javascript: combine array of arrays into an array of single multidimensional array

Question

I have N number of arrays in an array, for now N = 3

[[1,2,3, "a","b"] , [3,4,5,"c"], [6,7,8]...]

I would like for it to return

[[1,3,6], [2,4,7], [3,5,8], ...., ["a","c", ""], ["b", "",""]]

There are lot of solutions of combining two arrays but I want to handle any number of arrays. I would like for it to return above. I am looking to do this without underscore or jquery.

    Values=[];
    status = [[1,2,3, "a","b"] , [3,4,5,"c"], [6,7,8]];
    status_name = status[0]; //take the longest one always
    for (i = 0; i < status_name.length; ++i)
    {
        Values[i] =Array(status_name[i], status[i+1][i]);
    }

It is clear, but the question doesn't show any research effort. — elclanrs, Jul 25 '14 at 23:26
basically im stuck on how I can make it accept N number of arrays in the code I attempted. — user299709, Jul 25 '14 at 23:27
This might help http://stackoverflow.com/questions/4856717/javascript-equivalent-of-pythons-zip-function — elclanrs, Jul 25 '14 at 23:28
that works only for equal sizes, my question is about unequal sized arrays — user299709, Jul 25 '14 at 23:29
Did you try it and see what happens? You just need to add 4 characters to make it work http://jsbin.com/genaxe/1/edit — elclanrs, Jul 25 '14 at 23:30
I'd upvote the answer to the other question as I'd be basically copy/pasting that. And accept one of the answers below to close this question. — elclanrs, Jul 26 '14 at 01:03

Paul S. · Answer 1 · 2014-07-25T23:47:58.050

Conceptually, this is most easily achieved using nested loops;

var a = [[1, 2, 'x'], ['a', 'b'], ['+', '-', 'y', 'z']]; // input array

function transform(a) {
    var i, j, // two loops = two variables to iterate with
        foundSomething, // a flag so we know to stop
        b = []; // output array

    i = 0;
    do { // loop for i
        foundSomething = false;
        b[i] = [];
        for (j = 0; j < a.length; ++j) { // loop for j
            if (i in a[j]) { // something exists at this level of i
                foundSomething = true;
                b[i].push(a[j][i]);
            } else { // nothing here
                b[i].push(''); // insert empty string
            }
        }
        ++i;
    } while (foundSomething);
    b.length = b.length - 1; // snip empty end
    return b;
}

transform(a); // [[1, "a", "+"], [2, "b", "-"], ["x", "", "y"], ["", "", "z"]]

Sorting required?

a.sort(function (a, b) {return a.length < b.length;}); // bigger arrays first
transform(a); // [["+", 1, "a"], ["-", 2, "b"], ["y", "x", ""], ["z", "", ""]]

If you're always doing this sort, you can further optimise transform i.e. because you know that a[x].length >= a[y].length for all x < y, you don't need to flag whether you found something

this is pretty close the output is not quite what I expected. IT should always take the longest array as the primary item, and if an array is missing that nth item, it needs to be an empty string "". `[+,1,a], [-,2,b], [y,x,''], [z, '','']` — user299709, Jul 25 '14 at 23:32
You will need to sort your array by the `.length` of it's items (look up [`Array.prototype.sort`](https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/sort)); if you're doing this already then this code works, if you can assume that it's sorted then there will be other optimisations you can make, e.g. using a `for` instead of the `do..while` with a flag as you know the maximum depth from the first item's length, cutting the loop short when you know there will be no others that have the item, etc — Paul S., Jul 25 '14 at 23:36
@downvoter please comment so I might address the issue you have with this answer — Paul S., Jul 25 '14 at 23:48

score 0 · Answer 2 · answered Jul 25 '14 at 23:31

why not nest loops ?

sth. like this warning pseudocode

Values=[];
status = [[1,2,3, "a","b"] , [3,4,5,"c"], [6,7,8];
status_name = status[0]; 


for (i = 0; i < status.length; ++i)
{
    for (k = 0; k < status[i].length; ++i)
    {
        Values[i] =Array(status[i][k], status[k+1][k]);
    }

}

Aguardientico · Answer 3 · 2014-07-25T23:45:55.617

var originalArray = [[1, 2, 3, "a", "b"], [3, 4, 5, "c"], [6, 7, 8]];
var theValues = new Array();
var actualArray = originalArray.pop();
var maxLength = originalArray.length;
while (actualArray) {
  for (var i = 0; i < actualArray.length; i++){
    var newArray = theValues[i];
    if (!newArray) {
      theValues[i] = new Array();
    }
    theValues[i].push(actualArray[i])
  }
  actualArray = originalArray.pop();
}
for (var i = 0; i < theValues.length; i++) {
  for (var j = maxLength - theValues[i].length; j >= 0; j--) {
    theValues[i].push("");
  }
}

javascript: combine array of arrays into an array of single multidimensional array

3 Answers3