how to get RGB color?

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JMS
JMS 2014-5-9
评论： Oliver Woodford 2014-5-12
How do I get RGB color? Or what I got using 'pcolor()' can be considered to be as RGB?
Vx = V_x1 + V_x2;
Vy = V_y1 + V_y2;
ang = atan2(Vy, Vx);
pcolor(x, y, ang);
hold on;

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回答（2 个）

Image Analyst
Image Analyst 2014-5-9
编辑：Image Analyst 2014-5-9
pcolor() does not return an RGB image. It displays one. If you display a matrix though, pcolor chops off one row and one column so you're better off using image(). I never use pcolor for displaying images for that reason. I don't really know what you're doing because you didn't attach a screenshot. If you have certain kinds of data that pcolor is meant for, it might be fine, but I find it very deceptive for displaying 2D arrays of numbers.
If you do want what pcolor displays as an RGB image, I think you can use getframe().
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Image Analyst
Image Analyst 2014-5-11
Yeah, I always thought that was extremely confusing. For example let's look at this code where you send a 3 by 3 array into pcolor and it plots tiles in a 2 by 2 array:
clc;
m = [1 2 3;
4, 5, 6;
7, 8, 9]
pcolor(m) % Show 2 by 2 tiles.
cm = gray(9)
colormap(cm);
fr = getframe(gca);
theImage = fr.cdata;
[rows, columns, numberOfColorChannels] = size(theImage)
upperLeftColor = double(theImage(floor(rows/4), floor(columns/4), 1))/256
upperRightColor = double(theImage(floor(rows/4), floor(3*columns/4), 1))/256
lowerLeftColor = double(theImage(floor(3*rows/4), floor(columns/4), 1))/256
lowerRightColor = double(theImage(floor(3*rows/4), floor(3*columns/4), 1))/256
Be aware that the image is upside down with respect to the matrix.
Now, you can see that the lower left quadrant is supposed to be defined by a surface going between corner vertices that have values 1, 2, 4, and 5. So the surface is a tilted, warped surface. Wouldn't the color in that patch change as it goes from the lower values to the higher values? No, it's a single color? Okay, but which color does it use? Evidently it uses the color defined by the lower left vertex only, which for the lower left would be 1, and so the color is 0. The other vertices with values 2, 4, and 5 are apparently ignored when choosing a color. It's confusing. But I see a lot of people using pcolor() to display images.
JMS
JMS 2014-5-11
Ok, can we just forget about pcolor() command because I do not have much information about what you are talking about, sorry. Suppose we have the below equations '
x1 = -1; y1 = 0;x2 = 0; y2 = 0;
[x,y] = meshgrid(-5: .2: 5, -5: .2: 5);
V_x1 = ((y-y1)./((x-x1).^2+(y-y1).^2));
V_y1 = ((x-x1)./((x-x1).^2+(y-y1).^2));
V_x2 = (-(y-y2)./((x-x2).^2+(y-y2).^2));
V_y2 = (-(x-x2)./((x-x2).^2+(y-y2).^2));
Vx = V_x1 + V_x2;
Vy = V_y1 + V_y2;
quiver(x, y, Vx, Vy);
hold on
ang = atan2(Vy, Vx);
?
?
"
And I need to have 'RGB' image depending on the angle. How do I do that? I hope this is clear.

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Oliver Woodford
Oliver Woodford 2014-5-9
编辑：Oliver Woodford 2014-5-10
You can use the sc toolbox, from the File Exchange. In your case, try:
im = sc(ang, 'hsv');
However, there is also a colormap implemented for 2D vector flow fields, which encodes both direction and magnitude. Try:
im = sc(cat(3, Vx, Vy), 'flow');
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JMS
JMS 2014-5-12
Dear Oliver Woodford..I clicked on 'Get from GitHub' and downloaded the zip file to the "..\MATLAB" folder and unzipped and modified the simple code to be:
'
x1 = -1; y1 = 0;x2 = 0; y2 = 0;
[x,y] = meshgrid(-5: .2: 5, -5: .2: 5);
V_x1 = ((y-y1)./((x-x1).^2+(y-y1).^2));
V_y1 = ((x-x1)./((x-x1).^2+(y-y1).^2));
V_x2 = (-(y-y2)./((x-x2).^2+(y-y2).^2));
V_y2 = (-(x-x2)./((x-x2).^2+(y-y2).^2));
Vx = V_x1 + V_x2;
Vy = V_y1 + V_y2;
quiver(x, y, Vx, Vy);
hold on
ang = atan2(Vy, Vx);
im = sc(cat(3, Vx, Vy), 'flow');
'
But it doesn't work. Could you please tell what is wrong? Still say undefined function 'SC'.. Even I used 'im = sc(cat(3, Vx, Vy), 'flow');'. Thank you.
Oliver Woodford
Oliver Woodford 2014-5-12
The toolbox needs to be on your MATLAB search path .

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