JS Array Mix

[Brian Miller]

If I had 2 arrays that looked like this:

 

["1", "2", "3", "4", "5"]

and

["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K"]

 

and I wanted to make it end up like this:

["A", "1", "B", "2", "C", "3", "C", "4", "E", "5", "F", "G", "H", "I", "J", "K"]

 

What would be the easiest way to do it in JS, especially assuming that we may never know the length of the first two arrays?  It should also allow duplicates.

 

Thank you.

[Brian Miller]

I've found the solution:

 

var arr1 = ["1", "2", "3", "4", "5"],
    arr2 = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K"],
    arr3 = [],
    arrLen = Math.min(arr1.length, arr2.length);
    
for (i = 0; i < arrLen; i++) {
    result.push(arr1[i], arr2[i]);
}

arr3.push(...arr1.slice(arrLen), ...arr2.slice(arrLen));
console.log(arr3);

 

[pmrd]

Your code seems way too complicated for no reason. Here's a better solution if you don't mind modifying the source arrays

 

let array1 = ["1", "2", "3", "4", "5"];
let array2 = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K"];
let array3 = [];

while (array1.length > 0 || array2.length > 0)
{

	if (array1.length > 0)  array3.push(array1.shift())
	if (array2.length > 0)  array3.push(array2.shift())

}

 

 

Edited October 16, 2022 by PmRd

[Brian Miller]

  On 16/10/2022 at 03:23, PmRd said:

Your code doesn't work and seems way too complicated for no reason. Here's a better solution if you don't mind modifying the source arrays

 

let array1 = ["1", "2", "3", "4", "5"];
let array2 = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K"];
let array3 = [];

while (array1.length > 0 || array2.length > 0)
{

	if (array1.length > 0)  array3.push(array1.shift())
	if (array2.length > 0)  array3.push(array2.shift())

}

 

 

Expand  

Thank you. Your method is much more elegant

[pmrd]

  On 16/10/2022 at 03:30, Brian Miller said:

Thank you. Your method is much more elegant

Expand  

If you want the source arrays to remain intact you can do this instead

 

let array1 = ["1", "2", "3", "4", "5"];
let array2 = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K"];

function mix(arr1, arr2) {
  
    let array1clone = [...arr1];
    let array2clone = [...arr2];
  
    let result = [];
  
    while (array1clone.length > 0 || array2clone.length > 0)
    {

      if (array1clone.length > 0)  result.push(array1clone.shift())
      if (array2clone.length > 0)  result.push(array2clone.shift())

    }
  
    return result;
  
}

console.log(mix(array1, array2));

 

[Brian Miller]

  On 16/10/2022 at 03:35, PmRd said:

If you want the source arrays to remain intact you can do this instead

 

let array1 = ["1", "2", "3", "4", "5"];
let array2 = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K"];

function mix(arr1, arr2) {
  
    let array1clone = [...arr1];
    let array2clone = [...arr2];
  
    let result = [];
  
    while (array1clone.length > 0 || array2clone.length > 0)
    {

      if (array1clone.length > 0)  result.push(array1clone.shift())
      if (array2clone.length > 0)  result.push(array2clone.shift())

    }
  
    return result;
  
}

console.log(mix(array1, array2));

 

Expand  

I like that. Thank you

[+DonC Subscriber²]

Here's a fancy-pants version that supports an arbitrary number of arrays and also sparse arrays:

 

function mixArrays(...arrays) {

    const arrayLengths = arrays.map(array => array.length);
    const maxLength = Math.max(...arrayLengths);
    
    let result = [];

    for (let index = 0; index < maxLength; index++) {
        for (let array of arrays) {
            if (index in array) {
                result.push(array[index]);
            }
        }       
    }

    return result;
}

 

Example with two arrays:

 

let array1 = ["1", "2", "3", "4", "5"];
let array2 = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K"];
let array3 = ["a", "b", "c"];

mixArrays(array1, array2);
// ['1', 'A', '2', 'B', '3', 'C', '4', 'D', '5', 'E', 'F', 'G', 'H', 'I', 'J', 'K']

 

Three arrays:

 

mixArrays(array1, array2, array3);
// ['1', 'A', 'a', '2', 'B', 'b', '3', 'C', 'c', '4', 'D', '5', 'E', 'F', 'G', 'H', 'I', 'J', 'K']

 

And sparse arrays:

 

array1[8] = "8";

mixArrays(array1, array2);
// ['1', 'A', '2', 'B', '3', 'C', '4', 'D', '5', 'E', 'F', 'G', 'H', '8', 'I', 'J', 'K']

 

[Brian Miller]

  On 16/10/2022 at 12:39, DonC said:

Here's a fancy-pants version that supports an arbitrary number of arrays and also sparse arrays:

 

function mixArrays(...arrays) {

    const arrayLengths = arrays.map(array => array.length);
    const maxLength = Math.max(...arrayLengths);
    
    let result = [];

    for (let index = 0; index < maxLength; index++) {
        for (let array of arrays) {
            if (index in array) {
                result.push(array[index]);
            }
        }       
    }

    return result;
}

 

Example with two arrays:

 

let array1 = ["1", "2", "3", "4", "5"];
let array2 = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K"];
let array3 = ["a", "b", "c"];

mixArrays(array1, array2);
// ['1', 'A', '2', 'B', '3', 'C', '4', 'D', '5', 'E', 'F', 'G', 'H', 'I', 'J', 'K']

 

Three arrays:

 

mixArrays(array1, array2, array3);
// ['1', 'A', 'a', '2', 'B', 'b', '3', 'C', 'c', '4', 'D', '5', 'E', 'F', 'G', 'H', 'I', 'J', 'K']

 

And sparse arrays:

 

array1[8] = "8";

mixArrays(array1, array2);
// ['1', 'A', '2', 'B', '3', 'C', '4', 'D', '5', 'E', 'F', 'G', 'H', '8', 'I', 'J', 'K']

 

Expand  

Thank you, that is totally fancy pants!!!