pysparse-users

[Pysparse-users] transpose operation

[Toine B. <tb...@db...>]

Hi,

What is the most efficient way of transposing a matrix using PySparse?  
A search of the website doesn't yield any results and I don't see any  
transpose method in the list of methods. I must be missing something  
here; I can't imagine there is no fast transpose method available!

I've tried two different ways of calculating the transpose myself:

1) looping through the entries of a matrix M with .items() and then  
simply filling a new sparse matrix with the same values, but transposed.

     def T_ver1( self, M ):
         (rows, cols) = M.shape
         t = spmatrix.ll_mat(cols, rows, M.nnz)
         for (x, y), value in M.items():
             t[y, x] = M[x,y]
         return t

2) I've noticed that using the .dot() method to multiply a matrix with  
the identity matrix is a faster way of getting the transposed matrix M.

     def T_ver2( self, M ):
         (rows, cols) = M.shape
         I = spmatrix.ll_mat(rows, rows, rows)
         for i in xrange(0, rows):
             I[i, i] = 1
         return spmatrix.dot(M, I)):

Method 2 is faster, which I suspect is because the  .dot() function  
using either the C or Fortran code directly. But why is there no  
direct M.transpose() or M.T method to give me the transposed matrix?

Thanks in advance for the help!


Kind regards,
Toine Bogers

Re: [Pysparse-users] transpose operation

[<lu...@o2...>]

Toine Bogers <tb...@db...> writes:

> Hi,
>
> What is the most efficient way of transposing a matrix using PySparse? A search
> of the website doesn't yield any results and I don't see any transpose method
> in the list of methods. I must be missing something here; I can't imagine there
> is no fast transpose method available!
>
> I've tried two different ways of calculating the transpose myself:
>
> 1) looping through the entries of a matrix M with .items() and then simply
> filling a new sparse matrix with the same values, but transposed.
>
>     def T_ver1( self, M ):
>         (rows, cols) = M.shape
>         t = spmatrix.ll_mat(cols, rows, M.nnz)
>         for (x, y), value in M.items():
>             t[y, x] = M[x,y]
>         return t
>
> 2) I've noticed that using the .dot() method to multiply a matrix with the
> identity matrix is a faster way of getting the transposed matrix M.
>
>     def T_ver2( self, M ):
>         (rows, cols) = M.shape
>         I = spmatrix.ll_mat(rows, rows, rows)
>         for i in xrange(0, rows):
>             I[i, i] = 1
>         return spmatrix.dot(M, I)):
>
> Method 2 is faster, which I suspect is because the  .dot() function using
> either the C or Fortran code directly. But why is there no direct M.transpose()
> or M.T method to give me the transposed matrix?

Hi,

Mine is:

def transpose(a):
    b = ll_mat(a.shape[1], a.shape[0], a.nnz)
    v, r, c = a.find()
    b.put(v, c, r)
    return b

should be faster, though for the price of extra memory usage (v, r, c
arrays).

>
> Thanks in advance for the help!
>
>
> Kind regards,
> Toine Bogers
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Re: [Pysparse-users] transpose operation

[Dominique O. <dom...@gm...>]

2009/11/16 Łukasz Pankowski <lu...@o2...>

> Toine Bogers <tb...@db...> writes:
>
> > Hi,
> >
> > What is the most efficient way of transposing a matrix using PySparse? A
> search
> > of the website doesn't yield any results and I don't see any transpose
> method
> > in the list of methods. I must be missing something here; I can't imagine
> there
> > is no fast transpose method available!
> >
> > I've tried two different ways of calculating the transpose myself:
> >
> > 1) looping through the entries of a matrix M with .items() and then
> simply
> > filling a new sparse matrix with the same values, but transposed.
> >
> >     def T_ver1( self, M ):
> >         (rows, cols) = M.shape
> >         t = spmatrix.ll_mat(cols, rows, M.nnz)
> >         for (x, y), value in M.items():
> >             t[y, x] = M[x,y]
> >         return t
> >
> > 2) I've noticed that using the .dot() method to multiply a matrix with
> the
> > identity matrix is a faster way of getting the transposed matrix M.
> >
> >     def T_ver2( self, M ):
> >         (rows, cols) = M.shape
> >         I = spmatrix.ll_mat(rows, rows, rows)
> >         for i in xrange(0, rows):
> >             I[i, i] = 1
> >         return spmatrix.dot(M, I)):
> >
> > Method 2 is faster, which I suspect is because the  .dot() function using
> > either the C or Fortran code directly. But why is there no direct
> M.transpose()
> > or M.T method to give me the transposed matrix?
>
> Hi,
>
> Mine is:
>
> def transpose(a):
>    b = ll_mat(a.shape[1], a.shape[0], a.nnz)
>    v, r, c = a.find()
>    b.put(v, c, r)
>    return b
>
> should be faster, though for the price of extra memory usage (v, r, c
> arrays).


Hi Toine,

The find/put method is surely the fastest in general. The more fundamental
question is do you really need the explicit transpose? Often, algorithms
that need the transpose operate directly on the original matrix. If what you
need is matrix-vector products of the form A.T*x, you can always use the
matvec_transp() method. If you're using the higher-level PysparseMatrix
objects, you can do A*x and x*A. The latter actually computes A.T*x.

I hope this helps.

-- 
Dominique