ind2sub with arbitrary martix

9 vues (au cours des 30 derniers jours)
Thomas
Thomas le 16 Fév 2023
Commenté : Lucas Graham le 7 Juin 2023
Hi
I intend to use ind2sub to get the subscripts indexing from linear indexing and I will use if to read data in matrix of arbitrary size.
The problem with this is that you have to know the dimension of the matrix, so that you can call ind2sub with the corresponding number of outputs. What I would like is to get a output vector that has the corresponding number of elements. Is this possible to achieve?
Best regards
Thomas

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Stephen23
Stephen23 le 16 Fév 2023
Modifié(e) : Stephen23 le 16 Fév 2023
"In the example below: how to set the three output arguments [I1 I2 I3] "automatically" ?"
You use a comma-separated list:
A = rand(3,6,34);
X = 24;
C = cell(1,ndims(A));
[C{:}] = ind2sub(size(A),X);
Checking:
C{:}
ans = 3
ans = 2
ans = 2
  4 commentaires
Stephen23
Stephen23 le 16 Fév 2023
Modifié(e) : Stephen23 le 16 Fév 2023
"Strange that the output argument from int2sub is not automatically of size 1 x ndims(A) "
X can be any size, the outputs have exactly the same size as X. The proposed output size would conflict with that.
If you restrict X to scalar only, then the function is very easy to copy and modify.
Lucas Graham
Lucas Graham le 7 Juin 2023
I was about to try to gimmick some sort of iteratively generated string and then use eval(), this is so much better. Thanks.

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Plus de réponses (3)

Torsten
Torsten le 16 Fév 2023
You mean
A = rand(5,6);
s = size(A)
s = 1×2
5 6
?
  2 commentaires
Thomas
Thomas le 16 Fév 2023
[i1,i2,i3]=ind2sub([3,3,3],8)
i1 = 2
i2 = 3
i3 = 1
[i1,i2,i3,i4]=ind2sub([3,3,3,4],8)
i1 = 2
i2 = 3
i3 = 1
i4 = 1
But I would like to have a function my_ind2sub so that
I=my_ind2sub([3,3,3],8)
->I=[2,3,1]
and
I=my_ind2sub([3,3,3,3],8)
->I=[2,3,1,1]
Torsten
Torsten le 16 Fév 2023
I don't know how you automativcally get I according to the matrix size. I asked this question below. I think @Steven Lord will be able to answer this.

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dpb
dpb le 16 Fév 2023
Déplacé(e) : dpb le 16 Fév 2023
Use size(M) as the size argument to ind2sub where M is the matrix...the size argument doesn't have to be a constant expression.

Steven Lord
Steven Lord le 16 Fév 2023
You have to specify the size of the array you want to use the subscripts to index into because the same linear index could generate different subscripts based on the size. Consider two examples where I want the subscripts for element 4.
A1 = reshape(1:8, 2, 4)
A1 = 2×4
1 3 5 7 2 4 6 8
A2 = reshape(1:8, 4, 2)
A2 = 4×2
1 5 2 6 3 7 4 8
A1 and A2 have the same number of elements but different sizes. This means that element 4 of A1 is element (2, 2) and element 4 of A2 is element (4, 1).
[r1, c1] = ind2sub(size(A1), 4)
r1 = 2
c1 = 2
[r2, c2] = ind2sub(size(A2), 4)
r2 = 4
c2 = 1
If I just called ind2sub with the input 4, should the outputs be 2 and 2, 4 and 1, or something else?
But let's take a step back. When you say this:
What I would like is to get a output vector that has the corresponding number of elements.
do you just want to use linear indexing? How are you hoping or planning to use these indices with a "matrix of arbitrary size"?
B1 = A1.^2 % just so the elements aren't 1:8
B1 = 2×4
1 9 25 49 4 16 36 64
B2 = A2.^2
B2 = 4×2
1 25 4 36 9 49 16 64
[B1(4) B2(4)] % Both are 16
ans = 1×2
16 16
  1 commentaire
Torsten
Torsten le 16 Fév 2023
Say I have a matrix A I don't know the size of in advance.
How can I prescribe that the array of outputs I1,I2,...In from ind2sub as
matches the number of dimensions of A ?
In the example below: how to set the three output arguments [I1 I2 I3] "automatically" ?
A = rand(3,6,34);
s = size(A);
s = 1×3
3 6 34
[I1 I2 I3] = ind2sub(s,24)
I1 = 3
I2 = 2
I3 = 2

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