How to get the index from a matrix?

Hi. I have a problem where I need to update the index value with a larger index if elements in A greater than B..

A = [2;7.5;9;10.5;15;14]; % Also can be in random numbers     A and B are different in size
B = 5:5:50              % Fixed numbers
idx = ones(size(A,1),1; % Set to one if A less than B

for i = 1:size(A,1)
    [~,idx(i)] = ismembertol(A(i),B,0.05);
end

% Results
idx = [0;1;2;2;3;3]

%Which I want to get
idx = [1;2;2,3,4,3] 

Please help, and thanks in advance.

Answer 1

[Adam Danz]

Assuming B is in ascending order, idx(i) contains the index value of B that is closest to but not less than A(i).

A = [2;7.5;9;10.5;15;14]; 
B = 5:5:50; 
[~, idx] = max(A(:)'<B(:),[],1)

idx = 1×6

1 2 2 3 4 3

●

If it cannot be assumed that B is in ascending order,

A = [2;7.5;9;10.5;15;14]; 
B = [5, 15, 10, 20:5:50]; % note: not in ascending order
d = (A(:)'-B(:)); 
d(d>=0)= -inf;
[~, idx] = max(d,[],1)

idx = 1×6

1 3 3 2 4 2

●

Both solutions assume that at least 1 value in B is less than A(i). If this is not the case, you append inf to the end of B and if idx contains the index equal to the location of the final inf, you can replace it with 0. For example,

A = [99 ;7.5;9;10.5;15;14]; % Note; starts with 90
B = 5:5:50; 
[~, idx] = max(A(:)'<[B(:);inf],[],1);
idx(idx==numel(B)+1) = 0

idx = 1×6

0 2 2 3 4 3

●

[mrbond99]

Thank you very much Adam for this tremendous solution. It really solved my problem. Thank you again.