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how to change axis of a matix?

Asked by Asliddin Komilov on 13 Sep 2019
Latest activity Edited by Bruno Luong
on 15 Sep 2019 at 7:43
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?

  6 Comments

Rik
on 13 Sep 2019
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.
I also suspect that length(c)=y*z, still need to know how to do it.
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.

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2 Answers

Answer by Bruno Luong
on 15 Sep 2019 at 7:24
Edited by Bruno Luong
on 15 Sep 2019 at 7:43
 Accepted Answer

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)');
mapping3D.png

  0 Comments

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Answer by Walter Roberson
on 14 Sep 2019

[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

  8 Comments

Stephen Cobeldick on 15 Sep 2019 at 6:21
Please explain how you want to deal with the discontinuity in the middle of y./z:
>> plot(y./z)
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
on 15 Sep 2019 at 7:33
(y-z)/y is 1-z/y and since z and y both include 0, you still have nan and infinities.

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