Reshaping a 2 d matrix into a 3 d matrix row wise.

Hello, and thanks for taking the time to read this:
I have a 100 x 99 matrix of ratings. I would like to reshape the matrix into a 9x11x100 array where the first row represents the first 9 horizsontal items of the first row. Byu way of a smaller example, imagine the data were:
7 6 5 4 5 6
1 5 6 6 6 2
1 6 6 6 6 1
5 5 6 6 7 2
I would like the result to be:
765
456
156
662
556
672
Can anyone please help with this?

 Respuesta aceptada

the cyclist
the cyclist el 21 de Feb. de 2020
Editada: the cyclist el 21 de Feb. de 2020
For your small example, if A is your original matrix, then
permute(reshape(A',3,2,[]),[2 1 3])
For the larger one,
permute(reshape(A',9,11,[]),[2 1 3])

3 comentarios

Russ
Russ el 21 de Feb. de 2020
Great advice. The following alternative (which I would never have reached without your input) seemed to do the trick. I might haver described the problem incorrectly.
permute(reshape(A', 11,9,[]), [2 1 3])
Thanks!
Russ
the cyclist
the cyclist el 21 de Feb. de 2020
Yes, I see you effectively have these two conflicting statements in your question:
  • result is 9x11x100
  • first row is the first 9 items
These are conflicting because if the result is 9x11x100, then the first row has 11 items.
But it sounds like you got what you needed.
Russ
Russ el 21 de Feb. de 2020
Yep, now I understand where I went wrong. The first row was supposed to be the first 11 items, not the first 9 items. Thanks so much for solving this problem!

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Más respuestas (1)

Daniel Vieira
Daniel Vieira el 21 de Feb. de 2020
A=randi(10,100,99); % example of what your matrix might be
B=permute(reshape(A',9,11,[]),[2 1 3]) % solution

1 comentario

Russ
Russ el 21 de Feb. de 2020
Thanks! The following alternative seemed to do the trick.
permute(reshape(A', 11,9,[]), [2 1 3])
Thanks!
Russ

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