Thread Subject: re-positioning a Matrix

Hi all, during writing the code of an algorithms i have
falled into a problem.

The result of math calculation is a matrix, of this type:

Matrix1 = [0 0 0 1 5 ]
          [0 0 0 0.1 0.2]
          [0 0 0 0.5 0.4]
          [0 0 0 0.3 0.1]

Where the first row is similar an index.
I would re-positioning the matrix1 following the index of
first row, in this way:

Matrix2 = [1 0 0 0 5 ]
          [0.1 0 0 0 0.2]
          [0.5 0 0 0 0.4]
          [0.3 0 0 0 0.1]

(Where the number of first row is the number of column)
There is a way to do it? I have no idea in this moment : - )

Thanks all in advace

Emanuele

"Emanuele " <emanuelemignosa.nospam@mathworks.com> wrote in
message <g13865$nl6$1@fred.mathworks.com>...
> Hi all, during writing the code of an algorithms i have
> falled into a problem.
>
> The result of math calculation is a matrix, of this type:
>
> Matrix1 = [0 0 0 1 5 ]
> [0 0 0 0.1 0.2]
> [0 0 0 0.5 0.4]
> [0 0 0 0.3 0.1]
>
> Where the first row is similar an index.
> I would re-positioning the matrix1 following the index of
> first row, in this way:
>
> Matrix2 = [1 0 0 0 5 ]
> [0.1 0 0 0 0.2]
> [0.5 0 0 0 0.4]
> [0.3 0 0 0 0.1]
>
> (Where the number of first row is the number of column)
> There is a way to do it? I have no idea in this moment : - )
>
> Thanks all in advace
>
> Emanuele
>
>


it is not that trivial:

% Example data
  M = [ 0 0 0 1 5 ; 0 0 0 100 200 ; 0 0 0 10 20]

M2 = zeros(size(M)) ;
c = M(1,:)
M2(:,c(c>0)) = M(:,c>0)

hth
Jos

"Jos " <DELjos@jasenDEL.nl> wrote in message <g139di$727
$1@fred.mathworks.com>...
> "Emanuele " <emanuelemignosa.nospam@mathworks.com> wrote
in
> message <g13865$nl6$1@fred.mathworks.com>...
> > Hi all, during writing the code of an algorithms i have
> > falled into a problem.
> >
> > The result of math calculation is a matrix, of this
type:
> >
> > Matrix1 = [0 0 0 1 5 ]
> > [0 0 0 0.1 0.2]
> > [0 0 0 0.5 0.4]
> > [0 0 0 0.3 0.1]
> >
> > Where the first row is similar an index.
> > I would re-positioning the matrix1 following the index
of
> > first row, in this way:
> >
> > Matrix2 = [1 0 0 0 5 ]
> > [0.1 0 0 0 0.2]
> > [0.5 0 0 0 0.4]
> > [0.3 0 0 0 0.1]
> >
> > (Where the number of first row is the number of column)
> > There is a way to do it? I have no idea in this
moment : - )
> >
> > Thanks all in advace
> >
> > Emanuele
> >
> >
>
>
> it is not that trivial:
>
> % Example data
> M = [ 0 0 0 1 5 ; 0 0 0 100 200 ; 0 0 0 10 20]
>
> M2 = zeros(size(M)) ;
> c = M(1,:)
> M2(:,c(c>0)) = M(:,c>0)
>
> hth
> Jos
>
Hi Jos, wow, your solution works and it's very elegant and
fine,
I belived that solution probably would be a for cicle,
instead your solution is very fast.
It's very interesting, i want study it. : - )

I don't understand the last line of your code.
I don't know that Matlab accept code in his left part
"M2(:,c(c>0))" without for cicle.

However tahnks very much.

Emanuele

"Emanuele " <emanuelemignosa.nospam@mathworks.com> wrote in
message <g13b7p$t99$1@fred.mathworks.com>...
< SNIP .. trouble understanding (logical) indexing

Hi Emanuele,

If you break down the code "M2(:,c(c>0)) = M(:,c>0)" into
single lines, you might be able to follow

1) q = c > 0
returns a logical array where c is larger than zero

2) COLIND = c(q)
returns those elements in c that are larger than zero

3) M2(:,colind) selects only those columns specified by
COLIND (called "normal" indexing)

4) M(:,q) selects those columns for which q is true (called
logical indexing)

You can use indexing on both the lefthand or righthand side
of an assignment operator. For instance

X(3) = 2 sets the third element of X to 2

j = X(5) retrieves the fifth element of X

If you use them on the lefthand side you need to be sure
that the number of elements match:

X = [0 0 0 0] ;
X(2) = [2 3] ; error!

hth
Jos