Create a structure from discrete column vectors or a 3 x n matrix composed of column vectors

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I'm trying to build a structure consisting of elements which are column vectors representing x, y, z points. So far, I have written the following code, which indexes arrays representing x, y, z positions of points from a structure called 'data' (derived from a .ply 3d graphics file) and rearranges them into a 3 x n matrix:
% orders vertex data from structure into column vectors
x = data.vertex.x
y = data.vertex.y
z = data.vertex.z
% vertically concatenates x, y, z to form a 3 x n matrix stored in the
% structure, verts
verts = horzcat(x, y, z)
%Sequentially indexes PX PY PZ into a 3 x n matrix where columns are x y z
%components of position vectors
for i=1:length(x)
This creates a matrix of the form:
3.7427 3.74984 1.38903
4.7384 3.76483 8.76289
9.6389 9.62843 5.78290 .............
I want to create a structure where each column vector in this array is a seperate element. I thought the following might do the trick:
for k=1:size(x)
Vk = [vertsnew(1,k)];
vertz.(['VZ' num2str(k)])=Vk;
But this results in a structure with x, y, z arranged sequentially as individual elements. E.g:
vertz.VZ.1 = 3.7427
vertz.VZ.2 = 4.7384
vertz.VZ.3 = 9.6389
I am able to create individual column vectors using:
%Orders arrays, into discrete column vectors
for i=1:length(x)
assignin('base',['vertv' num2str(i)],[x(i);y(i);z(i)])
vertv1 = 3.7427
vertv2 = 4.7384
Though I am unsure how to obtain the desired structure from these seperate arrays. Any help would be greatly appreciated.
Thomas on 27 Nov 2012
Apologies, I forgot the bottom row is 0 0 0 1. Thanks for your advice Matt. I'll have to have a look up the bsxfun function as I'm not familiar with it. The reason I was to use a 4x4 matrix is that it is output by your absor() function: very useful to me for introducing arbitrerally oriented and scaled spatial data into a target coordinate system. Though I'm sure you know much more about that than I do! I may need the more efficient option though as I'm processing point datasets containing tens to hundreds of thousands of points (though I do have access to some pretty decent hardware to counter this).

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Accepted Answer

Azzi Abdelmalek
Azzi Abdelmalek on 27 Nov 2012
Edited: Azzi Abdelmalek on 27 Nov 2012
y=[3.7427 3.74984 1.38903
4.7384 3.76483 8.76289
9.6389 9.62843 5.78290]
for k=1:size(y,2)
Azzi Abdelmalek
Azzi Abdelmalek on 27 Nov 2012
I agree with him that is not the best way, but at the end you are the only person to decide how to do it. Also, whatever you do, there is always a better way!

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