Average Multiple Matrix Columns Based On Duplicate Entries In A Different ColumnVector

I have a column vector consisting of hundreds of unit64 values where duplicates appear and are always grouped together but groups appear only once. Each element of this vector corresponds to a row in a separate 2D matrix of double values. I would like to remove the duplicates from the row vector and then remove the corresponding rows in the matrix, replacing them with a single row where each column is average the removed elements in the column, while keeping the order stable. So with data like this:

      1
      1
      2 
 A=   2
      3
      3

      1 2 3
      2 3 4
      3 4 5
  B = 4 5 6
      5 6 7
      6 7 8

I would want the output to be

      1.5 2.5 3.5
  C = 3.5 4.5 5.5
      5.5 6.5 7.5

Some combination of unique and accumarray seems to be in order but I cannot figure out how to handle the multiple coulmn part of the problem.

Answer 1

[Guillaume]

So if I understood correctly, the shape of B does not matter and it is to be considered a column vector.

[~, ~, subs] = unique(A, 'stable');
C = accumarray(subs, B(:), [], @mean)

Note that whether or not the groups appear only once does not matter for the above. All the values with the same corresponding A still get averaged.

Also note that if A is already integer values from 1 to n with no gap and in the right order, then the call to unique is unnecessary and you can just pass A instead of subs.

[Jim Sculley]

No, the shape of B does matter. I've edited the example slightly to add a third row to B. With a third row, your solution gives an error:

"Error using accumarray Second input VAL must be a vector with one element for each row in SUBS, or a scalar."

A and B will always have the same number of rows. B will have more than one column. In any column of B, the values in the rows corresponding to the row indices of repeated values in A should be averaged and the end result should consist of a A with the duplicate rows removed and B with the rows corresponding to the duplicates in A replaced by a single row with the average.

[Guillaume]

I'm sorry I don't get it at all.

A and B will always have the same number of rows.

In your example, A has 6 rows, and B has 3.

the end result should consist of a A with the duplicate rows

That's easy to achieve:

unique(A, 'stable')

However, none of these values appear in your desired output.

and B with the rows corresponding to the duplicates in A replaced by a single row with the average

I don't understand that. I think you need to describe with an example all the steps leading to each value of C. What is C(1) the average of?

[Jim Sculley]

Yep. My first edit made it more confusing. I've made one more edit with the correct data. For the given data, the results should be each column of rows 1 and 2 in B are averaged and the result is row 1 in C. Columns of rows 3 and 4 in B are averaged and the result is row 2 in C and so on. Which rows are averaged are driven by the values in A.

[Guillaume]

Ok, got it now. This can be done awkwardly with accumarray but in this case, the newer splitapply is simpler. While we're at it we can use findgroups instead of unique:

g = findgroups(A);
C = splitapply(@mean, B, g)

[Jim Sculley]

This works great for my sample data above, but when I try it with my real data I am getting an error:

 Error using vertcat
 Dimensions of arrays being concatenated are not consistent.

 Error in splitapply>localapply (line 257)
       finalOut{curVar} = vertcat(funOut{:,curVar});

 Error in splitapply (line 132)
 varargout = localapply(fun,splitData,gdim,nargout);

My A vector is 1335 x 1 (unit64)

My B matrix is 1335 x 632 (double)

findgroups returns a 1335 x 1 (double) with indices from 1 to 516.

I'm not sure why splitapply would be having a problem with this.

[Guillaume]

C = splitapply(@(m) mean(m, 1), B, g);

should fix the problem. If a group only contain one row, the mean will be taken along the columns of the one row instead of across the rows as it happens for matrices. As a result, you'd get a scalar value instead of a row, hence the Dimensions of arrays being concatenated are not consistent. @(m) mean(m, 1) forces the mean to be taken across the rows regardless.