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Comparing elements of one vector with multiple values in a matrix of different size

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Benedict
Benedict on 28 Feb 2021
Answered: Steven Lord on 28 Feb 2021
Hi,
I have a m x 1 vector and a n x 2 matrix, e.g.,
v = [80; 99; 120; 150; 199; 204; 208; 239; 301; 402; 634; 689];
A = [0,100; 101,200; 201,300; 351,450; 451,550; 575,675; 681,780];
For each row in A, I want to know how many values in v fall between the two values of that row in A. So for the example above, I would like the result to be
output = [2; 3; 3; 1; 0; 1; 1]
Note that that there are gaps between the intervals defined by the rows of A and that there are values of v that don't fall into any of the intervals specified by the rows of A. In other words, sum(output) is not the same as length(v). However, the intervals always have the same size (100), i.e., A(:,2) - A(:,1) is a vector of 100s.
Best I have come up with so far is a for-loop
for i = 1:size(A,1)
output(i) = sum( (A(i,1) <= v) & (v < A(i,2)) );
end
But because my size(A,1) can be as large as 50,000 and I need to do this operation many times, it takes quite long. Is there a better way with indexing?
I was thinking something like
output = sum( (A(:,1) <= v) & (v < A(:,2)) )
but that doesn't work ("Matrix dimensions must agree.")
Any help is appreciated.
Thanks!
  5 Comments
Benedict
Benedict on 28 Feb 2021
Hi, thanks I had never used discretize before. I think the problem is that the intervals defines by the rows of A have gaps in between them. otherwise
discretize(v,mean(A,2))
might have worked.
@Mario Malic yes, unfortunately the values of the output vector are exactly what I'm interested in.
But thanks to both of you for your suggestions. I guess I'll stick to the for loop :(

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Answers (1)

Steven Lord
Steven Lord on 28 Feb 2021
If the rows in A are sorted and disjoint, this does seem like a job for histcounts. Just ignore the bins representing the spaces between the rows in A.
v = [80; 99; 120; 150; 199; 204; 208; 239; 301; 402; 634; 689];
A = [0,100; 101,200; 201,300; 351,450; 451,550; 575,675; 681,780];
edges = reshape(A.', 1, []);
values = histcounts(v, edges)
values(1:2:end)
results = table(A, values(1:2:end).', 'VariableNames', ["BinEdges", "Counts"])

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