how to sum cell elements in a dynamic cell table

Hi guys, I have a challenging problem here :

A{b,i} is a cell table with b lines and i columns. the problem is that b can be any integer (dynamic variable) while i is a fixed number. The problem is the following : how to sum elements by varying the column index of the first row while maintaining column indexes of other rows constant, and again fix column index of all the rows except second row and so on. Note that in case of try to use for loops to vary column indexes that we will need b for loops to be generated dynamically which is irrelevant. You will find more explanation in the following

for b = 1:L

for i = 1:16

A{b,i} = ones(1,8);

end

end

Question : how to sum table elements columnwise according to how many rows we are having. Example : in case of two rows we will have 16x16 sum values, in case of three rows we will have 16x16x16 sum values, etc...

[Jan]

The term "cell table" is confusing. "Cell" and "table" are specific types in Matlanb, but I do not know, what a "cell table" is. Your data look like a cell matrix, which contain vectors as elements.

This is not clear to me also: "how to sum elements by varying the column index of the first row while maintaining column indexes of other rows constant, and again fix column index of all the rows except second row and so on." A small example would be useful. Do you want to combine all elements of one vector with the elements of other vectors? If so, what exactly do you want to sum up?

[Akram RAYRI]

A =

3×8 cell array

{[1]} {[2]} {[3]} {[ 4]} {[ 5]} {[ 6]} {[ 7]} {[ 8]}

{[2]} {[4]} {[6]} {[ 8]} {[10]} {[12]} {[14]} {[16]}

{[3]} {[6]} {[9]} {[12]} {[15]} {[18]} {[21]} {[24]}

let's consider the following cell matrix, and taking into consideration suuming up in column wise, I want to have all possible sum value which will be 8x8x8 values.

Example : I want to sum elements of first colum while column indices of second and third rows are constant and while column index of first row is varying, 1+2+3, 2+2+3, 3+2+3,...8+2+3. then incrementing the column index of the second row by one and perform, 1+4+3, 2+4+3,...8+4+3 and then 1+6+3, 2+6+3,...8+6+3 until we will reach 8+16+3 then we will increment column index of third row by one and perform 1+2+6, 2+2+6...8+2+6 then 1+4+6, 2+4+6,...8+4+6....until we will reach 8+16+24. So basically I want to have all possibilities of summing values column wise.

[Jan]

In the question you show a cell matrix containing vectors. Now the cell matrix contains scalars. Do you want to get the combinations between the elements of the vectors, of of the rows of the cel matrix?

[Akram RAYRI]

I want to get the combination of the rows of the cell matrix regardless what do we have in the cell (scalars or vectors doesn't matter). In this example I give a cell matrix with 3x8 as a dimension.

[Image Analyst]

Can you just stop making us guess? And making us have to make up sample data which may or may not be the form you have? Or having to make some MATLAB code to generate the printout of A you gave above?

Simply attach your cell array in a .mat file with the paperclip icon.

save('answers.mat', 'A');

after you read this link:

TUTORIAL: How to ask a question (on Answers) and get a fast answer

Answer 1

[Jan]

Maybe:

A   = num2cell([1:6; 2:2:12; 3:3:18]);  % Some test input

S   = size(A);         % Size of the input
ini = ones(1, S(1));   % Start of virtual FOR loops
fin = ini * S(2);      % Stop of virtual FOR loops

n = numel(ini);
v = ini;               % Current index vector
Result = cell(repmat(S(2), 1, S(1)));
for k = 1:numel(Result)
   tmp = 0;             % [EDITED] Sum over elements from rows of A
   for r = 1:numel(v)
       tmp = tmp + A{r, v(r)};  % [EDITED]
   end
   Result{k} = tmp;
   
   for iv = 1:n          % Increase index vector:
      if v(iv) < fin(iv)
        v(iv) = v(iv) + 1;
        break;           % Stop "for iv" loop
      end
      v(iv) = ini(iv);   % Reset this index
   end
end

[Akram RAYRI]

By taking the following cell matrix

A =

2×3 cell array

{[1]} {[2]} {[3]}

{[2]} {[4]} {[6]}

I sould have as results the following :

3 5 7

4 6 8

5 7 9

this is : 1+2 1+4 1+6

2+2 2+4 2+6

3+2 3+4 3+6

Instead in the algorithm that you provided, if will take the same input matrix A = num2cell([1:3 ; 2:2:6]) I will get as results the following matrix :

Result =

3×3 cell array

{[2]} {[3]} {[3]}

{[3]} {[4]} {[4]}

{[3]} {[4]} {[4]}

[Jan]

@Akram RAYRI: I forgot the 1st index:

tmp = tmp + A{r, v(r)}
%             ^   inserted in my answer now
% and starting with tmp=0, because it looks simpler

The method works also, if the cell elements are vectors.

[Peter Perkins]

I have not read this whole thread, but:

Cell arrays are NOT a good way to store scalar numeric values. You are just making your life harder. If there's a reason you want to use a cell array, you should make that clear. Otherwise, don't do that.

Answer 2

[Voss]

Is this similar to what you want to do? (Using 5 columns here instead of 16, and using 3 rows.)

data = (1:5).*[1;7;20]

data = 3×5

1 2 3 4 5 7 14 21 28 35 20 40 60 80 100

[L,M] = size(data);
result = repmat({0},1,L);
c_data = num2cell(data.',1);
siz = ones(1,L);
for ii = 1:L
    for jj = 1:ii
        siz(jj) = M;
        result{ii} = result{ii}+reshape(c_data{jj},siz);
        siz(jj) = 1;
    end
end
disp(result)

    {5×1 double}    {5×5 double}    {5×5×5 double}

disp(result{1})

     1
     2
     3
     4
     5

disp(result{2})

     8    15    22    29    36
     9    16    23    30    37
    10    17    24    31    38
    11    18    25    32    39
    12    19    26    33    40

disp(result{3})

(:,:,1) =

    28    35    42    49    56
    29    36    43    50    57
    30    37    44    51    58
    31    38    45    52    59
    32    39    46    53    60


(:,:,2) =

    48    55    62    69    76
    49    56    63    70    77
    50    57    64    71    78
    51    58    65    72    79
    52    59    66    73    80


(:,:,3) =

    68    75    82    89    96
    69    76    83    90    97
    70    77    84    91    98
    71    78    85    92    99
    72    79    86    93   100


(:,:,4) =

    88    95   102   109   116
    89    96   103   110   117
    90    97   104   111   118
    91    98   105   112   119
    92    99   106   113   120


(:,:,5) =

   108   115   122   129   136
   109   116   123   130   137
   110   117   124   131   138
   111   118   125   132   139
   112   119   126   133   140

[Akram RAYRI]

disp(result{3}) (:,:,1) contains the combinations between row 1 and row2 while the first element in row3 is fixed. while in (:,:,2) the results are not completed. As mentioned before, we should have 5x5x5 combinations, as 5 is the number of columns.

[Voss]

Scroll down? It's 5x5x5.

[Akram RAYRI]

Thank you so much dear voss, your solution is working for matrix with integers, now what if we will have a cell matrix, and in each cell we have a vector instead of an integer, and in the end we want as a result a cell matrix that contains all possible sum combinations but this time the sum will be a vector instead of integer.For instance :

A =

3×3 cell array

{3×1 double} {3×1 double} {3×1 double}

{3×1 double} {3×1 double} {3×1 double}

{3×1 double} {3×1 double} {3×1 double}

> A{1,1}

ans =

1

1

1

>> A{2,1}

ans =

1

2

2

>> A{3,1}

ans =

1

3

3

[Akram RAYRI]

@Voss did you get the question ?