13x8 Matrix Cubic Interpolation

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I have a matrix [13x8]. I want to cubic interpolate it to [50x8]. I tried "csapi" but I couldn't manage to figure it out.
[0.0000 0.0000 0.0000 0.0000 0.0000 0.0890 0.2860 0.4380;
0.0177 0.0648 0.0888 0.1070 0.1640 0.3680 0.5720 0.7040;
0.0594 0.1778 0.2445 0.3140 0.4250 0.6140 0.7650 0.8540;
0.2652 0.4723 0.6038 0.7260 0.8250 0.8970 0.9500 0.9820;
0.5436 0.7547 0.8852 0.9570 0.9800 0.9910 0.9980 1.0000;
0.7409 0.9056 0.9870 1.0000 1.0000 1.0000 1.0000 1.0000
0.7710 0.9260 0.9980 1.0000 1.0000 1.0000 1.0000 1.0000
0.7695 0.9260 0.9980 1.0000 1.0000 1.0000 1.0000 1.0000
0.6399 0.8297 0.9401 0.9710 0.9800 0.9850 0.9900 0.9920
0.3369 0.5788 0.7226 0.7780 0.8020 0.8270 0.8510 0.8770
0.0771 0.2463 0.3493 0.3890 0.4070 0.4300 0.4720 0.5360
0.0200 0.1019 0.1577 0.1770 0.1840 0.1940 0.2290 0.2990
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0200 0.0510];

采纳的回答

Chris
Chris 2021-10-26
编辑:Chris 2021-10-26
Like this?
V = [0.0000 0.0000 0.0000 0.0000 0.0000 0.0890 0.2860 0.4380;
0.0177 0.0648 0.0888 0.1070 0.1640 0.3680 0.5720 0.7040;
0.0594 0.1778 0.2445 0.3140 0.4250 0.6140 0.7650 0.8540;
0.2652 0.4723 0.6038 0.7260 0.8250 0.8970 0.9500 0.9820;
0.5436 0.7547 0.8852 0.9570 0.9800 0.9910 0.9980 1.0000;
0.7409 0.9056 0.9870 1.0000 1.0000 1.0000 1.0000 1.0000
0.7710 0.9260 0.9980 1.0000 1.0000 1.0000 1.0000 1.0000
0.7695 0.9260 0.9980 1.0000 1.0000 1.0000 1.0000 1.0000
0.6399 0.8297 0.9401 0.9710 0.9800 0.9850 0.9900 0.9920
0.3369 0.5788 0.7226 0.7780 0.8020 0.8270 0.8510 0.8770
0.0771 0.2463 0.3493 0.3890 0.4070 0.4300 0.4720 0.5360
0.0200 0.1019 0.1577 0.1770 0.1840 0.1940 0.2290 0.2990
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0200 0.0510];
surf(V)
F = griddedInterpolant(V,'cubic');
X = linspace(1,13,50);
Y = 1:8;
[XX,YY] = ndgrid(X,Y);
interpolated = F(XX,YY);
surf(interpolated)
  2 个评论
Esat Akdöngel
Esat Akdöngel 2021-10-26
Yes thank you for the answer,how can i take the values from this model i need the values actually
Chris
Chris 2021-10-26
The values are in the matrix "interpolated."
Or use interp2.

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更多回答(2 个)

Image Analyst
Image Analyst 2021-10-26
You can use imresize() to do bicubic interpolation, if you have the image Processing Toolbox
resizedMatrix = imresize(m, [50,8])

John D'Errico
John D'Errico 2021-10-26
编辑:John D'Errico 2021-10-26
Simplest is to just read the help for interp2. It is silly to use a lower level tool when a high level tool does exactly what you want.

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