Plot checker board pattern where color of square is based on value of function.

Question

Logan 2013-3-23

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/68304-plot-checker-board-pattern-where-color-of-square-is-based-on-value-of-function

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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Answer 1

Image Analyst 2013-3-23

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/68304-plot-checker-board-pattern-where-color-of-square-is-based-on-value-of-function#answer_79629

编辑：Image Analyst 2013-3-23

在 MATLAB Online 中打开

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.

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Logan 2013-3-23

编辑：Logan 2013-3-24

在 MATLAB Online 中打开

Sorry about that. Here is the exact code that I am using for this bit. It is evaluating a system and coloring points based on their long term behavior.

clear all;
% Parametric Values
a1=0.2;  
k1=0.05;
k2=0.05;
% Initial Conditions for the Population
J(1)=0;
A(1)=4;
N(1)=4;
% Determines how many iterations the model will run for
T=1500;
% Vector specifying the birth rates to be evaluated over
% We are choosing 100 points for each parameter and thus creating a 100x100 grid to evaluate the long term behavior of our system over.
b=linspace(0,300,100);
a2 = linspace(0,1,100);
for j=1:length(b)
for i=1:length(a2)
    % Run the code for T iterations.
    for t=1:T      
        J(t+1)=a1*exp(-k1*J(t))*J(t)+b(j)*A(t);
        A(t+1)=a1*exp(-k1*J(t))*J(t)+a2(i)*exp(-k2*A(t))*A(t);
        N(t+1)=J(t+1)+A(t+1);
    end
% Discard the first 1000 iterations since we are interested in long term behavior. Assign color based on whether the points solution goes to 0, is an equilibria, or a periodic point.   
if (N(1001)==0)
    color = 'white'
elseif (N(1001)==N(1002))
    color = 'yellow'    
elseif N(1001)==N(1003) && N(1002)==N(1004)
    color = 'red'
elseif N(1001)==N(1004) && N(1002)==N(1005) && N(1003)==N(1006)
    color = 'blue'
elseif N(1001)==N(1005) && N(1002)==N(1006) && N(1003)==N(1007) && N(1004)==N(1008)
    color = 'magenta'    
elseif N(1001)==N(1006) && N(1002)==N(1007) && N(1003)==N(1008) && N(1004)==N(1009) && N(1005)==N(1010)
    color = 'green'
elseif N(1001)==N(1007) && N(1002)==N(1008) && N(1003)==N(1009) && N(1004)==N(1010) && N(1005)==N(1011) && N(1006)==N(1012)
    color = 'cyan'
%elseif N(1001)==N(1008) && N(1002)==N(1009) && N(1003)==N(1010) && N(1004)==N(1011) && N(1005)==N(1012) && N(1006)==N(1013) && N(1007)==N(1014)
%   color = 'cyan'
%elseif N(1001)==N(1009) && N(1002)==N(1010) && N(1003)==N(1011) && N(1004)==N(1012) && N(1005)==N(1013) && N(1006)==N(1014) && N(1007)==N(1015)
%    color = 'yellow'
else
    color = 'black'
end
figure(1)
hold on
if i==1 && j~=1
    fill(0,a2(j-1),b(i),a2(j-1),0,a2(j),b(i),a2(j),color)
elseif i~=1 && j==1
    fill(b(i-1),0,b(i),0,b(i-1),a2(j),b(i),a2(j),color)
elseif i==1 && j==1
    fill(0,0,b(i),0,0,a2(j),b(i),a2(j),color)
else
    fill(b(i-1),a2(j-1),b(i),a2(j-1),b(i-1),a2(j),b(i),a2(j),color)
end
end
end

Logan 2013-3-24

编辑：Logan 2013-3-24

This code is for an ecological model. J is the juvenile population, A is the adult population and N is the total population. I am discarding the first 1000 iterations because I want the system to settle into its long term behavior. My goal is to assign different colors for points that go extinction or 0, fall into an equilibrium ie N(t)=N(t+1), and periodic solutions. That is what the first if-elseif statements are for. It all works fine because the value of the variable color changes for successive iterations but it doesn't plot the colors. So for some reason the fill statements are faulty. I added a couple comments in the code to help make it easier to follow.

Image Analyst 2013-3-24

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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Plot checker board pattern where color of square is based on value of function.

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Plot checker board pattern where color of square is based on value of function.

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