[Numpy-discussion] Adding a 2D with a 1D array...
Ruben Salvador
rsalvador.wk@gmail....
Wed Sep 9 07:36:21 CDT 2009
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?
On Wed, Sep 9, 2009 at 1:20 PM, Citi, Luca <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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