[Numpy-discussion] Array concatenation performance

John Porter jporter@cambridgesys....
Thu Jul 15 10:05:55 CDT 2010


You're right - I screwed up the timing for the one that works...
It does seem to be faster.

I've always just built arrays using nx.array([]) in the past though
and was surprised
that it performs so badly.


On Thu, Jul 15, 2010 at 2:41 PM, Skipper Seabold <jsseabold@gmail.com> wrote:
> On Thu, Jul 15, 2010 at 5:54 AM, John Porter <jporter@cambridgesys.com> wrote:
>> Has anyone got any advice about array creation. I've been using numpy
>> for a long time and have just noticed something unexpected about array
>> concatenation.
>>
>> It seems that using numpy.array([a,b,c]) is around 20 times slower
>> than creating an empty array and adding the individual elements.
>>
>> Other things that don't work well either:
>>    numpy.concatenate([a,b,c]).reshape(3,-1)
>>    numpy.concatenate([[a],[b],[c]))
>>
>> Is there a better way to efficiently create the array ?
>>
>
> What was your timing for concatenate?  It wins for me given the shape of a.
>
> In [1]: import numpy as np
>
> In [2]: a = np.arange(1000*1000)
>
> In [3]: timeit b0 = np.array([a,a,a])
> 1 loops, best of 3: 216 ms per loop
>
> In [4]: timeit b1 = np.empty(((3,)+a.shape)); b1[0]=a;b1[1]=a;b1[2]=a
> 100 loops, best of 3: 19.3 ms per loop
>
> In [5]: timeit b2 = np.c_[a,a,a].T
> 10 loops, best of 3: 30.5 ms per loop
>
> In [6]: timeit b3 = np.concatenate([a,a,a]).reshape(3,-1)
> 100 loops, best of 3: 9.33 ms per loop
>
> Skipper
>
>> See the following snippet:
>> ---------------------------------------
>> import time
>> import numpy as nx
>> print 'numpy version', nx.version.version
>> t = time.time()
>> # test array
>> a = nx.arange(1000*1000)
>> print 'a ',time.time()-t
>> t = time.time()
>> # create array in the normal way..
>> b0 = nx.array([a,a,a])
>> print 'b0',time.time()-t
>> t = time.time()
>> # create using empty array
>> b1 = nx.empty((3,)+(a.shape))
>> b1[0] = a
>> b1[1] = a
>> b1[2] = a
>> print 'b1',time.time()-t
>> print nx.all((b0==b1))
>> -----------------------------------------
>> Produces the output:
>>   numpy version 1.3.0
>>   a  0.0019519329071
>>   b0 0.286643981934
>>   b1 0.0116579532623
>>   equal True
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