How to sepeperate a set of data to a new matrix from two set of matrix

Let,

A= [1 2 3 4 6 7 8 9; 10 11 12 13 14 15 16 17; 17 18 19 20 21 22 23 24]

B=[9 1; 17 1; 24 2]

I want to create a new set of matrix (say, C and D) consisting all the row values from matrix A based on the condition specified for matrix B:

Let [Row, Column 1] of Matrix B = B (:,1) = has different values;

[Row, Column 2] of Matrix B = B (:,2) = 1 or 2;

Column 2 of Matrix B has the same elements as Column 9 of Matrix A.

if [Row, Column 2] of Matrix B = B (:,2) = 1, then copy the entire row value of Matrix A corresponding to the [Row, Column 1] of Matrix B to a new Matrix C and so on...

and if [Row, Column 2] of Matrix B = B (:,2) = 2, then copy the entire row value of Matrix A corresponding to the [Row, Column 1] of Matrix B to a new Matrix D and so on...

[Geoff Hayes]

Column 2 of Matrix B has the same elements as Column 9 of Matrix A.

From your above example, it would seem that Column 1 of B has the same elements as column 9 of A.

Ravi - you need to clarify the rules and provide examples at each step so that it is clear to you and us what you are trying to do.

[Ravi]

Thank you Geoff...

Yes it is, Column 1 of B has the same elements as column 8 of A.

If Row1Column2 of Matrix B = 1, and it's corresponding Row1Column 1 of Matrix B = 9

then find the Row1Column1 value of Matrix B, (i.e, 9) in Matrix A (i.e., Row1Column8 of Matrix A) and copy the entire corresponding row of matrix A (i.e., row corresponding to value of 9 in Matrix A) to a new matrix (m).

Answer 1

[Guillaume]

Here is a generic way of splitting the original matrix into submatrices according to your condition:

function C = splitmatrix(A, B, indexcol)
   %indexcol: a scalar integer >=1
   %A: a M x N matrix, with N >= indexcol. the elements of A(:, indexcol)) must be unique
   %B: a P x 2 matrix. Column 1 of B must match one value of A(:, indexcol). Column 2 of B must be integer >= 1
   %C: cell array output of size max(B(:, 2) x 1. 
   %C{k} is a matrix whose rows are the rows of A for which A(:, indexrow) is found in B(1, B(:, 2) == k)

   %validate above preconditions
   validateattributes(indexcol, {'numeric'}, {'scalar', 'integer', 'positive'});
   validateattributes(A, {'numeric'}, {'2d'});
   assert(size(A, 2) >= indexcol, 'indexcol is greater than the number of columns of A');
   assert(numel(unique(A(:, indexcol))) == size(A, 1), 'some elements of A(:, indexcol) are non unique');
   validateattributes(B, {'numeric'}, {'2d', 'ncols', 2});
   assert(all(ismember(A(:, indexcol), B(:, 1))), 'some elements of B(:, 1) are not found in A:, indexcol)');
   validateattributes(B(:, 2), {'numeric'}, {'integer', 'positive'}, '', 'second column of B');

   %find the rows of A that match first column of B:
   Arows = arrayfun(@(indexval) find(A(:, indexcol) == indexval), B(:, 1));
   %concatenate Arows according to the index in B(:, 2):
   rowsdestination = accumarray(B(:, 2), Arows, [], @(v) {v});
   %split A according to the destination:
   C = cellfun(@(rows) A(rows, :), rowsdestination, 'UniformOutput', false);
end

[Ravi]

Thank you Guillaume.

I am getting an error while running the code.

"Error using splitmatrix (line 9)

Not enough input arguments".

I would be grateful if you could suggest me on this....

[Guillaume]

As documented in the code comments, the function needs three input arguments:

A = [1 2 3 4 6 7 8 9; 10 11 12 13 14 15 16 17; 17 18 19 20 21 22 23 24];
B = [9 1; 17 1; 24 2];
C = splitmatrix(A, B, 8);
celldisp(C)

returns:

C{1} =
    1     2     3     4     6     7     8     9
   10    11    12    13    14    15    16    17

C{2} =
   17    18    19    20    21    22    23    24

[Ravi]

It worked great, thank you so much Guillaume. I have a slightly more difficult problem related to this. It will take some time for me to simplify the problem. I would be grateful to you if you could have a look on that as well.

[Ravi]

Dear Guillaume, I posted the problem which I requested you to have a look just now in the answer section of the current topic. I would be grateful if you could have look on that as well...

[Ravi]

Thanks Guillaume.....Posted as a separate question...I also removed the question asked in the answer section under current topic to avoid any confusion....

Answer 2

[Geoff Hayes]

Ravi - try the following

 A = [1 2 3 4 6 7 8 9; 10 11 12 13 14 15 16 17; 17 18 19 20 21 22 23 24];
 B = [9 1; 17 1; 24 2];
 C = [];

 % loop over each row of B
 for k=1:size(B,1)
     % if the second column of row k is one
     if B(k,2) == 1
         % get the row and column indices of A which match on B(k,1)
         [ridx,cidx] = find(A == B(k,1));
         if ~isempty(ridx)
             % copy the row(s) of A to C
             C = [C ; A(ridx,:)];
         end
     end
 end

The result of running the above is

 C =

     1     2     3     4     6     7     8     9
    17    18    19    20    21    22    23    24
    10    11    12    13    14    15    16    17

We get three rows since 17 (the first element of the second column of B) is found in rows 2 and 3 of A.

[Ravi]

Thank you for this. Is it possible to restrict the search to column 8 only as I would like to get my data sorted based on that. The final results should be something like:

for B(k,2)==1

C =

     1     2     3     4     6     7     8     9
    10    11    12    13    14    15    16    17

and for B(k,2)==2,

D=

17 18 19 20 21 22 23 24

Column-8_Row-3 of Matrix A = 24, which is Column-1_Row-3 of Matrix B and Column-2_Row-3 of Matrix B corresponds to above to a value 2.