Kdjykuj

I'm trying to use reduce() combine a set of arrays in a "collated" order so items with similar indexes are together. For example:

input = [["one","two","three"],["uno","dos"],["1","2","3","4"],["first","second","third"]]



output = [ 'first','1','uno','one','second','2','dos','two','third','3','three','4' ]

It doesn't matter what order the items with similar index go as long as they are together, so a result of 'one','uno','1'... is a good as what's above. I would like to do it just using immutable variables if possible.

I have a way that works:

    const output = input.reduce((accumulator, currentArray, arrayIndex)=>{

        currentArray.forEach((item,itemIndex)=>{

            const newIndex = itemIndex*(arrayIndex+1);

            accumulator.splice(newIndex<accumulator.length?newIndex:accumulator.length,0,item);

        })

        return accumulator;

    })

But it's not very pretty and I don't like it, especially because of the way it mutates the accumulator in the forEach method. I feel there must be a more elegant method.

I can't believe no one has asked this before but I've tried a bunch of different queries and can't find it, so kindly tell me if it's there and I missed it. Is there a better way?

To clarify per question in comments, I would like to be able to do this without mutating any variables or arrays as I'm doing with the accumulator.splice and to only use functional methods such as .map, or .reduce not a mutating loop like a .forEach.

Maybe just a simple for... i loop that checks each array for an item in position i

var input = [["one","two","three"],["uno","dos"],["1","2","3","4"],["1st","2nd","3rd"]]



var output = 

var maxLen = Math.max(...input.map(arr => arr.length));



for (i=0; i < maxLen; i++) {

  input.forEach(arr => { if (arr[i] !== undefined) output.push(arr[i]) })

}



console.log(output)

Simple, but predictable and readable

Avoiding For Each Loop

If you need to avoid forEach, here's a similar approach where you could: get the max child array length, build a range of integers that would've been created by the for loop ([1,2,3,4]), map each value to pivot the arrays, flatten the multi-dimensional array, and then filter out the empty cells.

First in discrete steps, and then as a one liner:

var input = [["one","two","three"],["uno","dos"],["1","2","3","4"],["1st","2nd","3rd"]];

Multiple Steps:

var maxLen = Math.max(...input.map(arr => arr.length));

var indexes = Array(maxLen).fill().map((_,i) => i);

var pivoted = indexes.map(i => input.map(arr => arr[i] ));

var flattened = pivoted.flat().filter(el => el !== undefined);

One Liner:

var output = Array(Math.max(...input.map(arr => arr.length))).fill().map((_,i) => i)

               .map(i => input.map(arr => arr[i] ))

               .flat().filter(el => el !== undefined)

Use Array.from() to create a new array with the length of the longest sub array. To get the length of the longest sub array, get an array of the lengths with Array.map() and take the max item.

Then collect the non undefined items at the current index from each sub array with Array.reduceRight() or Array.reduce() (depending on the order you want), and use Array.flat() to get a single array.

const input = [["one","two","three"],["uno","dos"],["1","2","3","4"],["first","second","third"]]



const result = Array.from({ 

    length: Math.max(...input.map(o => o.length)) 

  }, (_, i) => input.reduceRight((r, o) => o[i] === undefined ? r : [...r, o[i]], ))

  .flat();



console.log(result);

Funny solution

1. add index as prefix on inner array


2. Flatten the array


3. sort the array


4. Remove the prefix

let input = [["one","two","three"],["uno","dos"],["1","2","3","4"],["first","second","third"]]

let ranked=input.map(i=>i.map((j,k)=>k+'---'+j)).slice()

console.log(ranked.flat().sort().map(i=>i.split('---')[1]));

Here I have provided a generator function that will yield the values in the desired order. You could easily turn this into a regular function returning an array if you replace the yield with a push to a results array to be returned.

The algorithm takes in all the arrays as arguments, then gets the iterators for each of them. Then it enters the main loop where it treats the iters array like a queue, taking the iterator in front, yielding the next generated value, then placing it back at the end of the queue unless it is empty. The efficiency would improve if you transformed the array into a linked list where adding to the front and back take constant time, whereas a shift on an array is linear time to shift everything down one spot.

function* collate(...arrays) {

  const iters = arrays.map(a => a.values());

  while(iters.length > 0) {

    const iter = iters.shift();

    const {done, value} = iter.next();

    if(done) continue;

    yield value;

    iters.push(iter);

  }

}