Borrar filtros
Borrar filtros

Create a cell array from matrices using for loop

2 visualizaciones (últimos 30 días)
MarshallSc
MarshallSc el 23 de Jun. de 2021
Comentada: MarshallSc el 16 de Jul. de 2021
Hello, I'm looking to derive a 10*10 cell array that each cell contains a 4*4 matrix within each cell. For example, by having four 10*10 matrices, I want to derive a 4*4 matrix for each cell:
x1=rand(10,10);
y1=rand(10,10);
z1=rand(10,10);
r1=rand(10,10);
x2=rand(10,10);
y2=rand(10,10);
z2=rand(10,10);
r2=rand(10,10);
I want to create a cell like:
L=cell(10,10);
In which I want to insert the sqrt of the ratio of the 4 matrices in each cell like:
L{1,1}=sqrt([x1(1,1)/x2(1,1), (x1(1,1)/y2(1,1)), x1(1,1)/z2(1,1), x1(1,1)/r2(1,1);...
y1(1,1)/x2(1,1), y1(1,1)/y2(1,1), y1(1,1)/z2(1,1), y1(1,1)/r2(1,1);...
z1(1,1)/x2(1,1), z1(1,1)/y2(1,1), z1(1,1)/z2(1,1), z1(1,1)/r2(1,1);...
r1(1,1)/x2(1,1), r1(1,1)/y2(1,1), r1(1,1)/z2(1,1), r1(1,1)/r2(1,1)])
So for each cell I need to calculate the cell arrays (L{1,2}, L{1,3} . . . L{10,10})
I want to use the for loop that can do the same procedure for all the cell array that each contain a 4*4 matrix that at the end I have 10*10 cell that each cell contain a 4*4 matrix. But my for loop doesn't give me the answer. Can someone please help me with writing the for loop? Thank you.

Respuesta aceptada

Stephen23
Stephen23 el 23 de Jun. de 2021
x1 = rand(10,10);
y1 = rand(10,10);
z1 = rand(10,10);
r1 = rand(10,10);
a1 = cat(3,x1,y1,z1,r1);
x2 = rand(10,10);
y2 = rand(10,10);
z2 = rand(10,10);
r2 = rand(10,10);
a2 = cat(4,x2,y2,z2,r2);
tmp = num2cell(sqrt(a1./a2),3:4);
fun = @(a)reshape(a,4,4);
out = cellfun(fun,tmp,'uni',0)
out = 10×10 cell array
{4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double} {4×4 double}
Compare:
out{1}
ans = 4×4
0.9472 2.1173 1.7208 0.7524 0.2632 0.5883 0.4782 0.2091 0.6565 1.4676 1.1927 0.5215 0.3798 0.8490 0.6900 0.3017
sqrt([x1(1,1)/x2(1,1), (x1(1,1)/y2(1,1)), x1(1,1)/z2(1,1), x1(1,1)/r2(1,1);...
y1(1,1)/x2(1,1), y1(1,1)/y2(1,1), y1(1,1)/z2(1,1), y1(1,1)/r2(1,1);...
z1(1,1)/x2(1,1), z1(1,1)/y2(1,1), z1(1,1)/z2(1,1), z1(1,1)/r2(1,1);...
r1(1,1)/x2(1,1), r1(1,1)/y2(1,1), r1(1,1)/z2(1,1), r1(1,1)/r2(1,1)])
ans = 4×4
0.9472 2.1173 1.7208 0.7524 0.2632 0.5883 0.4782 0.2091 0.6565 1.4676 1.1927 0.5215 0.3798 0.8490 0.6900 0.3017
  10 comentarios
MarshallSc
MarshallSc el 15 de Jul. de 2021
So sorry Stephen, I just realized what a silly mistake I made! I figured it out! Thank you brother!
MarshallSc
MarshallSc el 16 de Jul. de 2021
Stephen, I don't know if it's allowed to say it here, but I just wanted to thank you for all of your kind and invaluable helps with the codes. You personally hepled me out a lot so I wanted to show you my deepes gratitude; your kindness and the time you put into helping others don't go unnoticed man! Thank you!

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Matrix Indexing 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