Problems with intersect function

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Nithin
Nithin 2011-10-20
Hi everybody,
I have the following problem. I describe the following lists of points:
x1 = -1.5:h:1.5;
x2 = -1.5:h:1.5;
[x1 x2] = meshgrid(x1,x2);
x1 = x1(:);
x2 = x2(:);
X = [x1 x2];
I want to check wether the point some point (p, q) is part of the list. I do the following:
point = [-0.75, 0.55];
intersect(X, point, 'rows');
Matlab returns me an empty matrix. Whereas I know for certain that point is inside the list. I checked even manually by open X in the workspace!
Does anybody know what is going on. Is this is a bug? I need to perform set intersections for some problem.
regards,
nithin
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Fangjun Jiang
Fangjun Jiang 2011-10-24
Due to floating point comarison.
http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

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回答（1 个）

Sean de Wolski
Sean de Wolski 2011-10-20
A fix:
X(all(abs(bsxfun(@minus,X,point))<(10^-10),2),:)
Row with both absolute differences less than tolerance. Will only work as written for one point. That could be changed using the third dimension.
MORE
point = your_points;
X = your_matrix;
sz = size(X,1);
inpts = cell(sz,1)
for ii = 1:sz
inpts{ii} = expression_above_of_points(ii)
end
intersected_points = unique(cell2mat(inpts),'rows')
unique won't have the floating point issues since they'll all be extracted from X and thus actually equal.
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Nithin
Nithin 2011-10-24
Hi sean,
That is what I did, extend it to the third dimension. It works for a few points, but when you have a lot of points, making the 3D matrix takes too much time (I used repmat to copy X in the 3rd dimension).
regards,
nithin
Sean de Wolski
Sean de Wolski 2011-10-25
repmat is the wrong approach since it'll require creating that huge array twice (one too many times). permute and/or reshape is what you should be looking for, however, you'll still need that big matrix once.
I would just use a for-loop. See the edit it my above post.

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