Matrix manipulation and matrix dimensions

Hello Matlab community again,

I am still trying to figure out few steps with getting from one dimension to another in matlab matrices.

So here is what I have now, Two matrices; A(73400 by 30), the number of rows here(73400) is equal to 14680 of stacked (5*1) matrices/arrays, hence 14680*5=73400.

The second matrix B(14680,5)

My objective is to compute the sum of the product of the transpose of each (5*1) that is stacked in A & each row from B . So the sum of the product should yield one element in each column (1 through 30).

The final matrix C is then (14680 by 30)

I tired to use two loops:

for i=1:30;
for j=1:14680
   C(i,j)= sum(A(j:n,i)'*B(j,:));
end 
end

It did not work. But in words, every iteration from 1 through 30, pick up a (5*1) from A, transpose it so it is same dimensions as rows from B, then compute sum of the product of each row from (1:14680).

Matrix B is a constant input, so it doesn't change. The only change is the values from each column in A ( from 1 through 30).

I hope I explained this correctly this time.

Thank you so much guys!

Amine

Answer 1

[David Goodmanson]

Hello Amine, This is only an answer if the following idea is correct: For a given 5x1 (call it y) and if B is mx5, I think you are looking for the number

sum{k = 1 to m} sum((y.').*B(k,:))   which is the same as   sum(B*y).

If not, you can ignore what's next. Anyway, the following seems to work in this case:

%  A = m*5 x q
%  B = m x 5

D = reshape(A,5,m*q)   % line up 5x1's side by side
E = B*D                % all possible dot products of a
                       % transposed 5x1 from A and a row of B
F = sum(E)             % sum down rows of B
C = reshape(F,m,q)     % place the elements

or in cryptic form

C = reshape(sum(B*reshape(A,5,m*q)),m,q)

[Amine Ben Ayara]

Good morning David, Thank you so much for taking the time to help me with this. So I used the second portion of your script, assuming that the results will be all in matrix form that I can understand better, This is What I did: My matrix A(73400 by 30) and the 73400 rows are the (5by1) stacked up or better yet, that I have 14680 sets of (5by1) (14680*5=73400) and the matrix B(14680 by 5).

-------------Based on your script: ----- I gussed m=14680 and q=30

D = reshape(A,5,14680*30);   % line up 5x1's side by side
E = B*D;               % all possible dot products of a
                     % transposed 5x1 from A and a row of B
F = sum(E) ;            % sum down rows of B
C = reshape(F,14680,5);     % place the elements

I got an error message when I executed D = reshape(A,5,14680*30 >> D = reshape(A,5,14680*30); Error using reshape To RESHAPE the number of elements must not change. I would greatly appreciate any feedback on this, please! Kind Regards, Amine

[David Goodmanson]

Hello Amine, yes m = 14680, q = 30. If matrix A is 2d of size 73400 x 30, the reshape command for D pretty much has to work. I see from your comment to Jan's answer that you have extracted A from a 4d matrix. Could you check the size of A just before you try the reshape command? Thanks.

Answer 2

[Jan]

A = rand(73400, 30);
B = rand(14680, 5);

AA = reshape(A, 5, 14680 * 30);
C  = reshape(B * AA, 14680, 14680, 30);
Result = sum(CC, 3);

Do you mean something like this? If not, please explain the wanted result again. What is the desired size? A small example might be useful also.

[Amine Ben Ayara]

Good morning Jan, thank you for the reply, I really appreciate. So here is an example. The 2D matrix is A(73400 by 30) (total rows 73400=14680*5) [ 14680 of 5*1 arrays/matrices stacked horizontally, 73400=14680*5. The next matrix say; B(14680 by 5) ( where each row has a unique identifier, hence U need the operation to be exclusive between each row of B (i,:) with i=1:14680, and the transpose of each column(5*1) from A, which with transpose it becomes a row identical to that of B. First iteration suppose now, J=1:30 ( the 30 columns,2nd dimension from A), Every iteration j: Compute the sum of the product from transpose of the first column in A; the dimension of each array is then (1*5) and (1*5) so the result is only one element. then move down to the following row in B, and the next column in A (all the way through 14680, hence j=1:14680 and the result matrix will have the following dimension (14680 by 30) e.g: first 5 rows, in 1 st column from A is A11=A(1:,5,1)=[1;2;3;4;5]; First row from B is B1=B(1,:)=[1 1 1 1 1 1]; SumProduct= Sum(A11'.*B1)=15 ( this element will go in the final matrix in C(1,1), the next iteration will be for the following row in B(i=2) and the second set of 5 rows in A (6:10,1) ... until all the rows in the first column of A are done, then repeat the same process for the next column, hence (j=1:30, with j number of columns in A). I hope this helps? Sorry if It was long.

[Amine Ben Ayara]

Hello Jan, I just tried to run the script you provided and I got this error message: Error using * Requested 14680x440400 (48.2GB) array exceeds maximum array size preference. Creation of arrays greater than this limit may take a long time and cause MATLAB to become unresponsive. See array size limit or preference panel for more information.

[Amine Ben Ayara]

Here is an Example. My first matrix A (73400 by 30) ( I must mention that it was actually ( 5 by 5 by 14680 by 30) so 4D , but I converted it to 2D because I only need the first column (5*1) from each (5*5) matrix (14680 of them) in each set (30), so now and I stacked the (5*1) columns so the total rows became =14680*5=73400 and total number of columns is 30, hence 2D (73400*30). Second matrix is B(14680,5) My objective is to do the sum of the product of the transpose of each (5*1) from matrix A to each row from matrix B. The final product/Matrix should have this dimension (14680*30)

[Jan]

@Amine: Your descriptions are hard to read. I cannot follow you. Can you provide real values? Perhaps a 2x3 input instead of 73400x30 would clarify already, what you want to achieve. The error message shows, that you need a lot of RAM to claculate such large inputs efficiently. I do not know, if your machine has 120GB or only 4GB.