Transposing matrix using reshape
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Hello,
I am a student taking a class to learn matlab. For a project, our instructor is requiring us to transpose a function using the reshape command. Because of the way matlab reads matrixes, column-dominant, this is proving very difficult.
In essence, I need to perform the following using only reshape:
A = A'
or
A = [1,2,3;4,5,6] must become A = [1,4;2,5;3,6]
Thanks for your help.
10 个评论
the cyclist
2012-3-21
Are you able to post the full problem statement from your instructor?
Adam
2012-3-21
Adam
2012-3-21
the cyclist
2012-3-21
No, I just thought that perhaps there was something subtle in the problem student that you overlooked.
Geoff
2012-3-21
Well, are you allowed to use ANY other functions like size(), repmat() and colon()? =) What exactly is the restriction?
Adam
2012-3-21
Adam
2012-3-21
Geoff
2012-3-21
And no looping? She wants a one-liner?
Geoff
2012-3-21
Seems to me like your instructor is teaching you to hate MatLab.
Walter Roberson
2012-3-21
Is array indexing allowed? Are multiplication and addition allowed?
回答（5 个）
Walter Roberson
2012-3-21
1 个投票
reshape() by itself cannot be used to transpose a matrix unless the matrix happens to be a vector. If the matrix is not a vector then transpose alters the internal storage order of the elements, whereas reshape() never does.
For example, internally [1 2; 3 4] is stored in the order 1 3 2 4, and transpose of [1 2;3 4] would be [1 3;2 4] which would be stored in the order 1 2 3 4. You can see that the 2 and 3 have swapped internal places in the transpose. Reshape never swaps internal orderings.
James Tursa
2012-3-21
1 个投票
This is an ill-posed problem or something is missing from the problem statement. There are various ways to accomplish a transpose via indexing or permute etc as has already been pointed out. None of these involve the reshape function and as Walter points out reshape never alters the internal memory order (which is required for a general matrix transpose) so how the heck is reshape supposed to be involved in this in the first place?
Geoff
2012-3-21
Okay, got a solution. Your matrix (let's just use the example of A) can be indexed by the vector 1:6, but you need to translate this index to be row-wise instead of column-wise.
I = 1:size(A(:),1)
So first, work out how to generate your row and column indices so that you end up with something like:
r = [1 1 1 2 2 2]
c = [1 2 3 1 2 3]
Then use those to generate a transposed index for A, which will end up like this:
It = [1 3 5 2 4 6]
After you have that, it should be obvious what to do.
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