nested cell array into single cell array

A={<1x1 cell> <1x3 cell> 4 <1x4 cell> }

<1x1>= A{1, 1}{1, 1}{1, 1} <1x3 double> =[2 3 4]

<1x3>= A{1, 2}{1, 1} <0x0 double> = [ ]

A{1, 2}{1, 2} <1x2 double> = [ 3 4]

A{1, 2}{1, 3}{1, 1} <0x0 double> =[ ] A{1, 2}{1, 3}{1, 2} <1x3 double> =[3 8 13]

4= A{1, 3} <1x1 double> = [4]

<1x4>= A{1, 4}{1, 1} <1x2 double> = [9 4]

A{1, 7}{1, 2}{1, 1} <0x0 double> =[ ]

A{1, 7}{1, 2}{1, 2} <1x3 double> =[ 9 8 13 ]

A{1, 7}{1, 3} <0x0 cell> = [ ]

A{1, 7}{1, 4}{1, 1} <1x3 double> = [ 9 14 13 ] A{1, 7}{1, 4}{1, 2} <1x3 double> =[ 9 14 19 ]

and based on the output given above, i want my final answer to look like this:

B={ [2 3 4] [3 4; 3 8 13] [4] [9 4 ; 9 8 13 ; 9 14 13 ; 9 14 19] }

I hope my question is understandable

Thanks in advance

[Voss]

Try to create this matrix and see what happens:

[3 4; 3 8 13]

[Image Analyst]

Isn't this a duplicate of

https://www.mathworks.com/matlabcentral/answers/1617500-how-to-convert-array-of-cells-within-array-of-cells-in-single-cell-array?s_tid=srchtitle

[Mausmi Verma]

Thank you Benjamin for reply, and i got your point.

can i still get the desired output by the help of zero padding for matching the dimension

[Mausmi Verma]

@Image Analyst thank you for pointing out that but its exactly not the same because the pattern of the desired output is a bit different

Answer 1

[Voss]

As far as I can tell, the variable A has this structure (I'm going to assume those 7's in the question should actually be 4's):

A = { ...
    {{[2 3 4]}} ...
    {[] [3 4] {[] [3 8 13]}} ...
    4 ...
    {[9 4] {[] [9 8 13]} [] {[9 14 13] [9 14 19]}} ...
    };

To verify each entry (and leaving in redundant row index "1," throughout, for consistency with the notation in the question):

format('compact');
A{1,1}

ans = 1×1 cell array
    {1×1 cell}

A{1,1}{1,1}

ans = 1×1 cell array
    {[2 3 4]}

A{1,1}{1,1}{1,1}

ans = 1×3

2 3 4

A{1,2}

ans = 1×3 cell array
    {0×0 double}    {[3 4]}    {1×2 cell}

A{1,2}{1,1}

ans =
     []

A{1,2}{1,2}

ans = 1×2

3 4

A{1,2}{1,3}

ans = 1×2 cell array
    {0×0 double}    {[3 8 13]}

A{1,2}{1,3}{1,1}

ans =
     []

A{1,2}{1,3}{1,2}

ans = 1×3

3 8 13

A{1,3}

ans = 4

A{1,4}

ans = 1×4 cell array
    {[9 4]}    {1×2 cell}    {0×0 double}    {1×2 cell}

A{1,4}{1,1}

ans = 1×2

9 4

A{1,4}{1,2}

ans = 1×2 cell array
    {0×0 double}    {[9 8 13]}

A{1,4}{1,2}{1,1}

ans =
     []

A{1,4}{1,2}{1,2}

ans = 1×3

9 8 13

A{1,4}{1,3}

ans =
     []

A{1,4}{1,4}

ans = 1×2 cell array
    {[9 14 13]}    {[9 14 19]}

A{1,4}{1,4}{1,1}

ans = 1×3

9 14 13

A{1,4}{1,4}{1,2}

ans = 1×3

9 14 19

Below is a function you can use to do what you want. You can call it like this to get the B you want from the A you have:

