[Numpy-discussion] 3D array and the right hand rule

Dieter Van Eessen dieter.van.eessen at gmail.com
Tue Mar 17 04:11:23 EDT 2015


Sorry to disturb again, but the topic still bugs me somehow...
I'll try to rephrase the question:

- What's the influence of the type of N-array representation with respect
to TENSOR-calculus?
- Are multiple representations possible?
- I assume that the order of the dimensions plays a major role in for
example TENSOR product.
Is this assumption correct?

As I said before, my math skills are lacking in this area...
I hope you consider this a valid question.

kind regards,

On Fri, Jan 30, 2015 at 2:32 AM, Alexander Belopolsky <ndarray at mac.com>

> On Mon, Jan 26, 2015 at 6:06 AM, Dieter Van Eessen <
> dieter.van.eessen at gmail.com> wrote:
>> I've read that numpy.array isn't arranged according to the
>> 'right-hand-rule' (right-hand-rule => thumb = +x; index finger = +y, bend
>> middle finder = +z). This is also confirmed by an old message I dug up from
>> the mailing list archives. (see message below)
> Dieter,
> It looks like you are confusing dimensionality of the array with the
> dimensionality of a vector that it might store.  If you are interested in
> using numpy for 3D modeling, you will likely only encounter 1-dimensional
> arrays (vectors) of size 3 and 2-dimensional arrays  (matrices) of size 9
> or shape (3, 3).
> A 3-dimensional array is a stack of matrices and the 'right-hand-rule'
> does not really apply.  The notion of C/F-contiguous deals with the order
> of axes (e.g. width first or depth first) while the right-hand-rule is
> about the direction of the axes (if you "flip" the middle finger right hand
> becomes left.)  In the case of arrays this would probably correspond to
> little-endian vs. big-endian: is a[0] stored at a higher or lower address
> than a[1].  However, whatever the answer to this question is for a
> particular system, it is the same for all axes in the array, so right-hand
> - left-hand distinction does not apply.
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Dieter VE
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