finding distance within a range of distance

2 次查看（过去 30 天）
Mohammad Golam Kibria
[EDIT: 20110524 01:36 CDT - reformat, clarify - WDR]
Hi I have a matrix as follows.
I =
1 1 5 1 1 1 8
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 5 2 1 1 1
1 1 1 1 1 5 1
1 1 1 1 1 1 1
I need to know whether distances of value 5 with another 5 are in the range 3. Thanks in advance

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2011-5-24
ij5 = find(I==5);
[i5 j5] = ind2sub(size(I),ij5);
out1 = [nchoosek(ij5,2) bsxfun(@(x,y)sqrt(diff(nchoosek(x,2),[],2).^2....
+diff(nchoosek(y,2),[],2).^2),i5,j5)];
[non,indout]=min(out1(:,3));
out = out1(indout,:);
in vector "out" the first Two members - the indices of matrix "I", the third distance between them
EDIT
out = out1(out1(:,3) <= 3,:)
EDIT EDIT
The answer to the last Mohammad comment:
[ii,jj]=ind2sub(size(I),nchoosek(find(I==5),2));
out1 = [ii jj sqrt(sum(reshape(diff([ii;jj],[],2).^2,[],2),2))];
[non,indout]=min(out1(:,5)); % 1 variant
out = out1(indout,:);
out = out1(out1(:,5) <= 3,:); % 2 variant
  1 个评论
Mohammad Golam Kibria
Mohammad Golam Kibria 2011-5-25
your answer is correct.but I need indices in 2D form i.e. in the form as follows:
1 3 4 3 3 the first two is the row and column of first 5
second two is the row and column of second 5 and last one is the distance. or any other convenient way to get the indices in 2D form with distance

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更多回答（1 个）

Wolfgang Schwanghart
How about
[X,Y] = meshgrid(1:size(I,2),1:size(I,1));
ix = find(I == 5);
d = hypot(bsxfun(@minus,X(ix),X(ix)'),bsxfun(@minus,Y(ix),Y(ix)'));
[ixd,cc] = find(d<=3 & d>0);
II = false(size(I));
II(ix(ixd)) = true;
Cheers, W.
  2 个评论
Walter Roberson
Walter Roberson 2011-5-24
Wolfgang's output matrix II has a 1 in each position at which there was originally a 5 that was within a distance of 3 of another 5.
If that output is not what you wanted, then you need to specify the form of the output that you did want.

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