B = cell(size(A));
for i = 1:numel(A)
    B{i} = build_matrix(A{i});
end
display(B)

B = 1×4 cell array
    {[2 3 4]}    {2×3 double}    {[4]}    {4×3 double}

B{:}

ans = 1×3

2 3 4

ans = 2×3

3 4 0 3 8 13

ans = 4

ans = 4×3

9 4 0 9 8 13 9 14 13 9 14 19

function M = build_matrix(C,M)
    if nargin < 2
        M = [];
    end
    if iscell(C)
        for i = 1:numel(C)
            M = build_matrix(C{i},M);
        end
    elseif ~isempty(C)
        if ~isempty(M)
            sC = size(C,2);
            sM = size(M,2);
            if sM < sC
                M(:,end+1:sC) = 0;
            elseif sM > sC
                C(:,end+1:sM) = 0;
            end
        end
        M = [M; C];
    end
end

% function M = build_matrix(C,M)
%     if nargin < 2
%         M = [];
%     end
%     if iscell(C)
%         for i = 1:numel(C)
%             M = build_matrix(C{i},M);
%         end
%     else
%         if ~isempty(M) && size(M,2) ~= size(C,2)
%             M(:,end+1:size(C,2)) = 0;
%         end
%         M = [M; C];
%     end
% end

[Mausmi Verma]

Thanks a lot for ur help

[Voss]

No problem. By the way, I edited my answer to make an update to the function, which is to handle the situation where a shorter numeric array comes later in the 'parent' cell array than a longer numeric array, e.g., if A was like this:

A = { ...
    {{[2 3 4]}} ...
    {[] [3 8 13] {[] [3 4]}} ...
    4 ...
    {[9 8 13] {[] [9 4]} [] {[9 14 13] [9 14 19]}} ...
    };

The old version worked for the original A you posted, but it would've had an error with this A. The new version handles this A and the original A properly. (The old version is still there commented-out, for reference.)

[Mausmi Verma]

Thanks benjamin for ur support,

after applying the way mentioned above in my work i am getting my answer as:

[1,2] 2 [3,2,0;3,8,13] [4,3,2] [5,10,15] [6,1,2;6,7,2] [7,2,0;7,8,13;7,12,13] [8,3,2;8,7,2;8,13,0] <4x3 double>

[10,15] [11,12,13] [12,7,2;12,13,0;12,17,22] 13 [14,13;14,15] 15 [16,17,22;16,21,22] [17,12,13;17,18,13;17,22,0] [18,13,0;18,17,22;18,23,22] <4x3 double>

[20,15] [21,22] 22 [23,18,13;23,22,0] [24,23,22] [25,20,15]

where, <4x3 double> =

9 8 13

9 10 15

9 14 13

9 14 15

now here i have stuck with my further work as i want my answer to be like,

in place of <4x3> i want answer to be seen as [9,8,13;9,10,15;9,14,13;9,14,15]

and same goes for other <4x3 double> as [19,14,13; 19,14,15;19,18,13;19,20,15]

so finally i want my answer to look like this,

[1,2] 2 [3,2,0;3,8,13] [4,3,2] [5,10,15] [6,1,2;6,7,2] [7,2,0;7,8,13;7,12,13] [8,3,2;8,7,2;8,13,0] [9,8,13;9,10,15;9,14,13;9,14,15] [10,15] [11,12,13] [12,7,2;12,13,0;12,17,22] 13 [14,13;14,15] 15 [16,17,22;16,21,22] [17,12,13;17,18,13;17,22,0] [18,13,0;18,17,22;18,23,22] [19,14,13; 19,14,15;19,18,13;19,20,15] [20,15] [21,22] 22 [23,18,13;23,22,0] [24,23,22] [25,20,15]

[Voss]

Looks like that's just how MATLAB prints it to the command line, not a difference between the answer you get and the answer you want.