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I am working with a native class that represents a 2D image as a 1D array. If you want to change one pixel, for example, you need to now how to derive the index from the x,y coordinates.

So, let's say we have a 1D array array1d like this:

array1d = [ a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y ]

In the context of our program, array1d represents a 2D grid:

a b c d e
f g h i j
k l m n o
p q r s t
u v w x y

And we want to perform operations on array1d such as:

  • Get the value at x,y coordinates (in this example, 1,2 would give l)
  • Get any sub-grid using x,y,width,height (1,2,2,2 would give [l, m, q, r])
  • Set the value at any x,y coordinate (etc.)

How do we do these?

  • In Matlab, and thus likely math types (which spills into CS), to convert one matrix into another (be it a 1x12 into a 2x6 or a 2x6 into a 3x4) is known as "reshaping" mathworks.com/help/matlab/ref/reshape.html – user40980 Sep 28 '13 at 17:08
  • @MichaelT: the OP is not reshaping the grid. No mention of reshaping the 5x5 to anything else (which wouldn't make sense anyway). :) – IAbstract Feb 17 '16 at 12:21
  • @IAbstract that question was in revision 1 though. – user40980 Feb 17 '16 at 12:51
82

2D / 1D - mapping is pretty simple. Given x and y, and 2D array sizes width (for x-direction) and height (for y-direction), you can calculate the according index i in 1D space (zero-based) by

i = x + width*y;

and the reverse operation is

x = i % width;    // % is the "modulo operator", the remainder of i / width;
y = i / width;    // where "/" is an integer division

You can extend this easily to 3 or more dimensions. For example, for a 3D matrix with dimensions "width", "height" and "depth":

i = x + width*y + width*height*z;

and reverse:

x = i % width;
y = (i / width)%height;
z = i / (width*height);
  • @awashburn that is the traditional way to do it, it's even built into compilers for static 2D arrays – ratchet freak Sep 28 '13 at 18:26
  • @mtoast: I don't think so, its just basic integer math. – Doc Brown Sep 29 '13 at 19:11
  • This example is wrong for 3D. The word depth in the calculation should be height. – jiggunjer Jul 22 '15 at 11:15
  • @jiggunjer: thanks for the correction, changed my answer accordingly. – Doc Brown Jul 22 '15 at 12:47
  • 1
    @makakas: that is an exercise left to the reader ;-). Hint: you have to add/substract the lower bound as an offset at the right places. But before you try this, clarify to yourself which of the two arrays you mean, the 1D or the 2D array. – Doc Brown Jun 5 '16 at 18:13

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