Populating matrix with coordinates of each element

Hi Guys,

I am trying to populate a 100x100 matrix so that the first element is assigned to 1 and the last element is assigned to 100.

Any suggestions? I've already create the specified matrix using zeros and ones but I am guessing I have to use a nested for-loop with the outer looping over the rows and the inner looping over the columns to reassign each element?

Any assistance would be much appreciated.

[the cyclist]

You have not fully specified the matrix.

Let's take a smaller example: 3x3 matrix. Can you write out what you mean? For example, this is a matrix with the first element 1, and last element 3.

M = [1 0 0;
     0 0 0;
     0 0 3];

Is that what you mean?

[Will Pihir]

Sorry for the lack of clarification.

I meant something like this:

M =

[ 1 2 3 4 5 7 8 9

10 11 12 13 14

15 16 17 18 19

....... 100 ];

[Matt J]

That would be a 10x10 matrix, not 100x100.

[Will Pihir]

Yeah I just figured it out as I was typing.

Oops.

Same premise still applies

[Matt J]

Same premise still applies

Then, you should respond to the answer that you have been given by @DGM. Accept-click the answer if it does what you want and describe what is missing if it does not.

Answer 1

[DGM]

For your example using row-wise linear indexing:

s = [5 10]; % output size
reshape(1:prod(s),s(2),s(1)).'

ans = 5×10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

Obviously, you'd change s to suit your needs. To do the same thing columnwise:

s = [5 10]; % output size
reshape(1:prod(s),s(1),s(2))

ans = 5×10

1 6 11 16 21 26 31 36 41 46 2 7 12 17 22 27 32 37 42 47 3 8 13 18 23 28 33 38 43 48 4 9 14 19 24 29 34 39 44 49 5 10 15 20 25 30 35 40 45 50

[Will Pihir]

Thanks for the answer DGM. This was exactly what I was looking for.

Just so I can understand it better, what does the syntax .' at the end of the row-wise linear indexing code do?

[DGM]

.' transposes the array. For a 2D array, t's the same as doing

Atranspose = permute(A,[2 1]);

Since reshape() fills the result along columns, and the requirements need the indices ordered along rows, I have reshape generate the transpose of the desired array (hence the dimension order swap).

s = [5 10]; % output size
result = reshape(1:prod(s),s(2),s(1)) % with no transpose

result = 10×5

1 11 21 31 41 2 12 22 32 42 3 13 23 33 43 4 14 24 34 44 5 15 25 35 45 6 16 26 36 46 7 17 27 37 47 8 18 28 38 48 9 19 29 39 49 10 20 30 40 50

Transposing gives the correct orientation.

result = result.'

result = 5×10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50