Mailman 3 September 2013 - SciPy-User

Re: [SciPy-User] [Numpy-discussion] ANN: Scipy 0.13.0 beta 1 release
by David Cournapeau Sept. 2, 2013

Sept. 2, 2013

On Mon, Sep 2, 2013 at 5:46 PM, Pauli Virtanen <pav(a)iki.fi> wrote: > -----BEGIN PGP SIGNED MESSAGE----- > Hash: SHA1 > > Hi, > > 02.09.2013 18:30, David Cournapeau kirjoitti: > [clip] > > Ok, I think the commit 5935030f8cced33e433804a21bdb15572d1d38e8 is > > quite wrong. > > > > It conflates the issue of dealing with Accelerate brokenness and > > using g77 ABI. I would suggest reverting it for 0.13.0 (and > > re-disable single precision), as fixing this correctly may require > > quite some time/testing. > > I'm -1 on returning to the previous situation where many routines on > OSX are simply broken (which was the situation previously --- > "disabling single precision" left several things still broken). I'd > rather just postpone the 0.13.0 release until this issue is solved > properly. > I see, I missed that it was more than just reverting to slower versions. > > Can you say what exactly is wrong -- as far as I know, Accelerate on > OSX simply uses g77 ABI. There were some bugs previously where it in > places did things differently, but on recent OSX releases this is no > longer the case? > > I guess you are trying to link with MKL? In that case, I would rather > propose restoring the previous MKL wrappers (with trivial extensions > for the missing w* symbols). > The pb is specific to MKL, yes, but I've found a workaround: in the case of MKL, you just need to workaround the g77 ABI of the MKL vs gfortran ABI, but the LAPACK interface is the actual LAPACK, not CLAPACK. So for MKL, you need non dummy wrappers iff the function returns a complex. A dirty patch seems to confirm that this fixes the issue, I will prepare an actual patch tomorrow, David > > - -- > Pauli Virtanen > -----BEGIN PGP SIGNATURE----- > Version: GnuPG v1.4.12 (GNU/Linux) > > iEYEARECAAYFAlIkwPQACgkQ6BQxb7O0pWBnkwCfeKj32CYYCPdEWVcYYMtq/OZM > 32wAoIn9yG/HNtUeh+XwqAm2uAS9sVQ5 > =CKBN > -----END PGP SIGNATURE----- > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion(a)scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >

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ANN: Scipy 0.13.0 beta 1 release
by Ralf Gommers Sept. 2, 2013

Sept. 2, 2013

Hi all, I'm happy to announce the availability of the first beta release of Scipy 0.13.0. Please try this beta and report any issues on the scipy-dev mailing list. Source tarballs and release notes can be found at https://sourceforge.net/projects/scipy/files/scipy/0.13.0b1/. Windows and OS X installers will follow later (we have a minor infrastructure issue to solve, and I'm at EuroScipy now). Cheers, Ralf

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solve_banded usage (once again)
by Wagner Sebastian Sept. 2, 2013

