how to convert column cell to row cell?
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sumana
el 26 de En. de 2017
Comentada: jaweria kainat
el 16 de Jul. de 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.
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Respuesta aceptada
Jan
el 26 de En. de 2017
Editada: Jan
el 27 de En. de 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
Más respuestas (1)
Guillaume
el 26 de En. de 2017
cellfun(@transpose, yourcellarray, 'UniformOutput', false)
6 comentarios
Guillaume
el 27 de En. de 2017
Editada: Guillaume
el 27 de En. de 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).'
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