[Numpy-discussion] Adding a 2D with a 1D array...
Dag Sverre Seljebotn
dagss@student.matnat.uio...
Wed Sep 9 13:17:20 CDT 2009
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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Dag Sverre
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