Mailman 3 February 2012 - NumPy-Discussion

Masked array elements where mask = True?
by Howard Feb. 3, 2012

Feb. 3, 2012

Is there a method that gives an array of all the array indices of a masked array where the mask is True? I've been looking through the docs and don't see it yet... Thanks Howard -- Howard Lander <mailto:[email protected]> Senior Research Software Developer Renaissance Computing Institute (RENCI) <http://www.renci.org> The University of North Carolina at Chapel Hill Duke University North Carolina State University 100 Europa Drive Suite 540 Chapel Hill, NC 27517 919-445-9651

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Re: [Numpy-discussion] Trick for fast
by santhu kumar Feb. 3, 2012

Feb. 3, 2012

Hello all, Thanks for lovely solutions. I have sat on it for some time and wrote it myself : n =x.shape[0] ea = np.array([1,0,0,0,1,0,0,0,1]) inert = ((np.tile(ea,(n,1))*((x*x).sum(axis=1)[:,np.newaxis]) - np.hstack([x*x[:,0][:,np.newaxis],x*x[:,1][:,np.newaxis],x*x[:,2][:,np.newaxis]]))*mass[:,np.newaxis]).sum(axis=0) inert.shape = 3,3 Does the trick and reduces the time from over 45 secs to 3 secs. I do want to try einsum but my numpy is little old and it does not have it. Thanks Sebastian (it was tricky to understand your code for me) and Josef (clean). On Fri, Feb 3, 2012 at 12:00 PM, <numpy-discussion-request(a)scipy.org> wrote: > Send NumPy-Discussion mailing list submissions to > numpy-discussion(a)scipy.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://mail.scipy.org/mailman/listinfo/numpy-discussion > or, via email, send a message with subject or body 'help' to > numpy-discussion-request(a)scipy.org > > You can reach the person managing the list at > numpy-discussion-owner(a)scipy.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of NumPy-Discussion digest..." > > > Today's Topics: > > 1. Re: Trick for fast (josef.pktd(a)gmail.com) > 2. Re: Trick for fast (S?ren Gammelmark) > 3. Re: Trick for fast (Sebastian Berg) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Fri, 3 Feb 2012 09:10:28 -0500 > From: josef.pktd(a)gmail.com > Subject: Re: [Numpy-discussion] Trick for fast > To: Discussion of Numerical Python <numpy-discussion(a)scipy.org> > Message-ID: > <CAMMTP+B2yQ9Th5yhDEuogS6yWjuQpAnZZQ1GcU1GqRdvDk7itw(a)mail.gmail.com > > > Content-Type: text/plain; charset=ISO-8859-1 > > On Fri, Feb 3, 2012 at 8:44 AM, Alan G Isaac <alan.isaac(a)gmail.com> wrote: > > On 2/3/2012 5:16 AM, santhu kumar wrote: > >> x = nX3 vector. > >> mass = nX1 vector > >> inert = zeros((3,3)) > >> for i in range(n): > >> ? ? ? ?ri = x[i,:].reshape(1,3) > >> ? ? ? ?inert = inert + mass[i,]*(sum(ri*ri)*eye(3) - dot(ri.T,ri)) > >> > > > > > > This should buy you a bit. > > > > xdot = (x*x).sum(axis=1) > > for (massi,xi,xdoti) in zip(mass.flat,x,xdot): > > ? ? ? temp = -np.outer(xi,xi) > > ? ? ? temp.flat[slice(0,None,4)] += xdoti > > ? ? ? inert += massi*temp > > > > Alan Isaac > > maybe something like this, (self contained example and name spaces to > make running it easier) > > import numpy as np > n = 15 > x = np.arange(n*3.).reshape(-1,3) #nX3 vector. > mass = np.linspace(1,2,n)[:,None] #nX1 vector > inert = np.zeros((3,3)) > for i in range(n): > ri = x[i,:].reshape(1,3) > inert = inert + mass[i,]*(sum(ri*ri)*np.eye(3) - np.dot(ri.T,ri)) > print inert > > print np.diag((mass * x**2).sum(0)) - np.dot(x.T, mass*x) > > [[ 0. -16755. -17287.5] > [-16755. 0. -17865. ] > [-17287.5 -17865. 0. ]] > [[ 0. -16755. -17287.5] > [-16755. 0. -17865. ] > [-17287.5 -17865. 0. ]] > > Josef > > > > > > _______________________________________________ > > NumPy-Discussion mailing list > > NumPy-Discussion(a)scipy.org > > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > > ------------------------------ > > Message: 2 > Date: Fri, 3 Feb 2012 15:43:25 +0100 > From: S?ren Gammelmark <gammelmark(a)gmail.com> > Subject: Re: [Numpy-discussion] Trick for fast > To: Discussion of Numerical Python <numpy-discussion(a)scipy.org> > Message-ID: > <CAJO1x6qV0ME_2Xgr4oU8q=waekT2qXQ=iComzbbQNGzB_MPKkQ(a)mail.gmail.com > > > Content-Type: text/plain; charset="iso-8859-1" > > What about this? > > A = einsum("i,ij->", mass, x ** 2) > B = einsum("i,ij,ik->jk", mass, x, x) > I = A * eye(3) - B > > /S?ren > > On 3 February 2012 15:10, <josef.pktd(a)gmail.com> wrote: > > > On Fri, Feb 3, 2012 at 8:44 AM, Alan G Isaac <alan.isaac(a)gmail.com> > wrote: > > > On 2/3/2012 5:16 AM, santhu kumar wrote: > > >> x = nX3 vector. > > >> mass = nX1 vector > > >> inert = zeros((3,3)) > > >> for i in range(n): > > >> ri = x[i,:].reshape(1,3) > > >> inert = inert + mass[i,]*(sum(ri*ri)*eye(3) - dot(ri.T,ri)) > > >> > > > > > > > > > This should buy you a bit. > > > > > > xdot = (x*x).sum(axis=1) > > > for (massi,xi,xdoti) in zip(mass.flat,x,xdot): > > > temp = -np.outer(xi,xi) > > > temp.flat[slice(0,None,4)] += xdoti > > > inert += massi*temp > > > > > > Alan Isaac > > > > maybe something like this, (self contained example and name spaces to > > make running it easier) > > > > import numpy as np > > n = 15 > > x = np.arange(n*3.).reshape(-1,3) #nX3 vector. > > mass = np.linspace(1,2,n)[:,None] #nX1 vector > > inert = np.zeros((3,3)) > > for i in range(n): > > ri = x[i,:].reshape(1,3) > > inert = inert + mass[i,]*(sum(ri*ri)*np.eye(3) - np.dot(ri.T,ri)) > > print inert > > > > print np.diag((mass * x**2).sum(0)) - np.dot(x.T, mass*x) > > > > [[ 0. -16755. -17287.5] > > [-16755. 0. -17865. ] > > [-17287.5 -17865. 0. ]] > > [[ 0. -16755. -17287.5] > > [-16755. 0. -17865. ] > > [-17287.5 -17865. 0. ]] > > > > Josef > > > > > > > > > > _______________________________________________ > > > NumPy-Discussion mailing list > > > NumPy-Discussion(a)scipy.org > > > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > _______________________________________________ > > NumPy-Discussion mailing list > > NumPy-Discussion(a)scipy.org > > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > >

