Making a square figure with pcolor

9 vues (au cours des 30 derniers jours)
V.D-C
V.D-C le 12 Mar 2020
Commenté : darova le 18 Mar 2020
Hello,
I have 2 matrices for x and y coordinates, and a matrix of temperatures. From the matrices, I make 2 coordinate vectors of size 99000 x 1 from the 300 x 300 matrices.
When I plot with imagesc and the vectors, I have the linked result. When I plot with pcolor and the coordinates matrices, I have the other figure.
I would like to make the second figure (plotted with pcolor) square like the one plotted with imagesc. How could I achieve that ?
Thanks in advance !
  4 commentaires
darova
darova le 16 Mar 2020
Can you attach your code? Data?
V.D-C
V.D-C le 17 Mar 2020
Sorry again, I couldn't answer to your message earlier.
The code looks something like that :
%%% Pcolor
h = figure();
pcolor(x(50:250,50:250),y(50:250,50:250),Tair_mean(50:250,50:250));%% I frame the area of interest
shading flat
colorbar;
ylabel(colorbar,'Air Temprature [°C]');
title(['Mean Daily Air Temperature ' date]);
axis equal
set(gca, 'Color', 'none');
%%% Imagesc
% I transform my coordinates matrices in vectors in order to make them work with imagesc
x_vect = x(:);
y_vect = y(:);
h = figure();
imagesc(x_vect,y_vect,Tair_mean(50:250,50:250)) %% I frame the area of interest
How could I get a square figure like imagesc but with pcolor ?

Connectez-vous pour commenter.

Réponse acceptée

darova
darova le 17 Mar 2020
You can either use pcolor without X and Y vectors
newTair = flipud(Tair_mean(50:250,50:250)); % flip matrix upside-down
pcolor(newTair)
Or you can create rotate X and Y
a = 51;
R = [cosd(a) sind(a);-sind(a) cosd(a)];
V = R*[x(:) y(:)]';
x1 = reshape(V(1,:),size(x));
y1 = reshape(V(2,:),size(x));
pcolor(x1(50:250,50:250),y1(50:250,50:250),Tair_mean(50:250,50:250));%% I frame the area of interest
  5 commentaires
V.D-C
V.D-C le 18 Mar 2020
I will test it as soon as my matlab works !
One question though, how did you do to find the rotation value ?
darova
darova le 18 Mar 2020
  • One question though, how did you do to find the rotation value ?
Good question. I used my eyes to calculate it. But maybe it's not the best way
dx = x(2) - x(1);
dy = y(2) - y(1);
a = 90 + atan2d(dy,dx)
a =
50.9309

Connectez-vous pour commenter.

Plus de réponses (1)

J. Alex Lee
J. Alex Lee le 17 Mar 2020
Modifié(e) : J. Alex Lee le 17 Mar 2020
Your coordinate grid matrices x and y are not "aligned" with the axes. You can see this if you just do
plot3(x,y,zeros(size(x)),'.k')
view(2)
So I would say your pcolor code is giving you the correct rendering, and imagesc behavior seems anomalous, since your values of x and y (called x_vect and y_vect in your code), do not follow the requirements of imagesc:
per the documentation, x and y should either be vectors of respective lengths size(Tair_mean), or 2-element vectors specifying only the corner coordinates (implicitly assume on a rectangular canvas aligned with axes), or scalars specifying only 1 corner. As you supply the entire grid in a long vector, it must be assuming you only care about the first and last points, or something like that.
If you want to align with axes, I guess you have 2 options:
  • Forget about the actual coordinate values and simply issue
pcolor(Tair_mean(50:250,50:250));
  1 commentaire
V.D-C
V.D-C le 17 Mar 2020
Hello, thank you very much for your answer.
I think I will have to forget about the values because to transform my original grid I need a rotation matrix but also to take account of the deformation of the pixels because of the UTM conversion.
I will dig into the image processing toolbox though, as you suggested.

Connectez-vous pour commenter.

Catégories

En savoir plus sur Graphics Performance dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by