Error in interp2 (interpolation) command.

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Ivan Mich
Ivan Mich on 22 Oct 2022
Commented: Torsten on 24 Oct 2022
I woyld like to intepolate my data. I have an ascii file with 3 columns (x,y,z). Each column has 700 lines (number). I am using the following commands, in order to interpolate (smooth) my data:
clc
clear
filename1= 'mydata.csv';
[d1,tex]= xlsread(filename1);
y=d1(:,3);
x=d1(:,4);
z=d1(:,5);
[xq, yq] = meshgrid(...
linspace(min(x),max(x)),...
linspace(min(y),max(y)));
zq = interp2(x,y,z,xq,yq,'cubic');
[c,h]= contourf(xq,yq,zq);
but command window shows me:
Error using griddedInterpolant
The grid vectors must be strictly monotonically increasing.
Error in interp2>makegriddedinterp (line 228)
F = griddedInterpolant(varargin{:});
Error in interp2 (line 128)
F = makegriddedinterp({X, Y}, V, method,extrap);
could you please help me?
  3 Comments
Torsten
Torsten on 22 Oct 2022
No, the two inputs being provided to the INTERP2 are not vectors, but are in fact the (matrix) outputs from MESHGRID.
The inputs x,y and z to interp2 the OP provides are vectors (which is wrong). The query values xq and yq are matrices obtained from "meshgrid".

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Answers (2)

Torsten
Torsten on 22 Oct 2022
Edited: Torsten on 22 Oct 2022
Read about the requirements for input arguments x, y and z for interp2:
Paragraph "Input Arguments".
  4 Comments
Steven Lord
Steven Lord on 22 Oct 2022
Confirm that your x and y vectors are sorted.
issorted(x, 'strictascend')
issorted(y, 'strictascend')
One or both of those will return false. Once you know which one(s) are not strictly ascending, you need to determine why. Plotting that vector may help you identify where the flat, decreasing, or missing segments are.

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Torsten
Torsten on 22 Oct 2022
d1 = readmatrix("https://de.mathworks.com/matlabcentral/answers/uploaded_files/1165673/mydata.csv");
y=d1(:,3);
x=d1(:,4);
z=d1(:,5);
F = scatteredInterpolant(x,y,z,'nearest');
Warning: Duplicate data points have been detected and removed - corresponding values have been averaged.
[xq,yq] = meshgrid(linspace(min(x),max(x)),linspace(min(y),max(y)));
zq = F(xq,yq);
contourf(xq,yq,zq)
colorbar
  3 Comments
Torsten
Torsten on 24 Oct 2022
Your z-data are not continuous - they only take values 1 1.5 2 2.5 3 ... 9. So would continuous values as 3.1865 even make sense for your application ?
And if you look at the contour plot from your raw data above: the only thing you can learn is that the values for the south-east part are lower than those for the north-west part.
Everything else like further smoothing would not be appropriate in my opinion.

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