[Numpy-discussion] Multiple Boolean Operations

Andrea Gavana andrea.gavana@gmail....
Thu May 22 17:19:22 CDT 2008

Hi Chris and All,

On Thu, May 22, 2008 at 8:40 PM, Christopher Barker wrote:
> Andrea Gavana wrote:
>> By the way, about the solution Francesc posted:
>> xyzReq = (xCent >= xMin) & (xCent <= xMax) &  \
>>             (yCent >= yMin) & (yCent <= yMax) &  \
>>             (zCent >= zMin) & (zCent <= zMax)
>> xyzReq = numpy.nonzero(xyzReq)[0]
>> Do you think is there any chance that a C extension (or something
>> similar) could be faster?
> yep -- if I've be got this right, the above creates 7 temporary arrays.
> creating that many and pushing the data in and out of memory can be
> pretty slow for large arrays.
> In C, C++, Cython or Fortran, you can just do one loop, and one output
> array. It should be much faster for the big arrays.

Well, I have implemented it in 2 ways in Fortran, and actually the
Fortran solutions are slower than the numpy one (2 and 3 times slower
respectively). I attach the source code of the timing code and the 5
implementations I have at the moment (I have included Nathan's
implementation, which is as fast as Francesc's one but it has the
advantage of saving memory).

The timing I get on my home PC are:

Andrea's Solution: 0.42807561 Seconds/Trial
Francesc's Solution: 0.018297884 Seconds/Trial
Fortran Solution 1: 0.035862072 Seconds/Trial
Fortran Solution 2: 0.029822338 Seconds/Trial
Nathan's Solution: 0.018930507 Seconds/Trial

Maybe my fortran coding is sloppy but I don't really know fortran so
well to implement it better...

Thank you so much to everybody for your suggestions so far :-D


"Imagination Is The Only Weapon In The War Against Reality."
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