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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