13x8 Matrix Cubic Interpolation

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];

 Risposta accettata

Chris
Chris il 26 Ott 2021
Modificato: Chris il 26 Ott 2021
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 Commenti

Yes thank you for the answer,how can i take the values from this model i need the values actually
Chris
Chris il 26 Ott 2021
The values are in the matrix "interpolated."
Or use interp2.

Accedi per commentare.

Più risposte (2)

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 il 26 Ott 2021
Modificato: John D'Errico il 26 Ott 2021

1 voto

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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