Javascript: How to clear undefined values from an array

Question

I'm trying to loop through an array and delete and skip the elements until only one is existing. i've tried splicing but it messes up my loop because the element from arr[1] then becomes arr[0] etc.

Let's say there are 10 people. I'd like to remove person 1 then keep person 2 then remove person 3 and keep person 4. This pattern will go on until only one is left.

any kind of help will do.

Answer 1

you should not change the collection during the iterating, not just JavaScript but all language, define a new array and add those ones you want to delete in it, and iterate that one later to delete from first one.

Answer 2

Filter the falsy items:

var a=[1,2,"b",0,{},"",NaN,3,undefined,null,5];
var b=a.filter(Boolean); // [1,2,"b",{},3,5]

Answer 3

When you splice, just decrement your loop index.

There were lots of good suggestions, I'll post the code for the different options and you can decide which to use

Decrement index when splicing http://jsfiddle.net/mendesjuan/aFvVh/

var undef;
var arr = [1,2, undef, 3, 4, undef];
for (var i=0; i < arr.length; i++) {
    if ( arr[i] === undef ) {
        arr.splice(i,1);
        i--;
    }
}

Loop backwards http://jsfiddle.net/mendesjuan/aFvVh/1/

var undef;
var arr = [1,2, undef, 3, 4, undef];
for (var i=arr.length - 1; i >=0; i--) {
    if ( arr[i] === undef ) {
        arr.splice(i,1);
    }
}

Copy to new array http://jsfiddle.net/mendesjuan/aFvVh/2/

var undef;
var arr = [1,2, undef, 3, 4, undef];
var temp = [];
for (var i=0; i < arr.length; i++) {
    if ( arr[i] !== undef ) {
        temp.push(arr[i])
    }
}
arr = temp;

Use filter which is just a fancy way to create a new array

var undef;
var arr = [1,2, undef, 3, 4, undef];
arr = arr.filter(function(item){
    return item !== undef;
});

At the end of all those examples, arr will be [1,2,3,4]

Performance

IE 11, FF and Chrome agree that Array.splice is the fastest. 10 times (Chrome), 20 times (IE 11) as fast as Array.filter. Putting items into a new array was also slow when compared to Array.slice. See http://jsperf.com/clean-undefined-values-from-array2

I am really surprised to see IE lead the pack here, and to see Chrome behind FF and IE. I don't think I've ever run a test with that result.

Answer 4

Loop backwards. (Removing items will thus not affect the indexes of elements not yet processed.)

Answer 5

If by any chance you're using CoffeeScript then to remove undefined from Array do this

values = ['one', undefined]
values = (item for item in values when item != undefined) 

values
/* => ['one'] */

Answer 6

Surprisingly, nobody have answered the best and correct way:

1. Create new array
2. Iterate on the old array and only push the elements you want to keep to the new array

some credit goes to @nnnnnn comment

Answer 7

Not gathering exactly what you are trying to achieve, but I feel you are relying on the position index of an item in the array to continue with your program. I would in this case suggest a hashed array, i.e., a Key<>Value pair array.

In which case, arr["2"] always points at the item you had placed in it originally. Thus you can logically/numerically loop through, while not worrying about changes in position.

Beware of the Type Conversion risk and pitfalls!

Answer 8

Your best bet is to create a duplicate of the array, then splice from the original.

Or just go using a collection (key->value) and just delete the key eg

 People = {a: "Person A", b: "Person B", c:"Person C"};
 delete People.a;
 delete People.c;  //now the People collection only has 1 entry.

You can replace a,b,c with numbers just using it as an example,

 People = {0: "Person A", 1: "Person B", 2:"Person C"};
 delete People[0];
 delete People[1];

Answer 9

this is a sample for you

<script lanauge = "javascript">
    var arr = ["1","2","3","4"];
    delete arr[1];// arr[1] is undefined
    delete arr[2];// arr[2] is undefined

    // now arr.length is 4

    var todelete = [];
    for (i = 0 ; i < arr.length ;i++)
    {
        if (typeof arr[i] == 'undefined') todelete.push(i);
    }

    todelete.sort(function(a, b) { return b-a }); // make the indeies from big to small

    for (i = 0;i < todelete.length; i ++)
    {
        arr.splice(todelete[i],1);
    }

    // now arr.length is 2

</script>

Answer 10

This may not be what you want, but you can easily calculate what the final element at the end of this procedure will be, then just grab it. Assuming that the elements of the array are contiguous and start at arr[0], you can find:

var logBase2OfLength = Math.floor(Math.log(arr.length) / Math.log(2));
var finalElement = arr[(1 << logBase2OfLength) - 1];

Basically, if you take the integer power of 2 that is less than or equal to the number of elements in your array, that is the position of the element that will remain after all of the looping and deleting.

Answer 11

function removeUndefined(array)
{
    var i = 0;
    while (i < array.length)
        if (typeof array[i] === 'undefined')
            array.splice(i, i);
        else
            i++;
    return array;
}

EDIT: I wrote this based on the title. Looks like the question asks something completely different.

Answer 12

>If you are getting undefined during deletion of the key-pair, Then to prevent "undefined" you can try code given below to delete key-pair  


   1) test = ["1","2","3","4",""," "];
   2) var delete = JSON.stringify(test);   
      case1) delete = delete.replace(/\,""/g,'');  
         or  
      case2) delete = delete.replace(/\," "/g,'');  
         or  
      case3) delete = delete.replace(/\,null/g,'');  

   3) var result = JSON.parse(delete);  

Answer 13

  while(yourarray.length>1) //only one left
    {
    // your code
    }