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

3 Ansichten (letzte 30 Tage)
Russ
Russ am 21 Feb. 2020
Kommentiert: Russ am 21 Feb. 2020
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?

Akzeptierte Antwort

the cyclist
the cyclist am 21 Feb. 2020
Bearbeitet: the cyclist am 21 Feb. 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 Kommentare
the cyclist
the cyclist am 21 Feb. 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 am 21 Feb. 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!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Daniel Vieira
Daniel Vieira am 21 Feb. 2020
A=randi(10,100,99); % example of what your matrix might be
B=permute(reshape(A',9,11,[]),[2 1 3]) % solution
  1 Kommentar
Russ
Russ am 21 Feb. 2020
Thanks! The following alternative seemed to do the trick.
permute(reshape(A', 11,9,[]), [2 1 3])
Thanks!
Russ

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by