how to convert column cell to row cell?

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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 Comment
jaweria kainat
jaweria kainat on 16 Jul 2018
suppose i have 2 columns.how i can convert them into rows?

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Accepted Answer

Jan
Jan on 26 Jan 2017
Edited: Jan on 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

More Answers (1)

Guillaume
Guillaume on 26 Jan 2017
cellfun(@transpose, yourcellarray, 'UniformOutput', false)
  6 Comments
Guillaume
Guillaume on 27 Jan 2017
Edited: Guillaume on 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 on 27 Jan 2017
Thank you for explaining the dimensions. I guess my matrix had more than two dimensions.

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