MATLAB: 1D Interpolation on array without using loops.

Question

Hello.

I have been pondering this all day.

So, I have an 3D-array, X1, of size 48 x 232 x 61. Y1 is a row vector of size 1 x 232. Each row of X1 corresponds with Y1. I want to interpolate between each row of X1 and Y1 without using any loops. The output is a 48 x 21 x 61 array, Y2_array. With loops, my code is as follows:

% make Y1 array for interpolation
logz = log(10e-8):.10:log(1080);
Y1 = d + exp(logz);         % This is the 1 x 232 vector. 
Y2 = ones(48,21);           % Preallocating Y2 2D-array (size = 48 x 21). 
Y2_array = ones(48,21,61);  % Preallocating Y2 3D-array (size = 48 x 21 x 61)
% Interpolation
for S = 1:61
    for T = 1:48
        X1_row = X1(T,:,S);
        Y2 = interp1(X1_row, Y1, X2(T,:,S)); 
                                 % X2 is a 3D-array, size (48 x 21 x 61)
        Y2_array(T,:,S) = Y2;
    end
end

If I make the inputs of interp1 arrays, Matlab gives me the error message that X and Y need to be vectors.

I know there is a solution to this problem… any suggestion/thoughts about how to execute the interpolation without using loops?

Thanks!!

Answer 1

Try this instead of INTERP1:

function Yi = LeanInterp(X, Y, Xi)
X  = X(:);
Xi = Xi(:);
Y  = Y(:);
nY = numel(Y);
[dummy, Bin] = histc(Xi, X);  %#ok<ASGLU>
H            = diff(X);
if Bin(length(Bin)) >= nY
   Bin(length(Bin)) = nY - 1;
end
Ti = Bin + (Xi - X(Bin)) ./ H(Bin);
% Interpolation parameters:
Si = Ti - floor(Ti);
Ti = floor(Ti);
% Shift frames on boundary:
d     = (Ti == nY);
Ti(d) = Ti(d) - 1;
Si(d) = 1;
% Now interpolate:
Yi = Y(Ti) .* (1 - Si) + Y(Ti + 1) .* Si;

For [1x100] vectors this is about 3 times faster than INTERP1 of Matlab 2009a.