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];

 Akzeptierte Antwort

Chris
Chris am 26 Okt. 2021
Bearbeitet: Chris am 26 Okt. 2021

1 Stimme

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 Kommentare

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

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Weitere Antworten (2)

Image Analyst
Image Analyst am 26 Okt. 2021

2 Stimmen

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 am 26 Okt. 2021
Bearbeitet: John D'Errico am 26 Okt. 2021

1 Stimme

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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