determining whether a number lies in any of the intervals of a 2 column matrix

basically i have a matrix of numbers, single column, such as the following:

A = [2; 4; 17; 23; 30]

and i have another matrix with 2 columns, which represents an interval:

B =

1 &nbsp&nbsp&nbsp&nbsp&nbsp 3

5 &nbsp&nbsp&nbsp&nbsp&nbsp 10

15 &nbsp&nbsp&nbsp&nbsp 18

20 &nbsp&nbsp&nbsp&nbsp 22

29 &nbsp&nbsp&nbsp&nbsp 33

What I am tryin to do is have MATLAB go through each of the numbers in matrix A and have them run through each interval in matrix B to check whether or not it lies in any of those intervals.

For example,

A(1) = 2

This lies between 1 and 3 (the first interval in the list B, so that case would be a true.

The second case of A(2) = 4 does not lie in any of the intervals (1 to 3, nor 5 to 10, nor any of the ones below those two), so this would be a false for that value. I really am not sure how to do this.. I tried setting up a 'for' loop and got confused very quickly. Any help at all would be SO appreciated. thanks in advance!

Answer 1

[Sean de Wolski]

I have an elementary for-loop running twice as fast as the arrayfun method.

idx = false(size(A));
for ii = 1:length(A)
  idx(ii) = any((A(ii)>B(:,1))&(A(ii)<B(:,2)));
end

Answer 2

[Andrei Bobrov]

K = min(B(:)):max(B(:));
idx = arrayfun(@(x)find(K>=B(x,1) & K<=B(x,2)),1:length(A),'un',0);
out = A(ismember(A,K([idx{:}])));

[Ole Gunnar Nordli]

I am trying to find a interval where one number lies within in a 1.000 simulations.

Can I place the value I am looking for as A within the interval B? I do get errors, how do I fix this?

Thanks

Answer 3

[Malcolm Campbell]

What about:

bin_edges = sort(reshape(B,numel(B),1));

[~,~,bin] = histcounts(A,bin_edges);

idx = ~isEven(bin);

Works fast for me! (Much faster than the for loop in the accepted answer)