Best way to sort a nested array (returned value vs sort & math.max)

Question

I've been wondering... Is it possible to access a returned value from a function/code block that is nested inside another code block?

For example, I am trying to sort values in an array from largest to smallest using the .sort() method, and then take the largest from each nested array and push it to a new array:

function largestOfFour(arr) {
  for (var i = 0; i < arr.length; i++) {
    console.log(arr[i]);
    arr[i].sort(function(a,b) {
      return b - a; // trying to access this new, organized array
    });
    var newArr = [];
    newArr.push(arr[i][0]);
  }
}

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

.sort sorts the array in place, so arr[i][0] will work fine. Are you asking whether there is a shorter way to do this? FWIW, newArr is completely useless here. — Felix Kling
– Felix Kling, Commented Nov 17, 2015 at 21:58
Right now the sort method sorts the arrays just fine. The problem is accessing those sorted arrays. When I try and access them outside of the .sort() function, I get the original (unsorted) arrays. — Jose
– Jose, Commented Nov 17, 2015 at 22:00
I doubt that. arr[i][0] will give you the largest value. Again, the issue is rather that you are not doing anything with newArray. — Felix Kling
– Felix Kling, Commented Nov 17, 2015 at 22:01
arr[i][0] is currently resulting in: Cannot read property '0' of undefined. — Jose
– Jose, Commented Nov 17, 2015 at 22:04
No it doesn't: jsfiddle.net/tsgLj0sv . Your code executes fine. But again, you are not doing anything with newArr, and you are creating a new array in each iteration. — Felix Kling
– Felix Kling, Commented Nov 17, 2015 at 22:06

Andy · Accepted Answer · 2015-11-17 22:07:28Z

4

You don't even need to sort the array. You could use Math.max.

function largestOfFour(arr) {
  return arr.map(function (el) {
    return Math.max.apply(null, el);
  });
} // [ 5, 27, 39, 1001 ]

DEMO

answered Nov 17, 2015 at 22:07

Andy

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1 Comment

Victoria French Over a year ago

nice... you win (as long as it's a number... as long as it's a number). :-)

Victoria French · Accepted Answer · 2015-11-17 22:06:18Z

2

Like this?

function largestOfFour(arr) {
  var results = [];
  for (var i = 0; i < arr.length; i++) {
    console.log(arr[i]);
    arr[i].sort(function(a,b) {
      return b - a; // trying to access this new, organized array
    });
    results.push(arr[i][0]);
  }
  return results;
}

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

answered Nov 17, 2015 at 22:06

Victoria French

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1 Comment

Jose Over a year ago

That works, yep! Thanks a lot! I marked Andy's answer as the correct answer, however - only because it makes use of a shorter method!

Felix Kling · Accepted Answer · 2015-11-17 22:13:01Z

1

Is it possible to access a returned value from a function/code block that is nested inside another code block?

You can only get the return value if you call the function. You are not calling the .sort callback, it is called inside .sort, so you cannot get its return value.

I assume you are asking because the code you have doesn't work as you expect. Looking at your code, it appears obvious that, while you sort the array, you are not really doing anything with it after you sorted it. As I already said in my comments, the way you use .sort and access the largest value is perfectly fine.

I think you want to declare newArr outside the loop and return it from the function:

function largestOfFour(arr) {
  var newArr = [];
  for (var i = 0; i < arr.length; i++) {
    arr[i].sort(function(a, b) {
      return b - a;
    });
    newArr.push(arr[i][0]);
  }
  return newArr;
}

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

There is also a more streamlined way of writing this, that doesn't mutate the input array:

function largestOfFour(arr) {
  return arr.map(function(innerArray) {
    return innerArray.reduce(function(a, b) {
      return a > b ? a : b;
    });
  });
}

edited Nov 17, 2015 at 22:13

answered Nov 17, 2015 at 22:03

Felix Kling

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5 Comments

Jose Over a year ago

Unfortunately, that results in the same thing. I'm still only getting the original, unsorted arrays. return newArr doesn't output anything, either.

Felix Kling Over a year ago

I had a typo, but other than that it works perfectly fine. Try it again.

Jose Over a year ago

Thanks! I marked Andy's answer as the correct one, though. Only because it seems to be a more efficient method (without using sort).

Felix Kling Over a year ago

@Jose: In that case I suggest to update your question and reformulate it to ask what you really want to know. Because, while the accepted answer fixes your code, it doesn't seem to answer your question.

Andy Over a year ago

TBF I should write a bit more in my answer :) But it was really meant as an "As an aside..."

guest271314 · Accepted Answer · 2015-11-18 05:06:14Z

0

trying to sort values in an array from largest to smallest using the .sort() method, and then take the largest from each nested array and push it to a new array:

Try using while loop , .push() .sort() , .reverse() . largestOfFour returns two arrays , whose indexx can be toggled at return statement ; largest number from each array at first array , arrays in array sorted from largest to least at second array

function largestOfFour(arr) {
  var res = [], i = arr.length - 1;
  while(i > -1) {
    res.push(Math.max.apply(Math, arr[i].sort(function(a, b) {
      return Number(b) - Number(a)})
      )); 
  --i;
  };     
  return [res, arr.reverse()]
}

var input = [
    [4, 5, 1, 3],
    [13, 27, 18, 26],
    [32, 35, 37, 39],
    [1000, 1001, 857, 1]
  ];

function largestOfFour(arr) {
  var res = [], i = arr.length - 1;
  while(i > -1) {
    res.push(Math.max.apply(Math, arr[i].sort(function(a, b) {
      return Number(b) - Number(a)})
      )); 
  --i;
  };     
  return [res, arr.reverse()]
}

document.getElementsByTagName("pre")[0]
.textContent = JSON.stringify(largestOfFour(input), null, 2);

<pre></pre>

edited Nov 18, 2015 at 5:06

answered Nov 17, 2015 at 22:11

guest271314

1

5 Comments

Felix Kling Over a year ago

Why sort if you use Math.max anyway?

Andy Over a year ago

You don't need to sort the array as well as use Math.max.

guest271314 Over a year ago

@FelixKling "Why sort if you use Math.max anyway? " As interpreted here from OP "am trying to sort values in an array from largest to smallest using the .sort() method, and then take the largest from each nested array and push it to a new array:", two separate return results were expected; 1) sorted arrays, 2) max of each sorted array

Felix Kling Over a year ago

Ah I see. But in that case I would use return val[0]; instead of Math.max. (I assumed the OP only sorted in order to get the largest value).

guest271314 Over a year ago

@FelixKling "But in that case I would use return val[0]; instead of Math.max." Yes, that would be briefer . Have to ingrain to use array notation where possible . If single array was expected , could use [0] within first .map() at .sort() to avoid any further calls

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