Plot checker board pattern where color of square is based on value of function.
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This is part of a larger process so I have rewritten the problem so that it can stand on its own. If something is unclear or looks off I would be happy to clarify or take a look at it again.
So what I am doing is taking a 100 points along the x and y axis and evaluating a function, f(x,y), on each point created by that grid. So if (1,1) gives a value that I associate with blue then I want the square with the corners (0,0),(0,1),(1,0) and (1,1) to be colored blue.
for x=1:X
for y=1:Y
if [Conditions on the value of f(x,y)]
color='blue'
elseif ...
...
else
...
end
% Here is what I tried for the plotting the figure. The figure opens and runs, but no colors are plotted so I something is off.
figure(1)
hold on
if x==1 && y~=1
fill(0,Y(y-1),X(x),Y(y-1),0,Y(y),X(x),Y(y),color)
elseif x~=1 && y==1
fill(X(x-1),0,X(x),0,X(x-1),Y(y),X(x),Y(y),color)
elseif i==1 && j==1
fill(0,0,X(x),0,0,Y(y),X(x),Y(y),color)
else
fill(X(x-1),Y(y-1),X(x),Y(y-1),X(x-1),Y(y),X(x),Y(y),color)
end
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Réponses (1)
Image Analyst
le 23 Mar 2013
Modifié(e) : Image Analyst
le 23 Mar 2013
I don't understand. You have
for x=1:X
and then in the fills you have X(x). Exactly what is capital X? If it's a 1D array of 100 values, then what are you doing in the "for" line? Can you give the values of X and Y so we can run your code? It's not quite standalone right now because it does run as given. I want to see the range of values that X takes on.
By the way, there is a checkerboard() function in the Image Processing Toolbox.
6 commentaires
Image Analyst
le 24 Mar 2013
I don't have time to dive into the code and understand now. Maybe you can just use an image instead of the fill() function.
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