Color a grid by using an intensity matrix

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julio@sapienza
julio@sapienza 2011 年 3 月 29 日
Hi all! I need to color a grid according to the values of a specific matrix. In particular, I have a 5x5 matrix representing an intensity value (from 0 to 1) and each value A(i,j) of this matrix is related to a specific cell in the grid. The grid goes from 0 (meters) to 100 (meters)---> 0:20:100 for both axes. Thus, each 20m x 20m cell should have a color according to the matrix described above. I tried with pcolor(), imshow(), imagesc() and so on, but with bad results.
Can you help me? Thanks in advance!!! Giulio (Sapienza University of Rome)

採用された回答

Matt Tearle
Matt Tearle 2011 年 3 月 29 日
What do you mean "bad results"? Can you explain or show what you're doing?
Either way, I'm going to take a guess that the problem is related to the fact that you know the intensities are between 0 and 1, but imagesc always scales to the range of the data. But using image won't work either, because it uses actual values to index into the colormap. If that's the issue, then here are a couple of ways around it:
z = rand(5);
x = 10:20:90;
image(x,x,cat(3,z,z,z))
This makes a 3-D array out of z, then makes a true-color image. Or:
figure
colormap(gray(64))
image(x,x,0.5+z*64), colorbar
which sets the colormap to use 64 colors, then simply expands the possible data range to match that (1 to 64).

その他の回答 (4 件)

julio@sapienza
julio@sapienza 2011 年 3 月 29 日
Thanks for your reply! the first solution you proposed is what I was looking for... the problem now is that the y axis goes from 100 (bottom) to 0 (top) so I need to reverse it... how can I change the colormap? I tried colormap(autumn(3)) but anything changed... Thanks a lot!
  1 件のコメント
Matt Tearle
Matt Tearle 2011 年 3 月 29 日
The first problem is easily fixed with "axis xy"
The colormap is irrelevant if you are using a true-color image (which is what my first solution does). By making all three planes of the image equal to z, you make a grayscale image. If you want some other color scheme, you'll have to work out how to make the 3-D array. For example
image(x,x,cat(3,0*z,0.8*z,z))
But it might just be easier to use the second approach I showed.

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julio@sapienza
julio@sapienza 2011 年 3 月 29 日
I used the second approach and all the "problems" are now fixed! Thanks! I really appreciated your help!

julio@sapienza
julio@sapienza 2011 年 3 月 30 日
I have another simple question... what if is 10 x 10? (0:10:100 for both ax.) How do I modify this expression?
x = 10:20:90;
  2 件のコメント
julio@sapienza
julio@sapienza 2011 年 3 月 30 日
ok this is not a simple question but a stupid question :) the answer is 5:10:95...
Matt Tearle
Matt Tearle 2011 年 3 月 30 日
Not a stupid question, but glad you figured it out. For anyone else out there: the point is that image centers the pixels on the axis values, hence the x values have to be the *centers* of the squares, not the edges.

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julio@sapienza
julio@sapienza 2011 年 3 月 31 日
2D visualization is definitely ok, but what if I want to plot this matrix in 3D? I now have a 10x10 matrix with values in [0,1] but when I try 'surf(my_matrix)' MATLAB only displays a 9x9 grid. Do I have to use meshgrid command before? Thank you in advance!
  8 件のコメント
julio@sapienza
julio@sapienza 2011 年 4 月 19 日
Thanks for the answer...Can you give a practical example please?
Matt Tearle
Matt Tearle 2011 年 4 月 19 日
mesh(A,'FaceColor','texturemap','EdgeColor','none')
But I'm not sure that's what you're referring to. Can you explain what you'd like to see? Something more like this, perhaps?
bar3(A)

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