On Jul 19, 2011, at 3:15 PM, Chad Netzer wrote:
> %python
>>>> import timeit
>>>> import numpy as np
>
>>>> t=timeit.Timer('k = m - 0.5', setup='import numpy as np;m =
>>>> np.ones([8092,8092],float); k = np.zeros(m.size, m.dtype)')
>>>> np.mean(t.repeat(repeat=10, number=1))
> 0.58557529449462886
>
>>>> t=timeit.Timer('k = m - 0.5', setup='import numpy as np;m =
>>>> np.ones([8092,8092],float)')
>>>> np.mean(t.repeat(repeat=10, number=1))
> 0.53153839111328127
I am surprised that there is any difference between these two approaches at
all. I would have thought that in both cases a temporary array holding the
result of m-0.5 is created, which is then assigned to the variable k. However,
it seems that the second approach is about 10% faster (I see a similar
difference on my machine). Why would that be?
On the other hand, if I actually use the preallocated space for k, and replace
the assignment by
k[…] = m - 0.5
then this takes about 40% more time, presumably because the content of the
temporary array is copied into k (which must be initialized by m.shape instead
of m.size in this case).
Thanks,
Lutz
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