Error in interp2 (interpolation) command.

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Ivan Mich
Ivan Mich 2022 年 10 月 22 日
コメント済み: Torsten 2022 年 10 月 24 日
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 件のコメント
Ivan Mich
Ivan Mich 2022 年 10 月 22 日
I am uploading a sample of data file
Torsten
Torsten 2022 年 10 月 22 日
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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回答 (2 件)

Torsten
Torsten 2022 年 10 月 22 日
編集済み: Torsten 2022 年 10 月 22 日
Read about the requirements for input arguments x, y and z for interp2:
Paragraph "Input Arguments".
  4 件のコメント
Torsten
Torsten 2022 年 10 月 22 日
Your data are not gridded. Thus you will have to use "scatteredInterpolant":
F = scatteredInterpolant(x,y,z);
[xq,yq] = meshgrid(linspace(min(x),max(x)),linspace(min(y),max(y)));
zq = F(xq,yq)
Steven Lord
Steven Lord 2022 年 10 月 22 日
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 2022 年 10 月 22 日
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 件のコメント
Stephen23
Stephen23 2022 年 10 月 24 日
@Ivan Mich: you could try a thin-plate spline: https://www.mathworks.com/help/curvefit/tpaps.html
Torsten
Torsten 2022 年 10 月 24 日
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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