split matrix in diffrent sizes matrixes according to indicies vector

I have a matrix A (7x2) and a vector B. B represents the row indicies, where i want the matrix A to be sprlit.

e.g.

A = [2 6; 4 3; 3 3; 2 8;  9 1;  3 4;  7 5] B=[2; 3; 6]

output should be

A1 = [2 6; 4 3]
A2 = [3 3]
A3 = [2 8; 9 1; 3 4]
A4 = [7 5]

Do you have an idea how it could be done?

Thank you for help.

Answer 1

[Stephen23]

Do NOT create numbered variables. Indexing is simpler and much more efficient.

You can use mat2cell to split the matrix up into a cell array:

>> R = diff([0;B;size(A,1)]);
>> C = mat2cell(A,R,2);
>> C{:}
ans =
     2     6
     4     3
ans =
     3     3
ans =
     2     8
     9     1
     3     4
ans =
     7     5

and then you can trivially access the contents of that cell array using indexing:

https://www.mathworks.com/help/matlab/matlab_prog/access-data-in-a-cell-array.html

E.g.:

>> C{2}
ans =
     3     3
>> C{3}
ans =
     2     8
     9     1
     3     4   

[christine schaer]

Thanks for the answer. I use now the cell array but i have a bit trouble working with it:

I want to add a third row to each cell where the cumulative values of the second row are in it. My idea is smth like this (but i dont know how i use A(:,3) correclty in cell arays.. ):

A(:,3)=cellfun(cumsum,A(:,2));

[Stephen23]

"I want to add a third row to each cell ..."

This is confusing, because in your example you are adding a third column, not a row. Remember that matrices are defined by rows*columns, and this is how MATLAB indexing works too: (rows,columns,pages,...)

https://www.mathworks.com/help/matlab/math/array-indexing.html

In any case, you can simply use a loop, e.g.:

for k = 1:numel(C)
    C{k}(:,3) = cumsum(C{k}(:,2));
end

Giving:

>> C{:}
ans =
     2     6     6
     4     3     9
ans =
     3     3     3
ans =
     2     8     8
     9     1     9
     3     4    13
ans =
     7     5     5

Or, if you really want to use cellfun, something like this:

>> F = @(m)[m,cumsum(m(:,2))];
>> C = cellfun(F,C,'uni',0);
>> C{:}
ans =
     2     6     6
     4     3     9
ans =
     3     3     3
ans =
     2     8     8
     9     1     9
     3     4    13
ans =
     7     5     5

[christine schaer]

thank you.. i meant columne sorry

Answer 2

[the cyclist]

Stephen's solution is better, but I thought I would post this for loop version, which might be a bit clearer to you on what is happening:

A = [2 6;
     4 3;
     3 3;
     2 8;
     9 1;
     3 4;
     7 5];
 
B=[2; 3; 6];

C = [0; B; size(A,1)];

for nj = 1:numel(C)-1
    An{nj} = A(C(nj)+1:C(nj+1),:);
end

I would also advocate using cell arrays rather than dynamically named variables.