# Find the position within an array larger than 3D (New Features)

M...

for 2 D assuming d1 and d0 as the dimensions and a1 and a0 are the array index

the formula i2d = (a1-1) * d0 + (a0-1) would convert to 1D

3d d2, d1, d0 and a2, a1, a0 would be

i3d = (a2-1) * d1 * d0 + (a1-1) * d0 + (a0-1) - or -

i3d = (a2-1) * d1 * d0 + i2d

4d d3, d2, d1, d0 and a3, a2, a1, a0 would be

i4d = (a3-1) * d2 * d1 * d0 + (a2-1) * d1 * d0 + (a1-1) * d0 + (a0-1) - or -

i4d = (a3-1) * d2 * d1 * d0 + i3d

I think this would be a general (recursive) definition (in my own screwy notation.

i{N}d = (a{N-1} - 1) * PRODUCT(d{N-2} ... d{0}) + i{n-1}d

**Complete thread:**

- Find the position within an array larger than 3D -
**lrcvs**, 2012-05-01, 22:16- Find the position within an array larger than 3D -
**Jim**, 2012-05-02, 02:52- Find the position within an array larger than 3D -
**lrcvs**, 2012-05-05, 08:39- Find the position within an array larger than 3D -
**lrcvs**, 2012-05-05, 13:27

- Find the position within an array larger than 3D -

- Find the position within an array larger than 3D -

- Find the position within an array larger than 3D -