[Numpy-discussion] Simple problem. Is it possible without a loop?

josef.pktd@gmai... josef.pktd@gmai...
Wed Jun 9 10:37:48 CDT 2010


On Wed, Jun 9, 2010 at 11:31 AM, Vicente Sole <sole@esrf.fr> wrote:
> It gets even worse with data similar to those I will be using.
>
> With:
>
> x0 = numpy.linspace(-1,1, 10000)
> niter = 2000
>
> I get 24 seconds for option1 and 0.64 seconds for option2.
> Considering I expect between 5 and 50 times that number of iterations,
> the difference is already quite considerable.

but the two options don't produce the same result in general, the
cumsum version doesn't restart from zero, I think

try
x0 = np.random.randint(5,size=30).cumsum()
with delta=3

I don't see a way around recursive looping

Josef


> Armando
>
> Quoting Vicente Sole <sole@esrf.fr>:
>
>>>> ? Well a loop or list comparison seems like a good choice to me. It is
>>>> much more obvious at the expense of two LOCs. Did you profile the two
>>>> possibilities and are they actually performance-critical?
>>>>
>>>> cheers
>>>>
>>
>>
>> The second is between 8 and ten times faster on my machine.
>>
>> import numpy
>> import time
>> x0 = numpy.arange(10000.)
>> niter = 2000   # I expect between 10000 and 100000
>>
>>
>> def option1(x, delta=0.2):
>>      y = [x[0]]
>>      for value in x:
>>          if (value - y[-1]) > delta:
>>              y.append(value)
>>      return numpy.array(y)
>>
>> def option2(x, delta=0.2):
>>      y = numpy.cumsum((x[1:]-x[:-1])/delta).astype(numpy.int)
>>      i1 = numpy.nonzero(y[1:]>  y[:-1])
>>      return numpy.take(x, i1)
>>
>>
>> t0 = time.time()
>> for i in range(niter):
>>      t = option1(x0)
>> print "Elapsed = ", time.time() - t0
>> t0 = time.time()
>> for i in range(niter):
>>      t = option2(x0)
>> print "Elapsed = ", time.time() - t0
>>
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>
>
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