Sorting and averaging matrices

1. I have two matrices. One for example is A = 520 x 1 and it has numbers from 1-8 which are not in order. For every number in this matrix, I have corresponding 1000 samples in another matrix B 520 x 1000. I want to first sort A and arrange numbers from 1-8 for every 8 rows. So output should be (1,2,3,4,5,6,7,8,1,2,3,4,5 and so on) and arrange corresponding values in B also such that they match.
2. Once this is done, I want to average every 5 rows of B and store it in a new matrix which will be 104 x 1000.

Thanks for the help in advance!

[KALYAN ACHARJYA]

What you have tried so far?

[HT]

I kind of dealt with the point 1. So i concatenated A and B and sorted rows according to numbers from 1-8. I am looking to see if there is a better method to do this and point 2 which is averaging every 5 rows.

[Guillaume]

100 is not a multiple of 8. Isn't that a problem? Why average every 5 rows when you have groups of size 8?

[HT]

It will be a multiple of 8 always. I just took 100 as a random number. I will edit the main question, thank you.

Answer 1

[Matt J]

 [~,idx]=sort(  reshape(A,8,[])  ,1);
 
    [m,n]=size(idx);

  C=reshape(B,m,n,[]);
  
     [m,n,p]=size(C);
  
  idx=idx+(0:n-1)*m+reshape(0:p-1,1,1,p)*(m*n);
  
  C=mean(reshape(C(idx),5,[]));
  
  result=reshape( C  ,m*n/5,[]);
 

[Guillaume]

This can be done without a loop:

[~, order] = sort(reshape(A, 8, []), 1);
C = mean(reshape(B, 5, []), 1);
ngroups = size(order, 2);
nsamples = size(B, 2);
result = reshape(C(order + (0:ngroups-1)*8 + permute((0:nsamples-1)*8*ngroups, [1, 3, 2])), [], nsamples);

[HT]

Thank you @Matt J and @Guillaume for your inuputs.

I tried sorting using @Matt J's method but it did not sort the elements.

Also from the question, A and B actually have same number of rows so A will be 520 x 1. After sorting A & B there should be still 520 rows and then after taking mean result should be 104 x 1000. @Guillaume your method gives the result as 520 x 1000.

[Matt J]

I've updated the answer to handle the new format of B.

[HT]

Thanks @Matt J. This seems to work and I am getting the correct size of the result matrix but the mean values of result are different from those I calculated manually for a few rows. Is this method calculating mean for rows 1-5,6-10,11-15 and so on? if you can comment the steps that might be helpful to debug.

[Matt J]

Here is another way of writing the code that breaks it into two parts. In the first part, we compute the sorted version of B.

 [~,idx]=sort(  reshape(A,8,[])  ,1);
 
     [m,n]=size(idx);
    
  C=reshape(B,m,n,[]);
  
     [m,n,p]=size(C);
  
  idx=idx+(0:n-1)*m+reshape(0:p-1,1,1,p)*(m*n);
  
  Bsorted=reshape(C(idx),size(B));
  

and then we compute the mean of very 5 elements of Bsorted, giving the final result

  result= mean(  reshape(Bsorted,5,[]) ,1);
  
  result=reshape(result,m*n/5,[]);

[HT]

Thanks, I figured out and now getting the correct values.

[HT]

How can I save sorted A values in a variable?

[Matt J]

 [Asorted,idx]=sort(  reshape(A,8,[])  ,1);