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Trick for fast
by santhu kumar Feb. 3, 2012

Feb. 3, 2012

Hello all, I have tried to optimize most of my code but this ones seems to be the major bottleneck as it gets called many times. I have run out of ideas to make it more faster, can somebody please help me here. x = nX3 vector. mass = nX1 vector inert = zeros((3,3)) for i in range(n): ri = x[i,:].reshape(1,3) inert = inert + mass[i,]*(sum(ri*ri)*eye(3) - dot(ri.T,ri)) Any pythonic or numpy technique to make it faster. What I can think of is a sloppy nX9 matrix creation and then performing row wise summation and reshaping it. But is there any better way. Thanks santhosh

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Documentation question.
by Charles R Harris Feb. 3, 2012

Feb. 3, 2012

The macro PyArray_RemoveLargest has been replaced by PyArray_RemoveSmallest (which seems strange), but I wonder if this documentation still makes sense. diff --git a/doc/source/user/c-info.beyond-basics.rst b/doc/source/user/ c-info.beyond-basics.rs index 9ed2ab3..3437985 100644 --- a/doc/source/user/c-info.beyond-basics.rst +++ b/doc/source/user/c-info.beyond-basics.rst @@ -189,7 +189,7 @@ feature follows. PyArray_MultiIter_NEXT(mobj); } -The function :cfunc:`PyArray_RemoveLargest` ( ``multi`` ) can be used to +The function :cfunc:`PyArray_RemoveSmallest` ( ``multi`` ) can be used to take a multi-iterator object and adjust all the iterators so that iteration does not take place over the largest dimension (it makes that dimension of size 1). The code being looped over that makes use Chuck

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Heads up and macro deprecation.
by Charles R Harris Feb. 2, 2012

Feb. 2, 2012

Hi All, Two things here. 1) Some macros for threading and the iterator now require a trailing semicolon. This change will be reverted before the 1.7 release so that scipy 0.10 will compile, but because it is desirable in the long term it would be helpful if folks maintaining c extensions using numpy would try compiling them against current development and adding the semicolon where needed. The added semicolon will be backward compatible with earlier versions of numpy. 2) It is proposed to deprecate all of the macros in the old_defines.h file and require the use of their replacements. Numpy itself will have made this change after pull-189 <https://github.com/numpy/numpy/pull/189> is merged and getting rid of the surplus macros will help clean up the historical detritus that has built up over the years, easing maintenance, clarifying code, and making the eventual transition to 2.0 a bit easier. There is a sed script in the tools directory as part of the pull request that can be used to make the needed substitutions. Thoughts? Chuck

