, 2015 at 23:19

Pastel Belem

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3 Comments

Kirill Slatin Over a year ago

Math.max([1,2,3]) doesn't find the max aomng array elements. It returns NaN.

Pastel Belem Over a year ago

Yes! In fact, i've done a better research and Math.max only accepts numbers, not the array. If you pass sub[i] you will have the NaN. I suggest newarray.push(null, math.max.apply(sub[i]))

Kirill Slatin Over a year ago

There is no spread operator (...) in widely available javascript, i.e. in browsers. As mentioned in other answers it is a feature of ES6

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