Question

I want to create an n-dimensional array of doubles. At compile-time, the number of dimensions n is not known.

I ended up defining the array as a dictionary, with the key being an array of ints corresponding to the different axes (so in a 3-dimensional array, I’d supply [5, 2, 3] to get the double at (5, 2, 3) in the array.

However, I also need to populate the dictionary with doubles from (0, 0, … 0) to (m1, m2, … mn), where m1 to mn is the length of each axis.

My initial idea was to create nested for-loops, but as I still don’t know how many I’d need (1 for each dimension), I can’t do this at compile-time.

I hope I’ve formulated the question in an understandable manner, but feel free to ask me to elaborate parts.

Answer 1

A quick followup on this matter:

We used the Array.CreateInstance method with success, but as someone predicted, it was fairly inefficient, and additionally created readability problems.

Instead, we have developed a method, where the n-dimensional array is converted into a 1-dimensional (normal) array.

public static int NDToOneD(int[] indices, int[] lengths)
{
  int ID = 0;
  for (int i = 0; i < indices.Length; i++)
  {
    int offset = 1;
    for (int j = 0; j < i; j++)
{
      offset *= lengths[j];
}
    ID += indices[i] * offset;
  }
  return ID;
}

1DtoND(int[] indices, int[] arrayLengths)
{
  int[] indices = new int[lengths.Length];
  for (int i = lengths.Length - 1; i >= 0; i--)
  {
    int offset = 1;
    for (int j = 0; j < i; j++)
    {
      offset *= lengths[j];
    }
    int remainder = ID % offset;
    indices[i] = (ID - remainder) / offset;
    ID = remainder;
  }
  return indices;
}

This is essentially a generalisation on the conversion of cartesian coordinates to a single integer and back again.

Our testing is not formalized, so any speedup we have gained is entirely anecdotal, but for my machine, it has given about a 30-50% speedup, depending on the sample size, and the readability of the code has improved by a wide margin.

Hope this helps anyone who stumbles upon this question.