[Numpy-discussion] array slicing questions

Vlastimil Brom vlastimil.brom@gmail....
Tue Jul 31 11:20:27 CDT 2012


2012/7/31 eat <e.antero.tammi@gmail.com>:
> Hi,
>
> On Tue, Jul 31, 2012 at 5:01 PM, Vlastimil Brom <vlastimil.brom@gmail.com>
> wrote:
>>
>> 2012/7/31 eat <e.antero.tammi@gmail.com>:
>> > Hi,
>> >
>> > On Tue, Jul 31, 2012 at 10:23 AM, Vlastimil Brom
>> > <vlastimil.brom@gmail.com>
>> > wrote:
>> >>
>> >> 2012/7/30 eat <e.antero.tammi@gmail.com>:
>> >> > Hi,
>> >> >
>> >> > A partial answer to your questions:
>> >> >
>> >> > On Mon, Jul 30, 2012 at 10:33 PM, Vlastimil Brom
>> >> > <vlastimil.brom@gmail.com>
>> >> > wrote:
>> >> >>
>> >> >> Hi all,
>> >> >> I'd like to ask for some hints or advice regarding the usage of
>> >> >> numpy.array and especially  slicing.
>> >> >>
>> >> >> I only recently tried numpy and was impressed by the speedup in some
>> >> >> parts of the code, hence I suspect, that I might miss some other
>> >> >> oportunities in this area.
>> >> >>
>> >> >> I currently use the following code for a simple visualisation of the
>> >> >> search matches within the text, the arrays are generally much larger
>> >> >> than the sample - the texts size is generally hundreds of kilobytes
>> >> >> up
>> >> >> to a few MB - with an index position for each character.
>> >> >> First there is a list of spans(obtained form the regex match
>> >> >> objects),
>> >> >> the respective character indices in between these slices should be
>> >> >> set
>> >> >> to 1:
>> >> >>
>> >> >> >>> import numpy
>> >> >> >>> characters_matches = numpy.zeros(10)
>> >> >> >>> matches_spans = numpy.array([[2,4], [5,9]])
>> >> >> >>> for start, stop in matches_spans:
>> >> >> ...     characters_matches[start:stop] = 1
>> >> >> ...
>> >> >> >>> characters_matches
>> >> >> array([ 0.,  0.,  1.,  1.,  0.,  1.,  1.,  1.,  1.,  0.])
>> >> >>
>> >> >> Is there maybe a way tu achieve this in a numpy-only way - without
>> >> >> the
>> >> >> python loop?
>> >> >> (I got the impression, the powerful slicing capabilities could make
>> >> >> it
>> >> >> possible, bud haven't found this kind of solution.)
>> >> >>
>> >> >>
>> >> >> In the next piece of code all the character positions are evaluated
>> >> >> with their "neighbourhood" and a kind of running proportions of the
>> >> >> matched text parts are computed (the checks_distance could be
>> >> >> generally up to the order of the half the text length, usually less
>> >> >> :
>> >> >>
>> >> >> >>>
>> >> >> >>> check_distance = 1
>> >> >> >>> floating_checks_proportions = []
>> >> >> >>> for i in numpy.arange(len(characters_matches)):
>> >> >> ...     lo = i - check_distance
>> >> >> ...     if lo < 0:
>> >> >> ...         lo = None
>> >> >> ...     hi = i + check_distance + 1
>> >> >> ...     checked_sublist = characters_matches[lo:hi]
>> >> >> ...     proportion = (checked_sublist.sum() / (check_distance * 2 +
>> >> >> 1.0))
>> >> >> ...     floating_checks_proportions.append(proportion)
>> >> >> ...
>> >> >> >>> floating_checks_proportions
>> >> >> [0.0, 0.33333333333333331, 0.66666666666666663, 0.66666666666666663,
>> >> >> 0.66666666666666663, 0.66666666666666663, 1.0, 1.0,
>> >> >> 0.66666666666666663, 0.33333333333333331]
>> >> >> >>>
>> >> >
>> >> > Define a function for proportions:
>> >> >
>> >> > from numpy import r_
>> >> >
>> >> > from numpy.lib.stride_tricks import as_strided as ast
>> >> >
>> >> > def proportions(matches, distance= 1):
>> >> >
>> >> >     cd, cd2p1, s= distance, 2* distance+ 1, matches.strides[0]
>> >> >
>> >> >     # pad
>> >> >
>> >> >     m= r_[[0.]* cd, matches, [0.]* cd]
>> >> >
>> >> >     # create a suitable view
>> >> >
>> >> >     m= ast(m, shape= (m.shape[0], cd2p1), strides= (s, s))
>> >> >
>> >> >     # average
>> >> >
>> >> >     return m[:-2* cd].sum(1)/ cd2p1
>> >> > and use it like:
>> >> > In []: matches
>> >> > Out[]: array([ 0.,  0.,  1.,  1.,  0.,  1.,  1.,  1.,  1.,  0.])
>> >> >
>> >> > In []: proportions(matches).round(2)
>> >> > Out[]: array([ 0.  ,  0.33,  0.67,  0.67,  0.67,  0.67,  1.  ,  1.  ,
>> >> > 0.67,
>> >> > 0.33])
>> >> > In []: proportions(matches, 5).round(2)
>> >> > Out[]: array([ 0.27,  0.36,  0.45,  0.55,  0.55,  0.55,  0.55,  0.55,
>> >> > 0.45,
>> >> > 0.36])
>> >> >>
>> >> >>
>> >> >> I'd like to ask about the possible better approaches, as it doesn't
>> >> >> look very elegant to me, and I obviously don't know the implications
>> >> >> or possible drawbacks of numpy arrays in some scenarios.
>> >> >>
>> >> >> the pattern
>> >> >> for i in range(len(...)): is usually considered inadequate in
>> >> >> python,
>> >> >> but what should be used in this case as the indices are primarily
>> >> >> needed?
>> >> >> is something to be gained or lost using (x)range or np.arange as the
>> >> >> python loop is (probably?) inevitable anyway?
>> >> >
>> >> > Here np.arange(.) will create a new array and potentially wasting
>> >> > memory
>> >> > if
