how to convert column cell to row cell?

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sumana
sumana 2017 年 1 月 26 日
コメント済み: jaweria kainat 2018 年 7 月 16 日
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 件のコメント
jaweria kainat
jaweria kainat 2018 年 7 月 16 日
suppose i have 2 columns.how i can convert them into rows?

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採用された回答

Jan
Jan 2017 年 1 月 26 日
編集済み: Jan 2017 年 1 月 27 日
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
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sumana
sumana 2017 年 1 月 27 日
Thank you, the loop worked very well for me.

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その他の回答 (1 件)

Guillaume
Guillaume 2017 年 1 月 26 日
cellfun(@transpose, yourcellarray, 'UniformOutput', false)
  6 件のコメント
Guillaume
Guillaume 2017 年 1 月 27 日
編集済み: Guillaume 2017 年 1 月 27 日
" 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 2017 年 1 月 27 日
Thank you for explaining the dimensions. I guess my matrix had more than two dimensions.

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