How to interpolate 2D array from 3D array?

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piston_pim_offset
piston_pim_offset 2023-12-15
移动：Image Analyst 2023-12-29
A=[3 4 5 6 7];
numel(X)=23;
numel(Y)=42;
There are 5 different contour data with respect to X and Y like the one belove. I have 5 different contour data of 23x42, together a 3D array of 5x23x42. There are data for the given values of A, but l need data for interpolated values of A, for example in the case of A=4.7. Then l'll plot the data corresponding to the calculated value of A.
Thanks for your help!
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Torsten
Torsten 2023-12-15
编辑：Torsten 2023-12-15
@limon cu
I don't completely understand your question.
You have contour data on an x/y/z grid of size 5x23x42 and you have a value in between the 5 x-coordinates and you want to interpolate your 5x23x42 contour data to this value to get back a 23x42 matrix ?
piston_pim_offset
piston_pim_offset 2023-12-15
Yes, exactly. I have and contour plot of 23x42, and there are 5 of them wrt. 5 different values. I want to get interpolated interpolated data for every possible value of X.

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回答（3 个）

piston_pim_offset
piston_pim_offset 2023-12-15
编辑：piston_pim_offset 2023-12-15
Ran in:
AI Chat Playground gave me this answer with an error:
contourData = randn(5, 23, 42);
A = [3 4 5 6 7];
interpContourData = interp1(A, contourData, 4.7);
[X, Y] = meshgrid(1:size(interpContourData, 2), 1:size(interpContourData, 3));
contourf(X, Y, squeeze(interpContourData), 20);
Error using contourf
The size of X must match the size of Z or the number of columns of Z.
Dyuman Joshi
Dyuman Joshi 2023-12-15
移动：Image Analyst 2023-12-29
Ran in:
Try this. Though I have no idea if it does what you asked for or not.
contourData = randn(5, 23, 42);
A = [3 4 5 6 7];
interpContourData = interp1(A, contourData, 4.7);
[X, Y] = meshgrid(1:size(interpContourData, 2), 1:size(interpContourData, 3));
interpContourData = permute(interpContourData, [3 2 1]);
contourf(X, Y, squeeze(interpContourData), 20)
Torsten
Torsten 2023-12-15
移动：Torsten 2023-12-15
contourData = randn(5, 23, 42);
x = 1:5;
y = 1:23;
z = 1:42;
X = [2.6 4.7];
contourData_interpolated = zeros(numel(X),23,42);
for j = 1:23
for k = 1:42
contourData_interpolated(:,j,k) = interp1(x,squeeze(contourData(:,j,k)),X);
end
end

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