[Numpy-discussion] question about in-place operations

[Francesc Alted]

On 5/22/12 8:47 PM, Dag Sverre Seljebotn wrote:
> On 05/22/2012 04:54 PM, Massimo DiPierro wrote:
>> For now I will be doing this:
>>
>> import numpy
>> import time
>>
>> a=numpy.zeros(2000000)
>> b=numpy.zeros(2000000)
>> c=1.0
>>
>> # naive solution
>> t0 = time.time()
>> for i in xrange(len(a)):
>>       a[i] += c*b[i]
>> print time.time()-t0
>>
>> # possible solution
>> n=100000
>> t0 = time.time()
>> for i in xrange(0,len(a),n):
>>       a[i:i+n] += c*b[i:i+n]
>> print time.time()-t0
>>
>> the second "possible" solution appears 1000x faster then the former in my tests and uses little extra memory. It is only 2x slower than b*=c.
>>
>> Any reason not to do it?
> No, this is perfectly fine, you just manually did what numexpr does.

Yeah.  You basically re-discovered the blocking technique.  For a more 
general example on how to apply the blocking technique with NumPy see 
the section "CPU vs Memory Benchmark" in:

https://python.g-node.org/python-autumnschool-2010/materials/starving_cpus

Of course numexpr has less overhead (and can use multiple cores) than 
using plain NumPy.

-- 
Francesc Alted