I have a function of 3 variables, say A(x,y,z), that I represent as 3D object. (x,y,z are elements of a grid)
Suppose z is also function of x,y, say z(x,y) and I want to evaluate V(x,y) = A(x,y,z(x,y)). (Where z is the relevant index of the grid)
How do I do this? Obviously I can do this easily with a loop, but is there a better answer?
Btw, the way I am doing it now is like this
for i_y=1:Ny
for i_r = 1:Nr
for i_d =1:Nd
for i_w = 1:Nw
V(i_y,i_r,i_d,i_w)= A(i_y,i_r,i_d,i_w, Z(i_y,i_r,i_d,i_w));
end
end
end
end

2 Kommentare

Julien
Julien am 8 Okt. 2012
Bearbeitet: Julien am 8 Okt. 2012
If A=f(x,y,z(x,y)),then in fact A=f(x,y). Can you describe exactly what you want to do ? do you have the expression to calculate? the function that link z and x,y ?
Sergio, you need to look at meshgrid() function. This provides an easy way to get solution to the problem of vectorially calculating A(x,y,Z(x,y)) or in steps as Julien suggests.
[XX,YY] = meshgrid(x,y); %x,y can be 1D vectors here
Now XX,YY correspond to grid points (XX(1),YY(1)) etc.
Once you have meshgrid just directly evaluate your function, and it will have all the values across the X-Y grid.

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Andrei Bobrov
Andrei Bobrov am 8 Okt. 2012

1 Stimme

ii = cell(1,4);
[ii{:}] = ndgrid(1:Ny,1:Nr,1:Nd,1:Nw);
v = cellfun(@(x)x(:),ii,'un',0);
V = zeros(size(ii{1}));
V(v{:})= A(v{:}, Z(v{:}));

2 Kommentare

Thank you for your help. However, unless I am doing something wrong, this is no faster than using the loop! Actually, while I never ran out of memory by solving my problem with the loop, I am not managing to use your method even with very small dimension (i.e. Ny, Nr, Nd, Nw =4).
(??? Maximum variable size allowed by the program is exceeded.)
Any intuition about why this might be the case?
Andrei Bobrov
Andrei Bobrov am 9 Okt. 2012
Often, for large arrays, efficient use of cycle for .. end

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