Thread Subject: How to expand a matrix?

Hi,

I have the following matrix:

[ 1 5
  4 9]

What's the fastest way, and without a for loop, to expand it to:

[ 1 1 5 5
  1 1 5 5
  4 4 9 9
  4 4 9 9]

or to

[ 1 1 1 5 5 5
  1 1 1 5 5 5
  1 1 1 5 5 5
  4 4 4 9 9 9
  4 4 4 9 9 9
  4 4 4 9 9 9]
?

Thanks in advance!

Help kron

A=[1 4; 4 9]

kron(A,ones(3))

% Bruno

Thanks Bruno. You always answer all of my questions.

You are welcome. If you want to force your processor to multiplication and addition unnecessary, and to avoid KRON then here is another way:

[m n]=size(A)
mdup=4; ndup=3;

AX=repmat(A,[1 1 mdup ndup]);
AX=permute(AX,[3 1 4 2]);
AX=reshape(AX,m*mdup,n*ndup)

% Bruno

I'm a bit confused. Why would I want to do it the second way? Is it faster? Is it more advantages for any reason? Seems like kron does it fast and easy.

I'm not sure. KRON is more general purpose, it performs many matrix multiplications. In our case it is trivial scalar with 1. Whereas the second method just duplicate and moving elements around. I'm not sure which one is faster.

For sure KRON is more readable.

Up to you

Bruno

Just another method, using built-ins only:


% Data
A= [ 1 5;4 9]

% Engine
idx = cumsum(ones(2),2)
B = A(idx,idx)

Along the same line than Matt's solution using BSXOPS on FEX:

B = A(zeros(2,1)+(1:end),zeros(2,1)+(1:end))

% Bruno

Yet another method

[m n]=size(A)
mdup=4; ndup=3;

vxÎil((1:m*mdup)/mdup);
vyÎil((1:n*ndup)/ndup);

A(vx,vy)

Hmmm. Weird character translation problem...
 
 [m n]=size(A)
 mdup=4; ndup=3;
 
vxÎil((1:m*mdup)/mdup);
vyÎil((1:n*ndup)/ndup);

A(vx,vy)

"Matt " <xys@whatever.com> wrote in message <gtk66u$f3g$1@fred.mathworks.com>...
> Hmmm. Weird character translation problem...
>
> [m n]=size(A)
> mdup=4; ndup=3;
>
> vx?il((1:m*mdup)/mdup);
> vy?il((1:n*ndup)/ndup);
>
> A(vx,vy)

Well, the weird characters after vx and vy are supposed to be 'Îil'

vxÎil((1:m*mdup)/mdup);
vyÎil((1:n*ndup)/ndup);

"Matt " <xys@whatever.com> wrote in message
> Well, the weird characters after vx and vy are supposed to be '?il'
>
> vx?il((1:m*mdup)/mdup);
> vy?il((1:n*ndup)/ndup);


Still not showing up, I will try. Do you mean ceil?

vx = ceil((1:m*mdup)/mdup);
vy = ceil((1:n*ndup)/ndup);

"Matt Fig" <spamanon@yahoo.com> wrote in message <gtkeon$rhl$1@fred.mathworks.com>...

> Still not showing up, I will try. Do you mean ceil?
>
> vx = ceil((1:m*mdup)/mdup);
> vy = ceil((1:n*ndup)/ndup);

Yes. for some reason '= ceil' without the space generates these weird characters...

"Matt Fig" <spamanon@yahoo.com> wrote in message <gtio1e$dj3$1@fred.mathworks.com>...
> Just another method, using built-ins only:
>
>
> % Data
> A= [ 1 5;4 9]
>
> % Engine
> idx = cumsum(ones(2),2)
> B = A(idx,idx)

I like this method. It's simple. But, how would you do this for a 3x3 matrix or a 4x4 matrx?

"Kian " <kian.torab@utah.edu> wrote in message <gtq2qh$blq$1@fred.mathworks.com>...
> "Matt Fig" <spamanon@yahoo.com> wrote in message <gtio1e$dj3$1@fred.mathworks.com>...
> > Just another method, using built-ins only:
> >
> >
> > % Data
> > A= [ 1 5;4 9]
> >
> > % Engine
> > idx = cumsum(ones(2),2)
> > B = A(idx,idx)
>
> I like this method. It's simple. But, how would you do this for a 3x3 matrix or a 4x4 matrx?

My method is basically Matt Fig's idea, but generalized to arbitrary matrix size and arbitrary upsampling

 [m n]=size(A)
 mdup=4; ndup=3;
 
vx = ceil((1:m*mdup)/mdup);
vy = ceil((1:n*ndup)/ndup);

A(vx,vy)

"Kian " <kian.torab@utah.edu> wrote in message <gtq2qh$blq$1@fred.mathworks.com>...
> "Matt Fig" <spamanon@yahoo.com> wrote in message <gtio1e$dj3$1@fred.mathworks.com>...
> > Just another method, using built-ins only:
> >
> >
> > % Data
> > A= [ 1 5;4 9]
> >
> > % Engine
> > idx = cumsum(ones(2),2)
> > B = A(idx,idx)
>
> I like this method. It's simple. But, how would you do this for a 3x3 matrix or a 4x4 matrx?


Matt's method is more easily generalizable for arbitrary expansion in each direction. Nevertheless, here is a generalization of the method behind my example.

N = 3; % A is NxN
E = 4; % The expansion number.

A = round(rand(N)*10);
idx = cumsum(ones(E,N),2);
B = A(idx,idx)

"Matt Fig" <spamanon@yahoo.com> wrote in message <gtq4s5$6so$1@fred.mathworks.com>...
> "Kian " <kian.torab@utah.edu> wrote in message <gtq2qh$blq$1@fred.mathworks.com>...
> > "Matt Fig" <spamanon@yahoo.com> wrote in message <gtio1e$dj3$1@fred.mathworks.com>...
> > > Just another method, using built-ins only:
> > >
> > >
> > > % Data
> > > A= [ 1 5;4 9]
> > >
> > > % Engine
> > > idx = cumsum(ones(2),2)
> > > B = A(idx,idx)
> >
> > I like this method. It's simple. But, how would you do this for a 3x3 matrix or a 4x4 matrx?
>
>
> Matt's method is more easily generalizable for arbitrary expansion in each direction. Nevertheless, here is a generalization of the method behind my example.
>
> N = 3; % A is NxN
> E = 4; % The expansion number.
>
> A = round(rand(N)*10);
> idx = cumsum(ones(E,N),2);
> B = A(idx,idx)

matt - don't get sidetracked, don't procrastinate! you should be improving your FINDSUBMAT...

as ever with a :-)
urs

"us " <us@neurol.unizh.ch> wrote in message
> matt - don't get sidetracked, don't procrastinate! you should be improving your FINDSUBMAT...
>
> as ever with a :-)
> urs


The watchful eye of Urs! ;-).

Re find submatrix: Matt, a lazy hint: it takes only two lines to typecast the double/single array to respectively int64/int32, then compares with strfind. The NaN issue will be resolved in no time. You might continue to hang around here. ;-)

Bruno

"Bruno Luong" <b.luong@fogale.findmycountry> wrote in message <gtq8i9$lpv$1@fred.mathworks.com>...
> Re find submatrix: Matt, ... The NaN issue will be resolved in no time. You might continue to hang around here. ;-)

matt: bruno - the devil in disguise!...

www.youtube.com/watch?v=m3_Q96eJr1k

[lol]
urs

:-)
Cheers,
Bruno