Help needed adding features to this script.

2020 5 24
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更新时间：2020 5 25
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0 个投票

I have a script (shown below) that will generate a 12x12 matrix of all 0's and a single 1, which will replicate in a randomised order 7200 times. There are three features I want to add to this script but don't know how:
  1. Instead of generating a matrix in a complete random order I want digits to move either one place up, down, left or right from their position in the previous matrix.
  2. A shortcut that will create an imagesc() for all 7200 matrix's (like the image below).
Any help would be greatly appreciated.
Thanks,
Joe
count=0;
m=zeros(144,1);
rand_int=datasample(1:144,1);
m(rand_int)=1;
old_matrix=reshape(m,[12 12]);
for i=1:7200
m=zeros(144,1);
rand_int=datasample(1:144,1);
m(rand_int)=1;
new_matrix=reshape(m,[12 12]);
if sum(abs(old_matrix-new_matrix),'All')>0
count=count+1;
end
old_matrix=new_matrix
end

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 采纳的回答

Image Analyst
Image Analyst 2020-5-24

0 个投票

Here is one way:
m = zeros(12, 12);
numIterations = 7200;
row = ones(1, numIterations);
col = ones(1, numIterations);
m(row(1), col(1)) = 1;
imagesc(m);
axis('on', 'image');
for k = 2 : numIterations
row(k) = row(k-1) + randi([-1,1], 1);
while row(k) <= 0 || row(k) > size(m, 1)
row(k) = row(k-1) + randi([-1,1], 1);
end
col(k) = col(k-1) + randi([-1,1], 1);
while col(k) <= 0 || col(k) > size(m, 2)
col(k) = col(k-1) + randi([-1,1], 1);
end
m(row(k), col(k)) = 1; % Set new location
m(row(k-1), col(k-1)) = 0; % Clear old location
imagesc(m);
drawnow;
end

9 个评论
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Joseph Lee
Joseph Lee 2020-5-24
That works great thanks!
Can you think of any way to slow down the image creation? a delay of sorts.
Image Analyst
Image Analyst 2020-5-24
Put in a pause(0.5) to pause the display some time.
m = zeros(12, 12);
numIterations = 7200;
row = ones(1, numIterations);
col = ones(1, numIterations);
m(row(1), col(1)) = 1;
imagesc(m);
axis('on', 'image');
for k = 2 : numIterations
row(k) = row(k-1) + randi([-1,1], 1);
while row(k) <= 0 || row(k) > size(m, 1)
row(k) = row(k-1) + randi([-1,1], 1);
end
col(k) = col(k-1) + randi([-1,1], 1);
while col(k) <= 0 || col(k) > size(m, 2)
col(k) = col(k-1) + randi([-1,1], 1);
end
m(row(k), col(k)) = 1; % Set new location
m(row(k-1), col(k-1)) = 0; % Clear old location
imagesc(m);
drawnow;
pause(0.5);
end
Joseph Lee
Joseph Lee 2020-5-24
Great thanks!
Also could you tell me why the image will periodically turn a light blue colour and then back. Is this a processing issue due to the number of images being created?
Image Analyst
Image Analyst 2020-5-24
I didn't see that. Occasionally I'd see the screen flicker. Maybe it picks a colormap that's different if the array is all zeros with no ones. You might reverse the clearing and setting to make sure that there is always a 1 there in case it picked the same location:
Change
m(row(k), col(k)) = 1; % Set new location
m(row(k-1), col(k-1)) = 0; % Clear old location
to
m(row(k-1), col(k-1)) = 0; % Clear old location
m(row(k), col(k)) = 1; % Set new location
Joseph Lee
Joseph Lee 2020-5-24
Also one final question, is there a way of increasing the percentage chance that the 1 in the matrix wont change position when the matrix replicates? maybe a 1:2 ratio.
Again thanks for your speedy responses.
Joe
Image Analyst
Image Analyst 2020-5-24
Just skip the loop if the probability is less. And if trying a new place, make sure it's not the same as the old place.
m = zeros(12, 12);
numIterations = 7200;
row = ones(1, numIterations);
col = ones(1, numIterations);
m(row(1), col(1)) = 1;
imagesc(m);
axis('on', 'image');
percentage = 0.20; % 20% of time will be the same
for k = 2 : numIterations
if rand(1) < percentage
% 20% of the time it will stay in the same place.
row(k) = row(k-1);
col(k) = col(k-1);
else
% 80% of the time it will select a new place.
% First get a tentative new location.
row(k) = row(k-1) + randi([-1,1], 1);
col(k) = col(k-1) + randi([-1,1], 1);
while row(k) == row(k-1) && col(k) == col(k-1)
while row(k) <= 0 || row(k) > size(m, 1)
row(k) = row(k-1) + randi([-1,1], 1);
end
while col(k) <= 0 || col(k) > size(m, 2)
col(k) = col(k-1) + randi([-1,1], 1);
end
end
end
m(row(k-1), col(k-1)) = 0; % Clear old location
m(row(k), col(k)) = 1; % Set new location
imagesc(m);
grid on;
caption = sprintf('At Iteration %d, Row = %d, Column = %d', k, row(k), col(k));
title(caption, 'FontSize', 15);
drawnow;
pause(0.4)
end
Joseph Lee
Joseph Lee 2020-5-24
I get an error message when I run it and the matrix doesn't change
Image Analyst
Image Analyst 2020-5-24
Sorry, try this (actually tested this time):
m = zeros(12, 12);
numIterations = 7200;
row = ones(1, numIterations);
col = ones(1, numIterations);
m(row(1), col(1)) = 1;
imagesc(m);
axis('on', 'image');
percentage = 0.20; % 20% of time will be the same
for k = 2 : numIterations
if rand(1) < percentage
% 20% of the time it will stay in the same place.
row(k) = row(k-1);
col(k) = col(k-1);
else
% 80% of the time it will select a new place.
% First get a tentative new location.
row(k) = row(k-1) + randi([-1,1], 1);
col(k) = col(k-1) + randi([-1,1], 1);
while row(k) == row(k-1) && col(k) == col(k-1) || row(k) <= 0 || row(k) > size(m, 1) || col(k) <= 0 || col(k) > size(m, 2)
% Get a new location.
row(k) = row(k-1) + randi([-1,1], 1);
col(k) = col(k-1) + randi([-1,1], 1);
while row(k) <= 0 || row(k) > size(m, 1)
row(k) = row(k-1) + randi([-1,1], 1);
end
while col(k) <= 0 || col(k) > size(m, 2)
col(k) = col(k-1) + randi([-1,1], 1);
end
end
end
m(row(k-1), col(k-1)) = 0; % Clear old location
m(row(k), col(k)) = 1; % Set new location
imagesc(m);
grid on;
caption = sprintf('At Iteration %d, Row = %d, Column = %d', k, row(k), col(k));
title(caption, 'FontSize', 15);
drawnow;
% pause(0.4)
end
Joseph Lee
Joseph Lee 2020-5-25
Great thanks for your help.
Could you tell me how to assign colors to the numbers. white for zero, red for one? Is this possible to do using Colormaps?

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