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Finding position of an element in a two-dimensional array?

Well simple question here (maybe not a simple answer?)

Say I have a two dimensional array

[0] [1] [2]
[3] [4] [5]
[6] [7] [8]

Now suppose I want to get the position of the number 6

I know with a one-dimensional array i can use Array.indexOf() but what would my options be with 2-dimensional arrays?

Thanks!

like image 517
zeta Avatar asked Jul 15 '10 23:07

zeta


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2 Answers

I'd say something like this:

public static Tuple<int, int> CoordinatesOf<T>(this T[,] matrix, T value)
{
    int w = matrix.GetLength(0); // width
    int h = matrix.GetLength(1); // height

    for (int x = 0; x < w; ++x)
    {
        for (int y = 0; y < h; ++y)
        {
            if (matrix[x, y].Equals(value))
                return Tuple.Create(x, y);
        }
    }

    return Tuple.Create(-1, -1);
}
like image 130
Dan Tao Avatar answered Sep 22 '22 01:09

Dan Tao


Here is a method that should find an index in an array with an arbitrary rank.

... Added Upper/Lower bounds range per rank

public static class Tools
{
    public static int[] FindIndex(this Array haystack, object needle)
    {
        if (haystack.Rank == 1)
            return new[] { Array.IndexOf(haystack, needle) };

        var found = haystack.OfType<object>()
                          .Select((v, i) => new { v, i })
                          .FirstOrDefault(s => s.v.Equals(needle));
        if (found == null)
            throw new Exception("needle not found in set");

        var indexes = new int[haystack.Rank];
        var last = found.i;
        var lastLength = Enumerable.Range(0, haystack.Rank)
                                   .Aggregate(1, 
                                       (a, v) => a * haystack.GetLength(v));
        for (var rank =0; rank < haystack.Rank; rank++)
        {
            lastLength = lastLength / haystack.GetLength(rank);
            var value = last / lastLength;
            last -= value * lastLength;

            var index = value + haystack.GetLowerBound(rank);
            if (index > haystack.GetUpperBound(rank))
                throw new IndexOutOfRangeException();
            indexes[rank] = index;
        }

        return indexes;
    }
}
like image 31
Matthew Whited Avatar answered Sep 23 '22 01:09

Matthew Whited