Remove undefined from multidimensional array

Question

So I need to remove empty (undefined) items from the multidimensional array. Atm my code looks like this (it's a method I run so that's why i am using this:

f: function(arr) {
    var __ = this;
    arr = arr.filter(function(item) {
        return Array.isArray(item) ? __.f(item) : typeof(item) !== "undefined";
    });
return arr;
}

but if i run console.log(myObject.f([1, 2, , , , , 3, 4, [5, , , , , ], 6, , , , 8, 3, [[[], 9]]])); i get [ 1, 2, 3, 4, [ 5, , , , ], 6, 8, 3, [ [ [], 9 ] ] ] and that is kinda weird result. I goes pretty well for the first layer but removes only one undefined from inner layers. Also I would like to remove a subarray that consists of no items.

William B · Accepted Answer · 2016-06-08 20:47:28Z

2

Use reductions people! They are beautiful!

function stripUndefined (arr) {
  return arr.reduce(function (result, item) {
    result.push( Array.isArray(item) && !item.length ? stripUndefined(item) : item );
    return result;
  }, []);
}

edited Jun 8, 2016 at 20:47

answered Jun 8, 2016 at 20:27

William B

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6 Comments

Alex Bykov Over a year ago

Very elegant, thanks! So I take it reduce skips undefined items entirely and that's why it does not throw a syntax error regarding length check of undefined

William B Over a year ago

Correct. Strictly speaking Gilad's answer is also a reduction - the process of accumulating some result while looping over a collection - but Array.prototype.reduce does happen to exclude "holes" in the array

Alex Bykov Over a year ago

Yes, I understand that, but here result is a collector that all the items of an array are reduced into, and item is the next iteration of an item queue of an original array. What grabbed my attention is the idea that reduce goes down the array to reduce it, item by item. But why does not ternary throw an error when it reaches undefined item as you can not check the length of undefined?

William B Over a year ago

"reduce executes the callback function once for each element present in the array, excluding holes in the array"

William B Over a year ago

I added that after noticing that undefined values are skipped but null is not. Also because strings have a length.

|

Ankit Deshpande · Accepted Answer · 2016-06-08 20:21:55Z

2

How about something like this:

var arr = [1, 2, , , , , 3, 4, [5, , , , , ], 6, , , , 8, 3, [[[], 9]]];

var remove = function (array) {
  var result = [];

  array.forEach(function (item) {
    if (Array.isArray(item) && item.length!=0) {
      // Item is a nested array, go one level deeper recursively
      result.push(remove(item));
    }
    else if (typeof item !== 'undefined') {
      result.push(item);
    }
  });

  return result;
};

And then console.log(remove(arr)); gives us [1, 2, 3, 4, [5], 6, 8, 3, [[[], 9]]]

edited Jun 8, 2016 at 20:21

Ankit Deshpande

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answered Jun 8, 2016 at 20:00

Gilad Artzi

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2 Comments

Alex Bykov Over a year ago

Thanks a lot, simple but smart! Also I wonder, is there a technical possibility to cut out that null-length array [] (the deeper nested one)?

derekantrican Over a year ago

@AlexBykov See my answer below

derekantrican · Accepted Answer · 2019-07-09 18:55:58Z

0

Similar to Gilad's answer, but completely skipping arrays that result in an empty array (eg [,,,,,])

function cleanArray2d(array){
  var result = [];

  array.forEach(function (item) {
    if (Array.isArray(item) && item.length!=0) {
      // Item is a nested array, go one level deeper recursively

      var nestedArray = cleanArray2d(item);
      if (nestedArray.length > 0)
        result.push(cleanArray2d(item));
    }
    else if (typeof item !== 'undefined' && item != "") {
      result.push(item);
    }
  });

  return result;
}

answered Jul 9, 2019 at 18:55

derekantrican

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