Question

I am wondering whether I can do the following efficiently in Matlab. Writing a naive loop for the problem is pretty straightforward, but I am trying to find whether there are any specialized functions one could use (perhaps arrayfun / accumarray (?) — both of which I have great trouble in understanding!) Thanks in advance.

Let’s say I have two vectors as follows (in Matlab):

A = [15 4 9 6 7 5 11 3 14]; 
B = [2 7 13]; 

I would like to do the following:

1. Sort B, if not sorted already.
2. For every successive windows in B (i.e., [2,7], [7,13]), find the corresponding elements in A that lie within the window.
3. In this “partitioned” A, decrement by 1 n-times from those elements in A that lie in the n-th window of B.

Example: In the above case, the first window of B is [2,7]. The elements in A that lie within this window are [5,4,3,6]. As they lie within the first window of B, I need to decrement 1, one time from each of these elements. The new A will look like the following after this operation:
A = [15 3 9 5 7 4 11 2 14];

Can this problem be reduced to a few function calls in Matlab or one should go through the naive loop business anyway? Thanks!

Answer 1

This can be done quite easily by using the histc function to determine in what bin (what you called “window”) the values are.

A = [15 4 9 6 7 5 11 3 14];
B = [2 7 13];
B = sort(B); 
[~, bin] = histc(A, B);
A = A - bin;

edit:
I noticed my solution is different from yours, but I suspect you made a mistake in your calculation. Do you have to subtract 2 from the values within the second bin or do you leave them as-is? If you only want to change the values in the first bin, the last line should read A(bin==1) = A(bin==1) - 1.

A = [15 4 9 6 7 5 11 3 14]; % initial value of A
A = [15 3 9 5 7 4 11 2 14]; % your reference result
A = [15 3 7 5 5 4  9 2 14]; % my result (as above)
A = [15 3 9 5 7 4 11 2 14]; % my result with A(bin==1) = A(bin==1) - 1

To change in what bin a value on the edge of a bin should end up, you can try to add/subtract eps from B.