Hi, On Tue, Jul 31, 2012 at 5:01 PM, Vlastimil Brom <vlastimil.b...@gmail.com>wrote:
> 2012/7/31 eat <e.antero.ta...@gmail.com>: > > Hi, > > > > On Tue, Jul 31, 2012 at 10:23 AM, Vlastimil Brom < > vlastimil.b...@gmail.com> > > wrote: > >> > >> 2012/7/30 eat <e.antero.ta...@gmail.com>: > >> > Hi, > >> > > >> > A partial answer to your questions: > >> > > >> > On Mon, Jul 30, 2012 at 10:33 PM, Vlastimil Brom > >> > <vlastimil.b...@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 >
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