convert matrix

how can I convert a 6-dimensional matrix into a matrix of dimension 4 then 2

[Jan]

To let us understand what you are looking for, explain: SIZE(Inputs), kind of operation, SIZE(Outputs).

Answer 1

[Jan]

A 6-dimensional array (a "matrix" has 2 dimensions):

X = rand(2,3,4,5,6,7);

It is impossible to reshape this to the dimensions [4 x 2], which would be a matrix. So actual I believe that RESHAPE will do what you want, but a further explanaiotion of the wanted result is needed.

Answer 2

[Andrei Bobrov]

A = rand(6,1) % array dimension 6x1
out = reshape(A,4,2);

[Jan]

While we all agree, that the OP looks for RESHAPE, "6-dimensional" and "dimension 4 then 2" is questionable. Your interpretation seems to be the smartest yet (+1). It would be so nice if Samia cares about the tread and explains what was actually meant.

[samia]

if A is a matrix of size [8 8 3 3 32 32]how to convert to B of dimension 4 ,I dont want to use "RESHAPE"

[Jan]

What is "dimension 4"? Please try to use a proper Matlab terminology. See the commands SIZE and NDIMS.

I do not see any way to face the fact, that you do not "want" to use RESHAPE. Either RESHAPE creates the correct results, or you are looking for something else but did not explain it very well -- in both cases "wanting" does not matter.

Answer 3

[Igor]

6?? see

reshape()

Answer 4

[Laura Proctor]

The results below will reshape from a 6D to a 4D array. If you don't want to use reshape, is there something else you have in mind?

A = rand(8,8,3, 3, 32, 32);
B = reshape(A,[],64,8,6);

[samia]

So I simply give the problem:

I have a matrix A of size 256 * 256 I divided into blocks of 8 * 8 blocks are grouped into these 9 (3 * 3 blocks of 8 * 8) I hope that the trick is clear.

First of all I do these steps as follows:

a=3;

N=length(I);

B=[];

i=1;j=1;bi=1;bj=1;

while(i<=N-a+1)

while(j<=N-a+1)

B(:,:,bi,bj)=I(i:i+a-1,j:j+a-1);

j=j+a;

bj=bj+1;

end

i=i+a;

bi=bi+1;

j=1;

bj=1;

end

L=size(B,3);

C=size(B,4);

i=1;j=1;gri=1;grj=1;

while(i<=L-2)

while(j<=C-2)

gr9(:,:,:,:,gri,grj)=B(:,:,i:i+2,j:j+2);

j=j+1;

grj=grj+1;

end

i=i+1;

gri=gri+1;

j=1;

grj=1;

end

now how do the reverse path to return to A

[Jan]

Using FOR loops would increase the readability.

I'm surprised that you can divide a 256x256 matrix to 9 8x8 blocks.

[samia]

I put each 9 block of 8*8 together !!!!

[Jan]

@Samia: Sorry, it is not getting clearer. What does "put together" mean? 256*256 is simply not dividable by 9. Is it possible to replace your program by a single line containing RESHAPE and PERMUTE?

[Matt Fig]

I see this is more than a simple reshape. If size(I) = [256 256], then size(gr9) = [3 3 3 3 83 83], so that gr9 has 8.5 times _more_ elements than I! So it is not so simple as breaking a matrix into blocks.

[Jan]

@Matt: Do you understand how [3 3 3 3 83 83] and "put each 9 block of 8*8" match together?

[Matt Fig]

@Jan: No. I simply don't understand the way samia is doing this. The smallest I for which I can get the code to produce a 6D gr9 is:

I = round(rand(12,12)*200);

which produces a gr9 of size [3 3 3 3 2 2]. This I can actually trace through and look at how the elements of gr9 are taken, but I see no clear patter that I think would extend to I 256-by-256... With the latter I of course there is no chance of tracing through!

[samia]

I know it is not simple to understand! look: after devide A into block of 8*8 you get for example B; size(B) is [8 8 30 30] with last code now you take B and grouped each 9 small block together (small block mean block of 8*8) you gr9 of size= [8 8 3 3 30 30 ] you can test this result.

my problem is in the reverse path