[Numpy-discussion] array slicing questions
eat
e.antero.tammi@gmail....
Tue Jul 31 11:49:13 CDT 2012
Hi,
On Tue, Jul 31, 2012 at 7:20 PM, Vlastimil Brom <vlastimil.brom@gmail.com>wrote:
> 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:
>
It really shouldn't matter what the test case numbers are. We are just
calculating averages over of certain 'window', like:
In []: proportions(arange(1, 4, dtype= float))
Out[]: array([ 1. , 2. , 1.66666667])
In []: (0.+ 1.+ 2.)/ 3
Out[]: 1.0
In []: (1.+ 2.+ 3.)/ 3
Out[]: 2.0
In []: (2.+ 3.+ 0.)/ 3
Out[]: 1.6666666666666667
Regards,
-eat
>
> 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
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion@scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>
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