[SciPy-user] Python loops too slow

Ross Williamson Ross.Williamson@usap....
Wed Apr 8 18:36:21 CDT 2009


Actually I've got another puzzler:

I currently do the following to get an integrated vector from a 2d array:

for i in xrange(ngrid):
    for j in xrange(ngrid):
       itemp = binned_ell[j,i]
       nsamp[itemp] = nsamp[itemp] + mask[j,i]

where binned_ell is a set of indices and mask is just some numbers - Is 
there a more efficient way of doing this? I'm trying to work it out at 
the moment.

Cheers

Ross
> Hi Guys
>
> Thanks for all your help - My code is now super quick and I'll implement 
> meshgrid in other areas where I've got double loops.
>
> Cheers
>
> Ross
>   
>> Pauli Virtanen <pav <at> iki.fi> writes:
>>   
>>     
>>> i = arange(ngrid/2 + 1)
>>> j = arange(ngrid/2 + 1)
>>> result = 2*pi*hypot(i[None,:], j[:,None])/reso/ngrid
>>>
>>>     
>>>       
>> It seems Pauli's solution may need to be repeated in the final output as below.
>> Is that what you really want though because the submatrix isn't reflected
>> uniformly - i.e. the first row and column are missing.
>>
>> Anyway, timings are below - it seems using meshgrid is more efficient than
>> hypot.
>>
>> -Dave
>>
>> Python 2.5.4 (r254:67916, Dec 23 2008, 15:10:54) [MSC v.1310 32 bit (Intel)]
>> Type "copyright", "credits" or "license" for more information.
>>
>> IPython 0.9.1 -- An enhanced Interactive Python.
>>
>>   Welcome to pylab, a matplotlib-based Python environment.
>>   For more information, type 'help(pylab)'.
>>
>> In [2]: def f1(ngrid, reso):
>>    ...:         result = zeros([ngrid, ngrid])
>>    ...:     for i in arange((ngrid / 2)+1):
>>    ...:            for j in arange((ngrid / 2)+1):
>>    ...:               result[j,i] = 2. * pi * sqrt(i ** 2. + j ** 2) / reso /
>> (ngrid*1.0)
>>    ...:     for i in xrange(ngrid):
>>    ...:            result[i,ngrid / 2+1:] = result[i,1:(ngrid / 2)][::-1]
>>    ...:     for i in xrange(ngrid):
>>    ...:            result[ngrid / 2+1:,i] = result[1:(ngrid / 2),i][::-1]
>>    ...:     return result
>>    ...: #
>>    ...:
>>
>> In [3]: def f2(ngrid, reso):
>>    ...:         nsub = ngrid/2+1
>>    ...:     i = np.arange(nsub)
>>    ...:     j = np.arange(nsub)
>>    ...:     submatrix = 2*pi*np.hypot(i[None,:], j[:,None])/reso/np.float(ngrid)
>>    ...:     result = zeros([ngrid, ngrid])
>>    ...:     result[0:nsub,0:nsub] = submatrix
>>    ...:     result[0:nsub,nsub:] = np.fliplr(submatrix[:,1:-1])
>>    ...:     result[nsub:,:] = np.flipud(result[1:nsub-1,:])
>>    ...:     return result
>>    ...: #
>>    ...:
>>
>> In [4]: def f3(ngrid, reso):
>>    ...:         nsub = ngrid/2+1
>>    ...:     i, j = meshgrid(np.arange(nsub),np.arange(nsub))
>>    ...:     submatrix = 2*pi*np.sqrt(i**2 + j**2)/reso/np.float(ngrid)
>>    ...:     result = zeros([ngrid, ngrid])
>>    ...:     result[0:nsub,0:nsub] = submatrix
>>    ...:     result[0:nsub,nsub:] = np.fliplr(submatrix[:,1:-1])
>>    ...:     result[nsub:,:] = np.flipud(result[1:nsub-1,:])
>>    ...:     return result
>>    ...: #
>>    ...:
>>
>> In [5]: np.allclose(f1(100,1),f2(100,1))
>> Out[5]: True
>>
>> In [6]: np.allclose(f1(100,1),f3(100,1))
>> Out[6]: True
>>
>> In [7]:
>>
>> In [8]: timeit f1(100,1)
>> 10 loops, best of 3: 60.6 ms per loop
>>
>> In [9]:
>>
>> In [10]: timeit f2(100,1)
>> 1000 loops, best of 3: 840 µs per loop
>>
>> In [11]:
>>
>> In [12]: timeit f3(100,1)
>> 1000 loops, best of 3: 258 µs per loop
>>
>> In [13]:
>>
>>
>>
>>
>>
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>>   
>>     
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