Interpolating a 3D Matrix using interp1

2019 3 21
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更新时间：2019 3 26
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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
KSSV 2019-3-22

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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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Thomas Holmes
Thomas Holmes 2019-3-26
I keep getting this error when trying to spline column wise.
"Unable to perform assignment because the size of the left side is 14-by-1 and the size of the right side is 1-by-221" for this line
NewestA(:,g,h) = R;
N = 221 ;
[m,n,p] = size(A) ;
NewA = zeros(m,N,p) ;
xi = linspace(1,n,N) ;
for i = 1:m % Splining row wise
for j = 1:p
T = interp1(1:n,A(i,:,j),xi) ;
NewA(i,:,j) = T ;
end
end
A = NewA ;
[m,n,p] = size(A) ;
NewestA = zeros(m,N,p) ;
yi = linspace(1,m,N);
for g = 1:n
for h =1:p
R=interp1(1:m,A(:,g,h),yi);
NewestA(:,g,h) = R;
end
end
A=NewestA

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更多回答（1 个）

Matt J
Matt J 2019-3-22
编辑：Matt J 2019-3-22

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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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