how to change axis of a matix?

my matrix is M(x,y,z) where:

x=1:90
y=1:3:120
z=1:40

and

c=y./z

and I need to convert M into N(x,c).

Any ideas?

[KALYAN ACHARJYA]

Define M and N(x,c)?

[Rik]

There is not a unique way to determine y and z from c. Do you want to average over the line across your 3D matrix that have the same c? You will have to do some interpolation.

[Asliddin Komilov]

M is a parameter that depends on x,y,z variables so calculation gives M(x,y,z) 3d matrix. I want to represent (plot) M data as surface where M changes with x and y/z. as y and z are same size I thought its possible to find one value of M for each y/z which can be set to c, so I have M(x,c) which I would name N(x,c) so I don't loose M(x,y,z). so N=M(x,c).

[Rik]

consider a 2D example:

M=[1 2 3
    4 5 6
    7 8 9];

There are the same number of rows and columns ([1 2 3]). What would it mean if I reduce this to a single variable? Should M only keep 3 values? Should I have 9 options for the new parameter? These questions need to be answered for your situation as well.

[Asliddin Komilov]

I also suspect that length(c)=y*z, still need to know how to do it.

[Walter Roberson]

No, length of y and z are the same and y./z would have the same length.

Some of what you wrote does not seem to make sense until you start talking about grids of data, but then you have to ask about the sizes of the grids.

Answer 1

[Bruno Luong]

load('testdata.mat');

Ufun = @(X,Y,Z) X;
Vfun = @(X,Y,Z) Y./(Z+200);

[X,Y,Z] = ndgrid(x,y,z);
U = Ufun(X,Y,Z);
V = Vfun(X,Y,Z);
U = U(:);
V = V(:);

[umin,umax] = bounds(U);
nu = 21;
u = linspace(umin,umax,nu);
[vmin,vmax] = bounds(V);
nv = 21;
v = linspace(vmin,vmax,nv);
[~,~,I] = histcounts(U,u);
[~,~,J] = histcounts(V,v);
N = accumarray([J I], M(:), [nu nv]-1, @mean, NaN);
midfun = @(x) 0.5*(x(1:end-1)+x(2:end));
u = midfun(u);
v = midfun(v);

surf(u,v,N);
xlabel('u (=x)');
ylabel('v (=y/(200+z)');

Answer 2

[Walter Roberson]

[X, Y, Z]=ndgrid(1:90,1:3:120,1:40);
C = round(ceil(Y./Z));
N = accumarray([X(:), C(:)], M(:), [], @mean, nan);

surf(N, 'edgecolor', 'none')
xlabel('x') 
ylabel('c')

Or possibly N.' instead of N

[Asliddin Komilov]

thank you, it looks like what I was looking for but gives this error:

Error using accumarray

First input SUBS must contain positive integer subscripts.

I made all axis same size for the convinience and attached them

[Walter Roberson]

Did you change the definition of x, y, or z? I tested with what you posted before, and all of the subscripts are positive. It generates a 90 by 118 array.

[Asliddin Komilov]

this is my command window:

>> [X, Y, Z]=ndgrid(x,y,z);

>> C = round(ceil(Y./Z));

>> N = accumarray([X(:), C(:)], M(:));

Error using accumarray

First input SUBS must contain positive integer subscripts.

actually Z has negative values, I don't know it it is a problem.

[Walter Roberson]

You cannot create N. You have non-negative y values, and your z values include 0, so c = y./z includes c = non-zero / 0 which is infinite and therefore your N array would have to have infinite dimensions. Furthermore, y includes 0, so you have 0/0 which is nan, so some of your entries are at C values that do not exist.

C = round(ceil(300*Y./Z));  %0 <= Y <= 90, need to multiply by 300 or more to get best indices
C = C - min(C(:)) + 1;     %translate negative subscript to positive

There are now 255 unique finite C values, and 441 locations in C that are not finite.

badC = ~isfinite(C);
C(badC) = 10;         %turns out no C value comes out as 10
X = round(X*20 + 1);   %x is 0:0.05:1
N = accumarray([X(:), C(:)], M(:), [], @mean, nan);
N(:,10) = nan;        %zap locations corresponding to non-finite C

I think you will find N to be fairly useless. You might want to consider using inpaint_nan from the File Exchange

[Walter Roberson]

Note: there are 21 different locations in which there are 21 different M locations that have the same (X,C). It turns out that all 441 of those locations have the M value 1941637.

Have you considered using slice() or volumeViewer() or isosurface() to display M in 3D?

[Stephen23]

Please explain how you want to deal with the discontinuity in the middle of y./z:

>> plot(y./z)

[Asliddin Komilov]

I may use C=((y-z)./y); main thing is to obtain the value of M at x by some ratio of y and z, so I can make surface out of it.

[Walter Roberson]

(y-z)/y is 1-z/y and since z and y both include 0, you still have nan and infinities.