How to create N x M x ... matrix in javascript dynamically?

Question

I am working on something in javascript where I need to create a matrix with dynamic dimensions depending upon an array.

For example, I have an array with [3, 5, 6] - now I want to create a matrix dynamically that is 3 x 5 x 6, and in this matrix, all the values are initialized to 0.

I can seem to find a way to do this dynamically depending upon my initial array values. Anyone have any idea?

Just use a recursive function with for loops

Gpack
– Gpack

2021-06-23 01:03:14 +00:00
Commented Jun 23, 2021 at 1:03 — Gpack
– Gpack, Commented Jun 23, 2021 at 1:03

CertainPerformance · Accepted Answer · 2021-06-23 01:03:11Z

3

A recursive function that uses Array.from will do the trick.

const makeMatrix = (length, ...restDimensionLengths) => Array.from(
  { length },
  () => restDimensionLengths.length ? makeMatrix(...restDimensionLengths) : 0
);

console.log(makeMatrix(2, 3, 5));

answered Jun 23, 2021 at 1:03

CertainPerformance

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2 Comments

SnG Over a year ago

is there a way i can do this without arrow notation?

CertainPerformance Over a year ago

Just use function() { return instead of () => ??

Unmitigated · Accepted Answer · 2021-06-23 01:04:42Z

2

You can use recursion.

function createMatrix(dimensions) {
  return dimensions.length === 1 ? Array(dimensions[0]).fill(0) : Array.from({
    length: dimensions[0]
  }, () => createMatrix(dimensions.slice(1)));
}
console.log(JSON.stringify(createMatrix([3, 5, 6])));

answered Jun 23, 2021 at 1:04

Unmitigated

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6 Comments

SnG Over a year ago

is there a way i can do this without arrow notation?

Unmitigated Over a year ago

@SnG Yes, you can use function(){return createMatrix(dimensions.slice(1))}

SnG Over a year ago

this is an unrelated question, but in my example I can have N dimensions based on the input. Is there a way I can programmatically index the N dimensional matrix given an array of indexes?

Unmitigated Over a year ago

@SnG indexes.reduce((acc,i)=>acc[i], matrix)

Unmitigated Over a year ago

Get the second last element you need using reduce, and set the last index on that.

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