Interpolating a 3D Matrix using interp1
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Thomas Holmes
am 21 Mär. 2019
Kommentiert: Thomas Holmes
am 26 Mär. 2019
I have a 14x14x221 matrix, A. Using a loop over the third dimension, I'm trying to interpolate the 14x14 matrix at each iteration of this loop to produce a new matrix that has more data points. Then save the new matrix at each iteration to form a new 3D matrix that has X,Y dimensions much greater than 14.
My question is how to use interp1 to spline the 14x14 matrix using the third dimension of the 14x14x221 matrix.
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KSSV
am 22 Mär. 2019
A = rand(14,14,221) ;
% interpolation along row
N = 100 ;
[m,n,p] = size(A) ;
iwant = zeros(m,N,p) ;
xi = linspace(1,n,N) ;
for i = 1:m
for j = 1:p
T = interp1(1:n,A(i,:,j),xi) ;
iwant(i,:,j) = T ;
end
end
A = iwant ;
Repeat the same along columns.
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Matt J
am 22 Mär. 2019
Bearbeitet: Matt J
am 22 Mär. 2019
Using my KronProd class (Download)
m=14; p=221; A = rand(m,m,p) ;
M=40; %new XY dimension
B=interp1(eye(m),linspace(1,m,M),'spline' );
A_upsampled=KronProd({B,1},[1,1,2],[nan,nan, p])*A;
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