Spline interpolation of 4D matrix

Hello everyone,

I have a 4D matrix A of size 450x450x60x3. This matrix corresponds to images measured at different power levels, B. Where each pixel in A was measured with a different powerlevel. So I have a another matrix B of size 450x450x60x3. I want to perform a spline interpolation of A for power level [1.3, 1.7, 2.3]. What would be the best way to do this? Should I be treating this problem as a 3 times 1D interpolation or do I need to use gridded data? If so, how can I do this?

Thank you :)

Answer 1

[KSSV]

A = rand(450,450,50,3) ; 
B = zeros(size(A)) ;
for i = 1:450
    for j = 1:450
        for k = 1:50
            B(i,j,k,:) = interp1([1 2 3],squeeze(A(i,j,k,:)),[1.3 1.7 2.3]) ; 
        end
    end
end

[MM]

Thanks a lot KSSV. I forgot to mention in my question that actually each element A has a different value for B. So instead of a vector ([1 2 3]), I have a another 4D matrix, let's call it B1. So it would be:

for i = 1:450
    for j = 1:450
        for k = 1:50
            B(i,j,k,:) = interp1(squeeze(B1(i,j,k,:)),squeeze(A(i,j,k,:)),[1.3 1.7 2.3]) ; 
        end
    end
end

But this gives me an error

Error using griddedInterpolant

Interpolation requires at least two sample points in each dimension.

Error in interp1 (line 134)

F = griddedInterpolant(X,V(:,1),method);

[KSSV]

What is B1??? It seems in B1 the values are not in increasing order....they are supposed to be monotonically increasing.

[MM]

See attached for an example of B1. I only included B1(:,:,1:3,:) instead of all 50 for the 3rd dimension because otherwise it was too big.

Let me give a little bit more explanation on the data.

For the measurement of the data in A, I used B1 = [1.3, 1.7,2.3]. But the real value used by the machine is spatially dependent and so B1 is different for each x,y which is represented in the attached matrix. B1 is the same in the 3rd dimension.

I want to determine what the values in A would be if B1 was [1.3, 1.7, 2.3] for each x,y. So I want to interpolate the data in A for B1 = 1.3, 1.7 and 2.3.