[Numpy-discussion] Adding a 2D with a 1D array...

Ruben Salvador rsalvador.wk@gmail....
Thu Sep 10 03:40:02 CDT 2009


OK. I get the idea, but I can't see it. In both cases, as the print
statement shows, offspr is already created.

I need light :S

On Wed, Sep 9, 2009 at 8:17 PM, Dag Sverre Seljebotn <
dagss@student.matnat.uio.no> wrote:

> Ruben Salvador wrote:
> > Your results are what I expected...but. This code is called from my main
> > program, and what I have in there (output array already created for both
> > cases) is:
> >
> > print "lambd", lambd
> > print "np.shape(a)", np.shape(a)
> > print "np.shape(r)", np.shape(r)
> > print "np.shape(offspr)", np.shape(offspr)
> > t = clock()
> > for i in range(lambd):
> >     offspr[i] = r[i] + a[i]
> > t1 = clock() - t
> > print "For loop time ==> %.8f seconds" % t1
> > t2 = clock()
> > offspr = r + a[:,None]
> > t3 = clock() - t2
> > print "Pythonic time ==> %.8f seconds" % t3
> >
> > The results I obtain are:
> >
> > lambd 80000
> > np.shape(a) (80000,)
> > np.shape(r) (80000, 26)
> > np.shape(offspr) (80000, 26)
> > For loop time ==> 0.34528804 seconds
> > Pythonic time ==> 0.35956192 seconds
> >
> > Maybe I'm not measuring properly, so, how should I do it?
>
> Like Luca said, you are not including the creation time of offspr in the
> for-loop version. A fairer comparison would be

offspr[...] = r + a[:, None]
>
> Even fairer (one less temporary copy):
>
> offspr[...] = r
> offspr += a[:, None]
>
> Of course, see how the trend is for larger N as well.
>
> Also your timings are a bit crude (though this depends on how many times
> you ran your script to check :-)). To get better measurements, use the
> timeit module, or (easier) IPython and the %timeit command.
>
> >
> > On Wed, Sep 9, 2009 at 1:20 PM, Citi, Luca <lciti@essex.ac.uk
> > <mailto:lciti@essex.ac.uk>> wrote:
> >
> >     I am sorry but it doesn't make much sense.
> >     How do you measure the performance?
> >     Are you sure you include the creation of the "c" output array in the
> >     time spent (which is outside the for loop but should be considered
> >     anyway)?
> >
> >     Here are my results...
> >
> >     In [84]: a = np.random.rand(8,26)
> >
> >     In [85]: b = np.random.rand(8)
> >
> >     In [86]: def o(a,b):
> >       ....:     c = np.empty_like(a)
> >       ....:     for i in range(len(a)):
> >       ....:             c[i] = a[i] + b[i]
> >       ....:     return c
> >       ....:
> >
> >     In [87]: d = a + b[:,None]
> >
> >     In [88]: (d == o(a,b)).all()
> >     Out[88]: True
> >
> >     In [89]: %timeit o(a,b)
> >     %ti10000 loops, best of 3: 36.8 µs per loop
> >
> >     In [90]: %timeit d = a + b[:,None]
> >     100000 loops, best of 3: 5.17 µs per loop
> >
> >     In [91]: a = np.random.rand(80000,26)
> >
> >     In [92]: b = np.random.rand(80000)
> >
> >     In [93]: %timeit o(a,b)
> >     %ti10 loops, best of 3: 287 ms per loop
> >
> >     In [94]: %timeit d = a + b[:,None]
> >     100 loops, best of 3: 15.4 ms per loop
> >
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> >
> >
> >
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>
> --
> Dag Sverre
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