Sequential fill of an array of unknown dimensions

Question

I have an Array of unknown dimensions. E.g. it may be object[], object[,], object[,,,].

I want to fill it sequentially (e.g. for [2,2] this order: 0,0; 0,1, 1,0; 1,1):

Array arr = ... // input array
for (int i = 0; i < arr.Length; i++)
{
     arr.SetValue(stream.ReadNextObject(), ???); // convert i -> int[] indexes
}

I know that the conversation of i can be done with % operator but it's hard to imagine an exact algorithm for multiple dimensions. There is only an answer for two-dimensions: Converting index of one dimensional array into two dimensional array i. e. row and column

I could use Stack<int> to store indexes while walking an array but it seems that % will be more efficient (I really need to care about performance here). But I'm not sure about Stack<T> vs %.

What are you trying to achieve? You want to read objects from a stream and store them in an array? i is already going over every possible index in your array, why don't you use arr.SetValue(stream.ReadNextObject(), i);? — Audric de Schaetzen
– Audric de Schaetzen, Commented Mar 7, 2016 at 15:09
Sounds like switch/case. Why array is of unknown dimensions? How and from where you get it? — Sinatr
– Sinatr, Commented Mar 7, 2016 at 15:10
@Sinatr it's a serializer library. An array is created using reflection. — Vlad
– Vlad, Commented Mar 7, 2016 at 15:11
You can use recursion (because each dimension is a nested loop), check if item is of enumerable type (not sure if IEnumerable is the most basic) and if it is - then run same method recursively for this element. This will work for jagged array List<List<List<...>>>. Not sure what to do with multidimensional arrays though. — Sinatr
– Sinatr, Commented Mar 7, 2016 at 15:14

Community · Accepted Answer · 2020-06-20 09:12:55Z

5

Here's the algorithm that I believe you are looking for

public static int[] SingleIndexToMulti(int index, int[] dimentionSizes)
{
    var result = new int[dimentionSizes.Length];
    for (int i = dimentionSizes.Length - 1; i >=0; i--)
    {
        result[i] = index % dimentionSizes[i];
        index = index / dimentionSizes[i];
    }

    return result;
}

You would use it like this

Array myArray = Whatever();
int[] dimensionSizes = new int[myArray.Rank];
for(int i =0; i < myArray.Rank; i++)
    dimensionsSizes[i] = myArray.GetLength(i);
for (int i = 0; i < arr.Length; i++)
{
    arr.SetValue(stream.ReadNextObject(), SingleIndexToMulti(i, dimensionSizes)); 
}

To demonstrate the following code

for(int i=0; i < (2*3*4) ;i++)
    Console.WriteLine(string.Join(",", SingleIndexToMulti(i, new[] { 2, 3, 4 })));

Produces

0,0,0

0,0,1

0,0,2

0,0,3

0,1,0

0,1,1

0,1,2

0,1,3

0,2,0

0,2,1

0,2,2

0,2,3

1,0,0

1,0,1

1,0,2

1,0,3

1,1,0

1,1,1

1,1,2

1,1,3

1,2,0

1,2,1

1,2,2

1,2,3

edited Jun 20, 2020 at 9:12

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answered Mar 7, 2016 at 15:25

juharr

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1 Comment

Vlad Over a year ago

Yes, this is the algorithm I was looking for so I've accepted it. Until you posted this I couldn't even compare it with stack approach but now I see that using stack should be more efficient so I've chosen that way.

René Vogt · Accepted Answer · 2016-03-07 15:50:48Z

I have a recursive solution for you:

void Fill(Array a, Stream stream)
{
    int[] indices = new int[a.Rank]; // c# should initialize this with 0s
    FillRecursive(a, indices, 0, stream)    
}
void FillRecursive(Array a, int[] indices, int rank, Stream stream)
{
    for (int idx = 0; idx < a.GetLength(rank); idx++)
    {
        indices[rank] = idx;
        if (rank == a.Rank - 1)
            a.SetValue(stream.ReadNextObject(), indices);
        else
            FillRecursive(a, indices, rank + 1, stream);
    }
}

This should do the trick. I count up the individual dimensions and continously set the value. Note that Stream is kind of pseudo code (my Stream does not have a ReadNextObject).

I admit that I had no time to really test it. And I leave the performance measurements up to you, but I don't think there is a much faster alternative.

UPDATE: Test it, and it fills the "cells" in the correct order as you expected it. I testet it using an int[3,3,3] and a ReadNextValue that returns increasing integers. Result:

[0,0,0] => 0
[0,0,1] => 1
[0,0,2] => 2
[0,1,0] => 3
[0,1,1] => 4
...
[2,2,0] => 24
[2,2,1] => 25
[2,2,2] => 26

Note that this works not only for arbitrary dimensions, but also for arbitrary and different lengths of the individual dimensions.

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