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not sure how to approach this

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jenka
jenka 2013-2-1
I have a 2D matrix. Let's say A=80x100; I need a way to visit each cell in this matix only once. Moreover my problem is a little more involved than that. Not only do I want to visit each cell only once but the user also defines a window of certain size. Let's say WINDOW=3x3. Then let's say I randomly choose a location on A that corresponds to indecies [2,3]. Then this is considered a central location and in my algorithm I will also consider points around this central location as specified by the window. Hence in this example I will also consider points at indecies [1,3],[3,3],[1,2],[2,2],[3,2],[1,4],[2,4],[3,4]. So in the next round when I am trying to figure out which cell in A to visit, I will not consider these 9 points in my random draw. Any suggestions? Thanks!

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采纳的回答

Matt J
Matt J 2013-2-1
编辑：Matt J 2013-2-1
A=rand(80,100);
[m,n]=size(A);
winsize=3;
midwin=winsize/2+.5;
[dx,dy]=ndgrid(1:winsize);
jumps = sub2ind([m,n],dx,dy)-sub2ind([m,n],midwin,midwin);
B=A;
B(:)=1:numel(A);
B=B(midwin:end+1-midwin, midwin:end+1-midwin);
points=B(:);
numpoints=length(points);
while numpoints
randlocation=points(randi(numpoints));
window=randlocation+jumps;
%%%%%%%%%%%%do stuff to A(window)
points=setdiff(points,window(:));
numpoints=numel(points);
end

更多回答（2 个）

Sean de Wolski
Sean de Wolski 2013-2-1
idx = randperm(numel(A));
To give you a list of random linear indices to visit.
Then use a for-loop to loop over these. When you get to one index, use ind2sub() to convert it to row/column coordinates (or do this before hand for all of them and loop over that.)
doc ind2sub %for more info
Then make your window by adding and subtracting floor(window_size/2) to the indices and do whatever operation you have on this. You could also do this addition/subtraction before hand and loop over that instead.
jenka
jenka 2013-2-1
This is sooo great! I am so thankful to you!
  3 个评论
显示 1更早的评论
jenka
jenka 2013-2-1
Hi Matt,no, I did not know that. Will do this for sure. I have a quick question regarding border's effect of this problem. I am not sure how to quickly fix this? For example, assume
A=rand(4,5).
Then if randlocation is equal to 16.
The window is
window =
11 15 19
12 16 20
13 17 21
Which is not entirely correct. As in this case the window should be
window =
11 15 19
12 16 20
Any suggestions? Thanks
Matt J
Matt J 2013-2-1
编辑：Matt J 2013-2-1
In the code I gave you, randlocation will never be on the boundary. I excluded all locations without a full winsize x winsize neighborhood in this line
B=B(midwin:end+1-midwin, midwin:end+1-midwin);
If you're saying that you can't ignore boundary points, you could pre-embed your A matrix into a larger matrix, and modify as follows
A=rand(80,100);
winsize=3;
midwin=winsize/2+.5;
A0=A; %save original A
A=nan(size(A0)+winsize-1);
A(midwin:end+1-midwin, midwin:end+1-midwin)=A0; %enlarge
[m,n]=size(A);
[dx,dy]=ndgrid(1:winsize);
jumps = sub2ind([m,n],dx,dy)-sub2ind([m,n],midwin,midwin);
B=A;
B(:)=1:numel(A);
B=B(midwin:end+1-midwin, midwin:end+1-midwin);
points=B(:);
numpoints=length(points);
while numpoints
randlocation=points(randi(numpoints));
window=randlocation+jumps;
window(isnan(A(window)))=[]; %THROW AWAY NANS!!!!!!!!!!!!!
%%%%%%%%%%%%do stuff to A(window)
points=setdiff(points,window(:));
numpoints=numel(points);
end

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