adding elements in array in javascript

Question

I have an array of 5 elements ex. arr = [1,2,3,4,5]; I want to add elements between these elements, and place them in new array arr1 = [1,0,0,0,2,0,0,0,3,0,0,0,4,0,0,0,5,0,0,0];. When i alert arr1 i get the same array as it is, but when i alert arr1.length i get that it`s length is 5, when it is actually 20. Can you help me fix this, or tell me why do i get that result. Here is an example of the code i am using:

function niza(val,times){
  var arr = [];
  for (var i=0;i<times;i++) {
    arr.push(val);
  }
  return arr;
}

and then this:

var y1=0;
var arr= [];
var a = new Array();
for (var j=0;j<Niza1.length;j++) {
  y1 = Niza1[j];
  arr = y1 + "," + niza(0,11);
  a.push(arr);
}

where Niza1 holds the 5 elements mentioned before in arr, and a holds the elements mentioned in arr1.

You're creating a new Array that will be given 5 strings. I can't reconcile that with your desired result. Is this some sort of homework assignment? — cookie monster
– cookie monster, Commented Mar 7, 2014 at 15:13

Sebastien C. · Accepted Answer · 2014-03-07 15:16:01Z

5

I'm not sure to understand the code you wrote, but do you know you can add multiple elements at once with arr.push ?

var array1 = [1, 2, 3, 4, 5];

var array2 = [];
for(var i = 0 ; i < array1.length ; i++) {
    array2.push(array1[i], 0, 0, 0);
}

//array2 == [1, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 4, 0, 0, 0, 5, 0, 0, 0]

answered Mar 7, 2014 at 15:16

Sebastien C.

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4 Comments

Nicky Smits Over a year ago

Nice answer. At least this is clear code. Nicely formatted :)

Mark Walters Over a year ago

+1 - It's not a re-usable like p.s.w.g's answer but it is a simple and clear answer.

Sebastien C. Over a year ago

@user3242508 I saw what you tried to say (but you tried to edit my answer instead of adding a comment ...). If you want to add n zeroes and not strictly three, I need to know how do you determine how many zeroes you want to add ?

user3242508 Over a year ago

that is why i use the first function posted before

p.s.w.g · Accepted Answer · 2014-03-07 15:16:51Z

4

You question is a little hard to follow, but try something like this:

function inject(original, val, times) {
    var res = [];
    for(var i=0; i < original.length; i++){
        res.push(original[i]);

        for(var j = 0; j < times; j++){
            res.push(val);
        }
    }
    return res;
}

Demonstration

answered Mar 7, 2014 at 15:16

p.s.w.g

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2 Comments

user3242508 Over a year ago

I have tried this code, but it doesn`t just put n-zeros(which is the variable times) between the main numbers, but it multiplies the main numbers for "times" times. Hope i was clear.

p.s.w.g Over a year ago

@user3242508 Then you must have done something wrong when testing it. See the demonstration; the output is [1, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 4, 0, 0, 0, 5, 0, 0, 0].

krisk · Accepted Answer · 2014-03-07 15:47:10Z

3

Here's a solution that would work with an arbitrary list and length, and is therefore reusable:

function inject(original, values) {
    var result = [];
    for (var i = 0 ; i < original.length ; i++) {
        result.push(original[i]);
        result.push.apply(result, values);
    }
    return result;
}

console.log(inject([1,2,3], [0,0,0])); 
// output: [1, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0]

console.log(inject([1,1,1], [2,6,2,6])); 
// output: [1, 2, 6, 2, 6, 1, 2, 6, 2, 6, 1, 2, 6, 2, 6]

It leverages the the native apply (here's an explanation) function to execute the push with an arbitrary list of arguments, defined by the values array.

edited Mar 7, 2014 at 15:47

answered Mar 7, 2014 at 15:39

krisk

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