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Broadcasting doesn't work with divide after tile
by martin großhauser Feb. 2, 2012

Feb. 2, 2012

Hello, when I try in my script to divide a masked array by a scalar I get an error. The instruction is: >> sppa = sp / 100. sp is a masked array with ndim = 3. error is: Traceback (most recent call last): File "/media/nethome/Work/workspace/interimEnso/src/mlBudget.py", line 95, in <module> sppa = sp / 100. File "/usr/lib/pymodules/python2.7/numpy/ma/core.py", line 3673, in __div__ return divide(self, other) File "/usr/lib/pymodules/python2.7/numpy/ma/core.py", line 1072, in __call__ m |= ma ValueError: invalid return array shape The interesting thing is that this error only occurs after a tiling instruction: >> sp4d = N.tile(sp, (ninterf, 1, 1, 1)) If I do the division before the tiling I don't get an error. There's also no error if I do the division with N.divide(sp, 100.). Also printing the array sp after tiling doesn't work, while it works before. If I debug the script with eclipse/PyDev, in the variables window I get the message "Unable to get repr for <class 'numpy.ma.core.MaskedArray'>" for the sp array after tiling. The tiling operation shouldn't change the array, should it? Is this a bug, or is it expected behaviour? Regards, Martin Groszhauser If I try to print the

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histogram help
by Ruby Stevenson Feb. 2, 2012

Feb. 2, 2012

hi, all I am trying to figure out how to do histogram with numpy I have a three-dimension array A[x,y,z], another array (bins) has been allocated along Z dimension, z' how can I get the histogram of H[ x, y, z' ]? thanks for your help. Ruby

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Unexpected reorganization of internal data
by Mads Ipsen Feb. 2, 2012

Feb. 2, 2012

Hi, I am confused. Here's the reason: The following structure is a representation of N points in 3D space: U = numpy.array([[x1,y1,z1], [x1,y1,z1],...,[xn,yn,zn]]) So the array U has shape (N,3). This order makes sense to me since U[i] will give you the i'th point in the set. Now, I want to pass this array to a C++ function that does some stuff with the points. Here's how I do that void Foo::doStuff(int n, PyObject * numpy_data) { // Get pointer to data double * const positions = (double *) PyArray_DATA(numpy_data); // Print positions for (int i=0; i<n; ++i) { float x = static_cast<float>(positions[3*i+0]) float y = static_cast<float>(positions[3*i+1]) float z = static_cast<float>(positions[3*i+2]) printf("Pos[%d] = %f %f %f\n", x, y, z); } } When I call this routine, using a swig wrapped Python interface to the C++ class, everything prints out nice. Now, I want to apply a rotation to all the positions. So I set up some rotation matrix R like this: R = numpy.array([[r11,r12,r13], [r21,r22,r23], [r31,r32,r33]]) To apply the matrix to the data in one crunch, I do V = numpy.dot(R, U.transpose()).transpose() Now when I call my C++ function from the Python side, all the data in V is printed, but it has been transposed. So apparently the internal data structure handled by numpy has been reorganized, even though I called transpose() twice, which I would expect to cancel out each other. However, if I do: V = numpy.array(U.transpose()).transpose() and call the C++ routine, everything is perfectly fine, ie. the data structure is as expected. What went wrong? Best regards, Mads -- +-----------------------------------------------------+ | Mads Ipsen | +----------------------+------------------------------+ | Gåsebæksvej 7, 4. tv | | | DK-2500 Valby | phone: +45-29716388 | | Denmark | email: mads.ipsen(a)gmail.com | +----------------------+------------------------------+

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Curious behavior of __radd__
by Andreas Kloeckner Feb. 2, 2012

Feb. 2, 2012

Hi all, here's something I don't understand. Consider the following code snippet: --------------------------------------------------- class A(object): def __radd__(self, other): print(type(other)) import numpy as np np.complex64(1j) + A() --------------------------------------------------- In my world, this should print <type 'numpy.complex64'>. It does print <type 'complex'>. Who is casting my sized complex to a built-in complex, and why? It can be Python's type coercion, because the behavior is the same in Python 3.2. (And the docs say Python 3 doesn't support coercion.) (Please cc me.) Thanks, Andreas

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ufunc delegation to object method
by Warren Weckesser Feb. 2, 2012

Feb. 2, 2012

In the following code, numpy.sin() calls the object's sin() function: In [2]: class Foo(object): ...: def sin(self): ...: return "spam" ...: In [3]: f = Foo() In [4]: np.sin(f) Out[4]: 'spam' Is this, in fact, guaranteed behavior for a ufunc? It does not appear to be documented. This question came up in the discussion of SciPy pull request 138 ( https://github.com/scipy/scipy/pull/138), where the idea is to add numpy unary ufunc support to SciPy's sparse arrays. (Sorry if this email shows up twice. I sent it the first time while the Enthought servers were down, and eventually got an email back saying it had not been sent.) Warren

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