Auto-answer, sorry, Well, actually I made a mistake lower... that you may have noticed... On the faster (your) solution, even with a cleaner use of the out parameter, the fact that the all has then to be used with parameter axis=0 takes more time and makes it actually slower than the initial solution...
So I may go for the "multiplier" solution. Regards, Matthieu On Tuesday 20 March 2012 13:13:15 you wrote: > Hi Richard, > > Thanks for your answer and the related help ! > > In fact, I was hoping to have a less memory and more speed solution. > Something equivalent to a "raster calculator" for numpy. Wouldn't it make > sense to have some optimized function to work on more than 2 arrays for > numpy anyway ? > > At the end, I am rather interested by more speed. > > I tried first a code-sparing version : > array = numpy.asarray([(aBlueChannel < 1.0),(aNirChannel > aBlueChannel * > 1.0),(aNirChannel < aBlueChannel * 1.8)]).all() > > But this one is at the end more than 2 times slower than : > array1 = numpy.empty([3,6566,6682], dtype=numpy.bool) > numpy.less(aBlueChannel, 1.0, out=array1[0]) > numpy.greater(aNirChannel, (aBlueChannel * 1.0), out=array1[1]) > numpy.less(aNirChannel, (aBlueChannel * 1.8), out=array1[2]) > array = array1.all() > > (and this solution is about 30% faster than the original one) > > I could find another way which was fine for me too: > array = (aBlueChannel < 1.0) * (aNirChannel > (aBlueChannel * 1.0)) * > (aNirChannel < (aBlueChannel * 1.8)) > > But this one is only 5-10% faster than the original solution, even if > probably using less memory than the 2 previous ones. (same was possible > with operator +, but slower than operator *) > > Regards, > Matthieu Rigal > > On Monday 19 March 2012 18:00:02 [email protected] wrote: > > Message: 2 > > Date: Mon, 19 Mar 2012 13:20:23 +0000 > > From: Richard Hattersley <[email protected]> > > Subject: Re: [Numpy-discussion] Using logical function on more than 2 > > arrays, availability of a "between" function ? > > To: Discussion of Numerical Python <[email protected]> > > Message-ID: > > > > <CAP=RS9=UBOc6Kmtmnne7W093t19w=T=osrxuaw0wf8b49hq...@mail.gmail.com > > > > > Content-Type: text/plain; charset=ISO-8859-1 > > > > What do you mean by "efficient"? Are you trying to get it execute > > faster? Or using less memory? Or have more concise source code? > > > > Less memory: > > - numpy.vectorize would let you get to the end result without any > > intermediate arrays but will be slow. > > - Using the "out" parameter of numpy.logical_and will let you avoid > > one of the intermediate arrays. > > > > More speed?: > > Perhaps putting all three boolean temporary results into a single > > boolean array (using the "out" parameter of numpy.greater, etc) and > > using numpy.all might benefit from logical short-circuiting. > > > > And watch out for divide-by-zero from "aNirChannel/aBlueChannel". > > > > Regards, > > Richard Hattersley > RapidEye AG Molkenmarkt 30 14776 Brandenburg an der Havel Germany Follow us on Twitter! www.twitter.com/rapideye_ag Head Office/Sitz der Gesellschaft: Brandenburg an der Havel Management Board/Vorstand: Ryan Johnson Chairman of Supervisory Board/Vorsitzender des Aufsichtsrates: Robert Johnson Commercial Register/Handelsregister Potsdam HRB 24742 P Tax Number/Steuernummer: 048/100/00053 VAT-Ident-Number/Ust.-ID: DE 199331235 DIN EN ISO 9001 certified _______________________________________________ NumPy-Discussion mailing list [email protected] http://mail.scipy.org/mailman/listinfo/numpy-discussion
