Convert a bidimensional array to a 1D array alternating their values

Question

I have a bidimensional array like this:

let test2d = [
  ["foo", "bar"],
  ["baz", "biz"]
]

If I want to convert this 2D array into 1D array (not alternating their values), I can do it on two ways:

First way:

let merged = test2d.reduce( (prev, next) => prev.concat(next) )
console.log(merged) // ["foo", "bar", "baz", "biz"]

Second way:

let arr1d = [].concat.apply([], test2d)
console.log(arr1d) // ["foo", "bar", "baz", "biz"]

Question: How can I get a 1D array but with their values alternated? I mean like this:

["foo", "baz", "bar", "biz"]

score 3 · Answer 1 · edited May 23 '18 at 21:35

3

You can use the zip function defined here: Javascript equivalent of Python's zip function

let zip= rows => rows[0].map((_,c)=>rows.map(row=>row[c]))

So you can call:

let arr1d = [].concat.apply([], zip(test2d))

Here's the full code:

let test2d = [ ["foo", "bar"], ["baz", "biz"] ]

let zip = rows => rows[0].map((_, c) => rows.map(row => row[c]))

let arr1d = [].concat.apply([], zip(test2d))

console.log(arr1d)

edited May 23 '18 at 21:35

Ivan

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answered May 23 '18 at 21:32

CrociDB

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Very elegant this one! Thanks for your answer. – robe007 May 24 '18 at 17:00

score 2 · Accepted Answer · answered May 23 '18 at 21:33

2

You could take an array for each index and concat the arrays at the end.

var array = [["foo", "bar"], ["baz", "biz"]],
    result = [].concat(...array.reduce((r, a) => {
        a.forEach((v, i) => (r[i] = r[i] || []).push(v));
        return r;
    }, []));

console.log(result);

answered May 23 '18 at 21:33

Nina Scholz

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Hi @NinaScholz, if you don't mind can you elaborate `(r[i] = r[i] || []).push(v)`? I've followed your solution up until that line of code and lost – Isaac May 24 '18 at 00:56
2

you could make two lines of code of it: `r[i] = r[i] || [];` and `r[i].push(v)`. the first line checks if `r[i]` is truthy, like an array and if not it takes an array as assingment. the second pushes the item to this array. – Nina Scholz May 24 '18 at 07:00

score 1 · Answer 3 · answered May 23 '18 at 21:34

1

Why not use a regular for loop? This is efficient and easiest to read.

var arr1d = [];
for (let i = 0; i < test2d[0].length; i++) {
    for (let j = 0; j < test2d.length; j++) {
        arr1d.push(test2d[j][i]);
    }
}

answered May 23 '18 at 21:34

patstuart

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This solution works too, even with more rows and columns. Thanks ! – robe007 May 24 '18 at 16:47

Defacto · Answer 4 · 2018-05-23T22:08:08.980

1

This solution is easy to understand, it simply uses nested for-loops to get the desired output:

var data = [];  // This will be the new array
let test2d =    // This is your 2D array
[  
  ["foo", "bar"],
  ["baz", "biz"]
]

// Here we use nested for-loops to add the items to data[] in the desired order
for(var i = 0; i < test2d.length; i++) {
    for(var j = 0; j < test2d[0].length; j++) {
        data.push(test2d[j][i]);
    }
}

console.log(data); // Print the output :)

The data array is now equal to: [ "foo", "baz", "bar", "biz" ]

Hope this is helpful!

edited May 23 '18 at 22:08

answered May 23 '18 at 21:37

Defacto

141
7

Thanks for your answer, but this solution only works with my `test2d` array, but if I add more rows and columns to my array, it does not works. Why? Because the trick is to start the loop with the columns (`test2d[0].length`) not with the rows (`test2d.length`). Look on the **patstuart**'s answer. – robe007 May 24 '18 at 16:49

Convert a bidimensional array to a 1D array alternating their values

4 Answers4