how to convert column cell to row cell?

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sumana
sumana am 26 Jan. 2017
Kommentiert: jaweria kainat am 16 Jul. 2018
Hi,
For example I have a 3x1 cell matrix like this. {[1,2,3] [4,5,6] [7,8,9]}; where every element is a 1x3 matrix.
I want to convert the row cells to 3x1 column cells like this {[1;2;3] [4;5;6] [7;8;9]};
How do I do this ? Thank you.
  1 Kommentar
jaweria kainat
jaweria kainat am 16 Jul. 2018
suppose i have 2 columns.how i can convert them into rows?

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Jan
Jan am 26 Jan. 2017
Bearbeitet: Jan am 27 Jan. 2017
Or a loop:
for k = 1:numel(C)
C{k} = C{k}.';
end
[EDITED] Accroding to your comment:
M = [1 2 3; 4 5 6; 7 8 9];
C = mat2cell(M.', 3, [1,1,1]).';
{[1;2;3];
[4;5;6];
[7;8;9]}
Or again with a simple loop:
C = cell(3, 1);
for k = 1:3
C{k} = M(k, :).';
end
  1 Kommentar
sumana
sumana am 27 Jan. 2017
Thank you, the loop worked very well for me.

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Guillaume
Guillaume am 26 Jan. 2017
cellfun(@transpose, yourcellarray, 'UniformOutput', false)
  6 Kommentare
Guillaume
Guillaume am 27 Jan. 2017
Bearbeitet: Guillaume am 27 Jan. 2017
" have a 3x1 cell matrix like this. {[1,2,3] [4,5,6] [7,8,9]}" and "I ran that, and it returned all of the elements". One of these two statements directly contradict the other:
>>c = {[1,2,3] [4,5,6] [7,8,9]};
>>find(~cellfun(@ismatrix, c))
ans =
1x0 empty double row vector
If the arrays in your cell arrays have more than two dimensions, there's no way to transpose them since transposition is only defined for 2D matrices. If what you want to do is just swap 1st and 2nd dimension, leaving the others untouced:
cellfun(@(m) permute(m, [2 1 3:ndims(m)]), yourcellarray, 'UniformOutput', false)
"Can you make a normal matrix [...] to a cell matrix" There's no such thing as a cell matrix. To produce the cell array in your example:
m = [1 2 3; 4 5 6; 7 8 9];
c = num2cell(m.', 2).'
sumana
sumana am 27 Jan. 2017
Thank you for explaining the dimensions. I guess my matrix had more than two dimensions.

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