How to remove the blank element in javascript array?

Question

I've the following implementation.

Array.prototype.abc = function(condition, t){
    var arr = [];
    for( var i = 0; i < this.length; i++){
        arr.push(condition(this[i],t));         
    }
    return arr;
};

var a = [1,2,3,4];
var t = 2;

alert(a.abc( function(item,diviser){
    if(item % diviser === 0) { return item; }
},t));

The result should be [2,4]. Instead I'm getting [,2,,4].

I've tried the following conditions but it always returns the above result.

if(condition(this[i],t) !== false){ arr.push(condition(this[i],t)); }

This I'm doing when I do

if(item % diviser === 0) { return item; } else { return false; }

The false condition doesnot work for 'false' or true or 'true' or for that matter any value I return in else part. In the actual implementation I do not have the else part though. I know we can use splice or something to remove the blank sections. But I do not understand even if i don't return anything why blank in being created. And even after using if condition before arr.push it does not work. What exactly is happening and how to remove the blanks in the array?

Rajaprabhu Aravindasamy · Accepted Answer · 2016-04-14 10:35:28Z

1

You have to check whether the returned value is undefined or not from your callBack,

for( var i = 0; i < this.length; i++){
  var x = condition(this[i],t);
  if(typeof x !== "undefined") arr.push(x);         
}

You have made 2 different attempts to fix this but that will fail because,

!== false, your callBack will return either a number or undefined. So the strict equality will fails all the time and will be evaluated to false always since the operand's types are different.
if(item % diviser === 0) { return item; } else { return false; } This will return false when the condition fails and hence false will be pushed into the array. And it will not skip that pushing part.

edited Apr 14, 2016 at 10:35

answered Apr 14, 2016 at 10:28

Rajaprabhu Aravindasamy

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2 Comments

Sp0T Over a year ago

I tried comparing with !== false when i return else{ return false; }. But this doesn't work. why is that so?

Rajaprabhu Aravindasamy Over a year ago

@Sp0T Careful about this snippet, condition(this[i],t) && arr.push(this[i]); This would ignore 0 if it is there in your main array.

Nina Scholz · Accepted Answer · 2016-04-14 10:29:53Z

0

Move the cmondition outside of the push and place it before. The condition should only return true or false.

condition(this[i],t) && arr.push(this[i]);

answered Apr 14, 2016 at 10:29

Nina Scholz

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Comments

gurvinder372 · Accepted Answer · 2016-04-14 10:38:17Z

0

This is because you are doing a push regardless of what condition is returning.

Replace this

arr.push(condition(this[i],t));

with

var val = condition(this[i],t)
val = val !== false && arr.push(val);

Since your condition method is explicitly returing false, you can simply check for val !== false

DEMO

Array.prototype.abc = function(condition, t){
    var arr = [];
    for( var i = 0; i < this.length; i++){
       var val = condition(this[i],t); console.log(val);
       val !== false && arr.push(val);
    }
    return arr;
};

var a = [1,2,3,4];
var t = 2;

document.body.innerHTML += a.abc( function(item,diviser){
    if(item % diviser === 0) { return item; } else { return false; }
},t);

edited Apr 14, 2016 at 10:38

answered Apr 14, 2016 at 10:29

gurvinder372

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