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?

 Réponse acceptée

the cyclist
the cyclist le 21 Fév 2020
Modifié(e) : the cyclist le 21 Fév 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 commentaires

Russ
Russ le 21 Fév 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
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 le 21 Fév 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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Plus de réponses (1)

A=randi(10,100,99); % example of what your matrix might be
B=permute(reshape(A',9,11,[]),[2 1 3]) % solution

1 commentaire

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

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