JS multidimentionnal array sort list

Question

With this array:

var arr = [];
arr[0] = [1, 'Peter', 3];
arr[1] = [1, 'Mary', 2];
arr[2] = [0, 'David', 5];
arr[3] = [0, 'John', 4];
arr[4] = [0, 'Billy', 1];

This works fine:

arr.sort(function (a,b) {
    console.log(a[2]);

    if (a[2] > b[2]) return  1;
    if (a[2] < b[2]) return -1;
    return 0;
});

But with an array like this:

var arr = [];
arr[0] = [1, 1, 0, 0, 0];
arr[1] = ['Peter', 'Mary', 'David', 'John', 'Billy'];
arr[2] = [3, 2, 5, 4, 1];

A[2] gets 0-David.

I returned everything, I really cannot figure it out. Please, do you know how I could sort the second array according to arr[2] list?

You might be better off mapping it to the first and than converting it back to the second after the sort. — epascarello
– epascarello, Commented Apr 24, 2017 at 13:33
One thing to keep in mind is that in the first case you are sorting the array arr and in the second you are sorting a mash of the subarrays, so arr.sort will never do the trick because you are not trying to sort arr. As @epascarello said, you will be better mapping one to the other. — Marcelo Lazaroni
– Marcelo Lazaroni, Commented Apr 24, 2017 at 13:35
So, if I understand the problem correctly, you're trying to sort array1 and array2 based on the values of array3. This is not trivially implemented, and I suppose you're trying to do something that can be way easier if you didn't want to do it this way. I can suggest you transpose your matrix, sort it with the first implementation, then transpose it back. Or you can implement a very special sort, but by using the .sort() it's not really possible. — Alex
– Alex, Commented Apr 24, 2017 at 13:36
Thank you I appreciated your advice. I have distributed the table at the base. It is fascinating to see such a dynamic and voluntary community. Thank you all! — UnicornBuster
– UnicornBuster, Commented Apr 24, 2017 at 17:56

Nina Scholz · Accepted Answer · 2017-04-24 13:47:09Z

1

You could use another array with the indices, sort them as desired and map the result to the given array.

var array = [[1, 1, 0, 0, 0], ['Peter', 'Mary', 'David', 'John', 'Billy'], [3, 2, 5, 4, 1]],
    sortBy = array[2],
    indices = sortBy.map(function (_, i) { return i; });

indices.sort(function (a, b) { return sortBy[a] - sortBy[b]; });
array = array.map(function (a) {
    return indices.map(function (i) { return a[i]; });
});

console.log(array);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Apr 24, 2017 at 13:47

answered Apr 24, 2017 at 13:40

Nina Scholz

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Comments

Alex · Accepted Answer · 2017-04-24 13:42:25Z

Here's an ugly implementation for what you want to achieve, however, I feel that you're overcomplexifying something simple with this approach.

What I suggest is that you transpose your matrix, sort it with your first implementation, then transpose it back to the original layout.

var arr = [];
arr[0] = [1, 1, 0, 0, 0];
arr[1] = ['Peter', 'Mary', 'David', 'John', 'Billy'];
arr[2] = [3, 2, 5, 4, 1];

function transpose(array) {
    return array[0].map(function(col, i) {
        return array.map(function(row) {
            return row[i]
        })
    });
}

function twistedSort(matrix, sortingRowIndex) {
    var transposed = transpose(matrix);

    transposed.sort(function(a, b) {
        if (a[sortingRowIndex] > b[sortingRowIndex]) return 1;
        if (a[sortingRowIndex] < b[sortingRowIndex]) return -1;
        return 0;
    });

    return transpose(transposed);
}

twistedSort(arr, 2);

Once again, I suggest rethinking your problem, but if you're sure you need this complex of a solution for this problem, then here you go :)

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