CODEWITHSUNDEEP
javascriptsorting
genericUser
genericUser

Reputation: 7128

How to sort 2 dimensional array (matrix) in JavaScript?

I want to sort a 2 dimensional array of integers. What is the simplest and most readable way to achieve this?

input:

[
    [3,4,2],
    [5,1,3],
    [2,6,1],
]

output:

[
    [1,1,2],
    [2,3,3],
    [4,5,6]
]

Upvotes: 1

Views: 696

Answers (3)

Nina Scholz
Nina Scholz

Reputation: 386570

This is a general approach of nested array, which could have more nested arrays or different lengths.

It works by taking an array of indices to every value, a flat array of values and finally by assigning all values back to their place.

const
    getIndices = (value, index) => Array.isArray(value)
        ? value.flatMap(getIndices).map(array => [index, ...array])
        : [[index]],
    setValue = (array, keys, value) => {
        const last = keys.pop();
        keys.reduce((a, i) => a[i] ??= [], array)[last] = value;
        return array;
    }; 
    data = [[3, 4, 2], [5, 1, 3], [2, 6, 1]],
    references = data.flatMap(getIndices),
    result = data
        .flat()
        .sort((a, b) => a - b)
        .reduce((r, v, i) => setValue(r, references[i], v), []);

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

Upvotes: 0

Mr. Me
Mr. Me

Reputation: 41

When working with nested arrays, its common to work from the inner most level and then step out. Using this method, we start with the inner arrays:

// Sorting the inner most array
var result = [];
[
  [3,4,2],
  [5,1,3],
  [2,6,1],
].forEach(function(row) {  // a for loop could work as well, this is just shorter.
  result.push(row.sort(function(a, b) {
    return a-b; // ascending 
    // return b-a; // descending 
  }));
});

Then, you sort the outer array. You cannot sort the outer array as a single number, so its necessary to decide on a method. Examples being: average of the array, lowest value, or highest value.

// Using the first value to sort the outer array.
[
  [2,3,4],
  [1,3,5],
  [1,2,6],
].sort(function(a, b) {
  return a[0]-b[0];
});

Alternatively

The wanted output "[[1,1,2],[2,3,3],[4,5,6]]" disregards the inner arrays so it doesn't seem practical. To do this however, we'd reconstruct all inner arrays into one, sort, and then rebuild a nested array assuming each new array is to have 3 values.

  var oneDimensionalArray = [];
  var finalArray = [];
  [
    [3,4,2],
    [5,1,3],
    [2,6,1],
  ].forEach(function(row) {
    oneDimensionalArray = oneDimensionalArray.concat(row);
  });
  oneDimensionalArray.sort();
  for (var i=0; i<oneDimensionalArray.length; i++) {
    if (i%3==0) {
      temp = [];
    } 
    temp.push(oneDimensionalArray[i]);
    if (i%3==2) {  // 3 per line (0-2)
      finalArray.push(temp);
    }
  }

Again, I don't see this having practical usefulness. It would be easier to leave them as a regular array or start with a different setup if all the data is to be used in a way that disregards grouping/array.

Upvotes: 1

0stone0
0stone0

Reputation: 43962

If you'd need the deeper arrays to be in order as well, I'd tackel it like so:

  1. Flatten the arrays using flat(), so get just a regular list

    input.flat()

  2. Sort them using a custom integer sort function

    .sort((a, b) => a - b)

  3. Re-create the second dimension

    array_chunks(sortedArray, 3);

    (Function used taken from this answer)

const input = [
    [3,4,2],
    [5,1,3],
    [2,6,1],
];
const array_chunks = (array, chunk_size) => Array(Math.ceil(array.length / chunk_size)).fill().map((_, index) => index * chunk_size).map(begin => array.slice(begin, begin + chunk_size));

let result = array_chunks(input.flat().sort((a, b) => a - b), 3);
console.log(result);

[
  [ 1, 1, 2 ], 
  [ 2, 3, 3 ], 
  [ 4, 5, 6 ]
]

Upvotes: 3

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