Sept. 2, 2013

Dear Scipy-Users, I'm currently working with Spare Matrices, in particular diagonal spare matrices. When searching for solve-Functions I found a function called solve_banded and some information about it in the docs (very little and that's why I searched for more), StackOverflow (http://stackoverflow.com/questions/12978518/scipy-sparse-dia-matrix-solver), a ticket on this problem (https://github.com/scipy/scipy/issues/2285) and this mailinglist. But nowhere I found any information about how to use it with a diagonal sparse matrix. The mentioned thread on this is at http://thread.gmane.org/gmane.comp.python.scientific.user/10308, which states that "the docstring of solve_banded seems to be not correct/up-to-date, or at least unclear." Which is totally correct (the Mail is from 2007!) and until now nothing has changed. Due to lack of documentation I tried to get it running, and got it running partly with contiuous offsets, like this way: dia = scipy.sparse.dia_matrix(([[1,1,3,0,0],[1,1,1,1,1],[0,2,1,3,2]],[-2,0,1]), shape=(5,5)) offsets = [np.abs(dia.offsets[0]), dia.offsets[-1]] data = np.flipud(dia.data) scipy.linalg.solve_banded(offsets, data, b) This gives be the same solutions as linalg.solve and sparse.linalg.spsolve. But with uncontinuous offsets I couldn't, I thought of inserting a row of zeros for the missing diagonals: dia = scipy.sparse.dia_matrix(([[1,1,3,0,0],[1,1,1,1,1],[0,2,1,3,2]],[-2,0,1]), shape=(5,5)) l = np.abs(dia.offsets[0]) u = dia.offsets[-1] data = np.empty((l+u+1,dia.data.shape[1]),dtype=int) missingCounter = 0 for i, d in enumerate(range(-l, u+1)): if d in dia.offsets: data[i] = dia.data[-(i+1-missingCounter)] # flip the array else: data[i] = np.zeros(dia.data.shape[1]) missingCounter += 1 For me it looks fine, but it gives not the same answer as linalg.solve and sparse.linalg.spsolve. Can anybody give me an advice of how to use solve_banded correctly? Regards, Sebastian

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solve_banded usage (once again)
by Wagner Sebastian Sept. 2, 2013

Sept. 2, 2013

Dear Scipy-Users, I'm currently working with Spare Matrices, in particular diagonal spare matrices. When searching for solve-Functions I found a function called solve_banded and some information about it in the docs (very little and that's why I searched for more), StackOverflow (http://stackoverflow.com/questions/12978518/scipy-sparse-dia-matrix-solver), a ticket on this problem (https://github.com/scipy/scipy/issues/2285) and this mailinglist. But nowhere I found any information about how to use it with a diagonal sparse matrix. The mentioned thread on this is at http://thread.gmane.org/gmane.comp.python.scientific.user/10308, which states that "the docstring of solve_banded seems to be not correct/up-to-date, or at least unclear." Which is totally correct (the Mail is from 2007!) and until now nothing has changed. Due to lack of documentation I tried to get it running, and got it running partly with contiuous offsets, like this way: dia = scipy.sparse.dia_matrix(([[1,1,3,0,0],[1,1,1,1,1],[0,2,1,3,2]],[-2,0,1]), shape=(5,5)) offsets = [np.abs(dia.offsets[0]), dia.offsets[-1]] data = np.flipud(dia.data) scipy.linalg.solve_banded(offsets, data, b) This gives be the same solutions as linalg.solve and sparse.linalg.spsolve. But with uncontinuous offsets I couldn't, I thought of inserting a row of zeros for the missing diagonals: dia = scipy.sparse.dia_matrix(([[1,1,3,0,0],[1,1,1,1,1],[0,2,1,3,2]],[-2,0,1]), shape=(5,5)) l = np.abs(dia.offsets[0]) u = dia.offsets[-1] data = np.empty((l+u+1,dia.data.shape[1]),dtype=int) missingCounter = 0 for i, d in enumerate(range(-l, u+1)): if d in dia.offsets: data[i] = dia.data[-(i+1-missingCounter)] # flip the array else: data[i] = np.zeros(dia.data.shape[1]) missingCounter += 1 For me it looks fine, but it gives not the same answer as linalg.solve and sparse.linalg.spsolve. Can anybody give me an advice of how to use solve_banded correctly? Regards, Sebastian

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Stick intersection path algorithm
by Josè Luis Mietta Sept. 1, 2013

Sept. 1, 2013

Hi experts! I wanna study the intersection between line segments (sticks). I wrote a algorithm that generate a matrix, M, with N rows and N columns. The M-element Mij is 1 if stick number 'i' intersect stick number 'j' (obviously M is symmetric). Given two arbitrary sticks, i need a simple and effective algorithm that determinate if that two sticks are conected by a 'intersected-sticks' path. Any idea for that? Thanks a lot!

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