>> >> > it's not otherwise used. IMO nothing wrong looping with xrange(.) (if
>> >> > you
>> >> > really need to loop ;).
>> >> >>
>> >> >> Is there some mor elegant way to check for the "underflowing" lower
>> >> >> bound "lo" to replace with None?
>> >> >>
>> >> >> Is it significant, which container is used to collect the results of
>> >> >> the computation in the python loop - i.e. python list or a numpy
>> >> >> array?
>> >> >> (Could possibly matplotlib cooperate better with either container?)
>> >> >>
>> >> >> And of course, are there maybe other things, which should be made
>> >> >> better/differently?
>> >> >>
>> >> >> (using Numpy 1.6.2, python 2.7.3, win XP)
>> >> >
>> >> >
>> >> > My 2 cents,
>> >> > -eat
>> >> >>
>> >> >> Thanks in advance for any hints or suggestions,
>> >> >>    regards,
>> >> >>   Vlastimil Brom
>> >> >> _______________________________________________
>> >> >> NumPy-Discussion mailing list
>> >> >> NumPy-Discussion@scipy.org
>> >> >> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>> >> >
>> >> Hi,
>> >> thank you very much for your suggestions!
>> >>
>> >> do I understand it correctly, that I have to special-case the function
>> >> for distance = 0 (which should return the matches themselves without
>> >> recalculation)?
>> >
>> > Yes.
>> >>
>> >>
>> >> However, more importantly, I am getting a ValueError for some larger,
>> >> (but not completely unreasonable) "distance"
>> >>
>> >> >>> proportions(matches, distance= 8190)
>> >> Traceback (most recent call last):
>> >>   File "<input>", line 1, in <module>
>> >>   File "<input>", line 11, in proportions
>> >>   File "C:\Python27\lib\site-packages\numpy\lib\stride_tricks.py",
>> >> line 28, in as_strided
>> >>     return np.asarray(DummyArray(interface, base=x))
>> >>   File "C:\Python27\lib\site-packages\numpy\core\numeric.py", line
>> >> 235, in asarray
>> >>     return array(a, dtype, copy=False, order=order)
>> >> ValueError: array is too big.
>> >> >>>
>> >>
>> >> the distance= 8189 was the largest which worked in this snippet,
>> >> however, it might be data-dependent, as I got this error as well e.g.
>> >> for distance=4529 for a 20k text.
>> >>
>> >> Is this implementation-limited, or could it be solved in some
>> >> alternative way which wouldn't have such limits (up to the order of,
>> >> say, millions)?
>> >
>> > Apparently ast(.) does not return a view of the original matches rather
>> > a
>> > copy of size (n* (2* distance+ 1)), thus you may run out of memory.
>> >
>> > Surely it can be solved up to millions of matches, but perhaps much
>> > slower
>> > speed.
>> >
>> >
>> > Regards,
>> > -eat
>> >>
>> >>
>> >> Thanks again
>> >>   regards
>> >>     vbr
>> >>
>> >> _______________________________________________
>> >> NumPy-Discussion mailing list
>> >> NumPy-Discussion@scipy.org
>> >> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>> >
>>
>> Thank you for the confirmation,
>> I'll wait and see, whether the current speed isn't actually already
>> acceptable for the most cases...
>> I could already gain a speedup by using the array.sum()  and other
>> features, maybe I will find yet other possibilities.
>
> I just cooked up some pure pyhton and running sum based solution, which
> actually may be faster and it scales quite well up to millions of matches:
>
> def proportions_p(matches, distance= 1):
>
>     cd, cd2p1= distance, 2* distance+ 1
>
>     m, r= [0]* cd+ matches+ [0]* (cd+ 1), [0]* len(matches)
>
>     s= sum(m[:cd2p1])
>
>     for k in xrange(len(matches)):
>
>         r[k]= s/ cd2p1
>
>         s-= m[k]
>
>         s+= m[cd2p1+ k]
>
>     return r
>
>
> Some verification and timings:
> In []: a= arange(1, 100000, dtype= float)
> In []: allclose(proportions(a, 1000), proportions_p(a.tolist(), 1000))
> Out[]: True
>
> In []: %timeit proportions(a, 1000)
> 1 loops, best of 3: 288 ms per loop
> In []: %timeit proportions_p(a.tolist(), 1000)
> 10 loops, best of 3: 66.2 ms per loop
>
> In []: a= arange(1, 1000000, dtype= float)
> In []: %timeit proportions(a, 10000)
> ------------------------------------------------------------
> Traceback (most recent call last):
> [snip]
> ValueError: array is too big.
>
> In []: %timeit proportions_p(a.tolist(), 10000)
> 1 loops, best of 3: 680 ms per loop
>
>
> Regards,
> -eat
>>
>>
>> regards,
>>
>>     vbr
>> _______________________________________________
>> NumPy-Discussion mailing list
>> NumPy-Discussion@scipy.org
>> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>
>
Thanks for further assistance;
I hope, I am not misunderstanding something in the code, but this
calculation of proportions is supposed to be run over an array of
either 0 or 1, rather than a range;
a test data would be something like:

import random
test_lst = [0,1]*500
random.shuffle(test_lst)

For this data, I am not getting the same results like with my
previously posted python function.

Thanks and regards
 vbr


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