Transposing matrix using reshape

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.

[the cyclist]

Are you able to post the full problem statement from your instructor?

[Adam]

Unfortunately, no. It's a segment at the end of a long pdf that would make no sense without context. I've found two or three other ways of doing this, but none involve reshape.

[Adam]

is there some detail I can clear up for you?

[the cyclist]

No, I just thought that perhaps there was something subtle in the problem student that you overlooked.

[Geoff]

Well, are you allowed to use ANY other functions like size(), repmat() and colon()? =) What exactly is the restriction?

[Adam]

size or basic data analysis functions like max or mean would be the only other allowed functions.

[Adam]

also sort() and sortrows(), but those don't seem like they'd be much help.

[Geoff]

And no looping? She wants a one-liner?

[Geoff]

Seems to me like your instructor is teaching you to hate MatLab.

[Walter Roberson]

Is array indexing allowed? Are multiplication and addition allowed?

Answer 1

[Walter Roberson]

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.

Answer 2

[James Tursa]

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?

Answer 3

[Image Analyst]

Are you sure he didn't mean permute?

permute(A, [2 1])

ans =
     1     4
     2     5
     3     6

[Adam]

I'm sorry, but no. She explicitly said reshape. Thanks for trying to help, though.

Answer 4

[Geoff]

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.

Answer 5

[Jan]

As long as the problem is ill-posed, weird solutions are valid:

B = feval(reshape(['tno'; 'rss'; 'ape'], 1, 9), A);

[Jan]

The similarities between ...rssape and reshape are magic.