assign the same vector to be the same cell
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ha ha
el 21 de Nov. de 2018
Comentada: ha ha
el 24 de Nov. de 2018
Let's say, I have the matrix:
A=[x,y]=[1 2;1.1 2;1.2 2;1 3;1.1 3;1.2 3;1 4;1.1 4;1.2 4];
If i wanna group all vector having the same value of y coordinate. How can I do that?
Example, the result like that:
Cell1=[1 2
1.1 2
1.2 2]
Cell2=[1 3
1.1 3
1.2 3]
Cell3=[1 4
1.1 4
1.2 4]
2 comentarios
madhan ravi
el 23 de Nov. de 2018
People here put some efforts to help you and you mercilessly close the question without clarifying how rude
Respuesta aceptada
madhan ravi
el 21 de Nov. de 2018
Editada: madhan ravi
el 21 de Nov. de 2018
A=[1 2;1.1 2;1.2 2;1 3;1.1 3;1.2 3;1 4;1.1 4;1.2 4]';
group=reshape(A,2,3,3);
CELL=cell(1,3);
for i = 1:size(group,3)
CELL{i}=group(:,:,i)';
end
celldisp(CELL)
command window:
>>
CELL{1} =
1.0000 2.0000
1.1000 2.0000
1.2000 2.0000
CELL{2} =
1.0000 3.0000
1.1000 3.0000
1.2000 3.0000
CELL{3} =
1.0000 4.0000
1.1000 4.0000
1.2000 4.0000
>>
Más respuestas (2)
Andrei Bobrov
el 21 de Nov. de 2018
Cell = mat2cell(A,accumarray(findgroups(A(:,2)),1),size(A,2));
0 comentarios
Guillaume
el 21 de Nov. de 2018
Editada: Guillaume
el 21 de Nov. de 2018
Can be done easily with findgroups (or the older unique) and splitapply (or the older accumarray), in just one line:
A = [1 2;1.1 2;1.2 2;1 3;1.1 3;1.2 3;1 4;1.1 4;1.2 4];
C = splitapply(@(rows) {A(rows, :)}, (1:size(A, 1))', findgroups(A(:, 2)));
celldisp(C)
with unique and accumarray, you need two lines as you need the 3rd return value of unique:
A = [1 2;1.1 2;1.2 2;1 3;1.1 3;1.2 3;1 4;1.1 4;1.2 4];
[~, ~, id] = unique(A(:, 2));
C = accumarray(id, (1:size(A, 1))', [], @(rows) {A(rows, :)});
celldisp(C)
0 comentarios
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