Compare 2 time series, array and map to third.

Hi,

I have a random column vector A sorted from high to low. I have to map the position of its each entry on column vector B (again sorted high to low), and use that indexing to write a new function C.

Here's an example:

B = [3, 2.5, 2.0, 1.5, 1.0]';
%% this BB column vector represents values from intervals: 0 to 1.0, 1.0 to 1.5, 1.5 to 2.0, 2.0 to 2.5, 2.5 to 3.0

A = [3.3, 2.4, 2.2, 1.3, 0.5]'

idx_A = zeros(5,1); % ??? finds the position of A on the point intervals of B.

% I'm lookimg to use a formula or a faster way to get the following result. 
idx_A(1,1) = 1; % as A(1,1) is greater than the 1st value of B. 
idx_A(2,1) = 3; % as A(2,1) is greater than the 3rd value of B but smaller than 2nd value of B 
idx_A(3,1) = 3; % as A(3,1) is greater than the 3rd value of B but smaller than 2nd value of B 
idx_A(4,1) = 5; % as A(4,1) is greater than the 5th value of B but smaller than 4nd value of B 
idx_A(5,1) = 5; % A(5,1) is lower than the least value of B so should be equal to the lowest value of B, which is 5.


% after finding idx_A, there is another column vector D which is used to find C as:

C = D(idx_A,1);

display(D)

many thanks for your help.

Answer 1

[Adam Danz]

This should work for you. The last value of B is changed to -inf so that all values other than NaN will be greater than or equal to the last one.

B(end) = -inf; 
idx_A = arrayfun(@(x)find(x>=B,1),A); 

Result

idx_A =
     1
     3
     3
     5
     5

*Assumes B is sorted in descending order as described by OP.

[J Yadav]

Thank you very much.

[Adam Danz]

Glad I could help!