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ind2sub with arbitrary martix

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Thomas
Thomas 2023-2-16
评论： Lucas Graham 2023-6-7
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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采纳的回答

Stephen23
Stephen23 2023-2-16
编辑：Stephen23 2023-2-16
Ran in:
"In the example below: how to set the three output arguments [I1 I2 I3] "automatically" ?"
You use a comma-separated list:
https://www.mathworks.com/matlabcentral/answers/1656435-tutorial-comma-separated-lists-and-how-to-use-them
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 个评论
显示 2更早的评论
Stephen23
Stephen23 2023-2-16
编辑：Stephen23 2023-2-16
"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 2023-6-7
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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更多回答（3 个）

Torsten
Torsten 2023-2-16
Ran in:
You mean
A = rand(5,6);
s = size(A)
s = 1×2
5 6
?
  2 个评论
Thomas
Thomas 2023-2-16
Ran in:
[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 2023-2-16
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 2023-2-16
移动：dpb 2023-2-16
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 2023-2-16
Ran in:
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 个评论
Torsten
Torsten 2023-2-16
Ran in:
@Steven Lord
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
[I1,I2,...,In] = ind2sub(sz,ind)
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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