Borrar filtros
Borrar filtros

How can I find the maximum element of each column of a cell array

1 visualización (últimos 30 días)
I have a cell array like this:
if true
ca={'x' 1 1 9 3;'y' 5 1 1 1; 'z' 5 7 5 9};
end
I have to find the max number of each column then divide all element of that column to maximum I wrote this code but it had errors:
if true
for j=2:3
max_column=max(ca(:,j));
for i=1:5
ca(i,j)= ca(i,j)/max_column;
end
end
disp(ca)
end
Please help me to edit this code. thanks in advance

Respuesta aceptada

Gopalkrishna Bhat
Gopalkrishna Bhat el 5 de Mayo de 2015
Hi Afsane,
The first column of char elements have to be removed & then the cell needs to converted to a matrix. Then you can run the code.
ba={'x' 1 1 9 3;'y' 5 1 1 1; 'z' 5 7 5 9};
temp = [ba(:,2),ba(:,3),ba(:,4),ba(:,5)];
ca = cell2mat(temp)
for j=1:4
max_column=max(ca(:,j));
for i=1:3
ca(i,j)= ca(i,j)/max_column;
end
end disp(ca)
  7 comentarios
Guillaume
Guillaume el 5 de Mayo de 2015
No, I mean:
ca(:, 2:end) = num2cell(ba);
for loops are often unnecessary in matlab. You can operate on the whole matrix / cell array at once.

Iniciar sesión para comentar.

Más respuestas (1)

Jos (10584)
Jos (10584) el 5 de Mayo de 2015
Life would be much easier if you didn't have your data arranges like this in the first place. How did you get this cell array? I strongly suggest you reconsider your approach.
  1 comentario
Stephen23
Stephen23 el 6 de Mayo de 2015
Editada: Stephen23 el 6 de Mayo de 2015
Yes!
Strangely it seems to be the people with the most awkward and bizarre nested data arrangements that insist that "this is the only way" to arrange it, and that "it has to be this way..."... usually followed up with some painful loops and a whole brace [sic] of braces, parentheses and brackets...
Sigh

Iniciar sesión para comentar.

Categorías

Más información sobre Creating and Concatenating Matrices en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by