Mailman 3 August 2006 - NumPy-Discussion

Backward compatibility plans
by Travis Oliphant Aug. 4, 2006

Aug. 4, 2006

For backward-compatibility with Numeric and Numarray I'm leaning to the following plan: * Do not create compatibility array objects. I initially thought we could sub-class in order to create objects that had the expected attributes and methods of Numeric arrays or Numarray arrays. After some experimentation, I'm ditching this plan. I think this would create too many array-like objects floating around and make unification even harder as these objects interact in difficult-to-predict ways. Instead, I'm planning to: 1) Create compatibility functions in oldnumeric and numarray sub-packages that create NumPy arrays but do it with the same function syntax as the old packages. 2) Create 4 scripts for assisting in conversion (2 for Numeric and 2 for Numarray). a) An initial script that just alters imports (to the compatibility layer) and fixes method and attribute access. b) A secondary script that alters the imports from the compatibility layer and fixes as much as possible the things that need to change in order to make the switch away from the compatibility layer to work correctly. While it is not foolproof, I think this will cover most of the issues and make conversion relatively easy. This will also let us develop NumPy without undue concern for compatibility with older packages. This must all be in place before 1.0 release candidate 1 comes out. Comments and criticisms welcome. -Travis

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a**2 60 times slower than a*a - ONLY for int32
by Sebastian Haase Aug. 4, 2006

Aug. 4, 2006

Hi, >>> a=N.random.poisson(N.arange(1e6)+1) >>> U.timeIt('a**2') 0.59 >>> U.timeIt('a*a') 0.01 >>> a.dtype int32 my U.timeIt function just returns the difference of time in seconds before and after evaluation of the string. For >>> c=N.random.normal(1000, 100, 1e6) >>> c.dtype float64 i get .014 seconds for either c*c or c**2 (I averaged over 100 runs). After converting this to float32 I get 0.008 secs for both. Can the int32 case be speed up the same way !? Thanks, Sebastian Haase

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Mean of n values within an array
by Phil Ruggera Aug. 4, 2006

Aug. 4, 2006

Tweek2 is slightly faster, but does not produce the same result as the regular python baseline: regular python took: 11.997997 sec. numpy convolve took: 0.611996 sec. numpy convolve tweek 1 took: 0.442029 sec. numpy convolve tweek 2 took: 0.418857 sec. Traceback (most recent call last): File "G:\Python\Dev\mean.py", line 57, in ? numpy.testing.assert_equal(reg, np3) File "C:\Python24\Lib\site-packages\numpy\testing\utils.py", line 130, in assert_equal return assert_array_equal(actual, desired, err_msg) File "C:\Python24\Lib\site-packages\numpy\testing\utils.py", line 217, in assert_array_equal assert cond,\ AssertionError: Arrays are not equal (mismatch 17.1428571429%): Array 1: [ 0.0000000000000000e+00 6.5000000000000002e-01 1.3000000000000000e+00 ..., 1.7842500000000002e+03 1.785550000... Array 2: [ 0.0000000000000000e+00 6.5000000000000002e-01 1.3000000000000000e+00 ..., 1.7842500000000002e+03 1.785550000... Code: # mean of n values within an array import numpy, time def nmean(list,n): a = [] for i in range(1,len(list)+1): start = i-n divisor = n if start < 0: start = 0 divisor = i a.append(sum(list[start:i])/divisor) return a def testNP(code, text): start = time.clock() for x in range(1000): np = code(t,50) print text, "took: %f sec."%(time.clock() - start) return np t = [1.3*i for i in range(1400)] reg = testNP(nmean, 'regular python') t = numpy.array(t,dtype=float) def numpy_nmean_conv(list,n): b = numpy.ones(n,dtype=float) a = numpy.convolve(list,b,mode="full") for i in range(n): a[i] /= i + 1 a[n:] /= n return a[:len(list)] np1 = testNP(numpy_nmean_conv, 'numpy convolve') def numpy_nmean_conv_nl_tweak1(list,n): b = numpy.ones(n,dtype=float) a = numpy.convolve(list,b,mode="full") a[:n] /= numpy.arange(1, n+1) a[n:] /= n return a[:len(list)] np2 = testNP(numpy_nmean_conv_nl_tweak1, 'numpy convolve tweek 1') def numpy_nmean_conv_nl_tweak2(list,n): b = numpy.ones(n,dtype=float) a = numpy.convolve(list,b,mode="full") a[:n] /= numpy.arange(1, n + 1) a[n:] *= 1.0/n return a[:len(list)] np3 = testNP(numpy_nmean_conv_nl_tweak2, 'numpy convolve tweek 2') numpy.testing.assert_equal(reg, np1) numpy.testing.assert_equal(reg, np2) numpy.testing.assert_equal(reg, np3) On 8/3/06, Charles R Harris <charlesr.harris(a)gmail.com> wrote: > Hi Scott, > > > On 8/3/06, Scott Ransom <sransom(a)nrao.edu> wrote: > > You should be able to modify the kernel so that you can avoid > > many of the divides at the end. Something like: > > > > def numpy_nmean_conv_nl2(list,n): > > b = numpy.ones(n,dtype=float) / n > > a = numpy.convolve (c,b,mode="full") > > # Note: something magic in here to fix the first 'n' values > > return a[:len(list)] > > > Yep, I tried that but it wasn't any faster. It might help for really *big* > arrays. The first n-1 values still need to be fixed after. > > Chuck > > > I played with it a bit, but don't have time to figure out exactly > > how convolve is mangling the first n return values... > > > > Scott > > > > > > > > On Thu, Aug 03, 2006 at 09:38:25AM -0600, Charles R Harris wrote: > > > Heh, > > > > > > This is fun. Two more variations with 1000 reps instead of 100 for > better > > > timing: > > > > > > def numpy_nmean_conv_nl_tweak1(list,n): > > > b = numpy.ones(n,dtype=float) > > > a = numpy.convolve(list,b,mode="full") > > > a[:n] /= numpy.arange(1, n + 1) > > > a[n:] /= n > > > return a[:len(list)] > > > > > > def numpy_nmean_conv_nl_tweak2(list,n): > > > b = numpy.ones(n,dtype=float) > > > a = numpy.convolve(list,b,mode="full") > > > a[:n] /= numpy.arange(1, n + 1) > > > a[n:] *= 1.0/n > > > return a[:len(list)] > > > > > > Which gives > > > > > > numpy convolve took: 2.630000 sec. > > > numpy convolve noloop took: 0.320000 sec. > > > numpy convolve noloop tweak1 took: 0.250000 sec. > > > numpy convolve noloop tweak2 took: 0.240000 sec. > > > > > > Chuck > > > > > > On 8/2/06, Phil Ruggera <pruggera(a)gmail.com> wrote: > > > > > > > >A variation of the proposed convolve routine is very fast: > > > > > > > >regular python took: 1.150214 sec. > > > >numpy mean slice took: 2.427513 sec. > > > >numpy convolve took: 0.546854 sec. > > > >numpy convolve noloop took: 0.058611 sec. > > > > > > > >Code: > > > > > > > ># mean of n values within an array > > > >import numpy, time > > > >def nmean(list,n): > > > > a = [] > > > > for i in range(1,len(list)+1): > > > > start = i-n > > > > divisor = n > > > > if start < 0: > > > > start = 0 > > > > divisor = i > > > > a.append(sum(list[start:i])/divisor) > > > > return a > > > > > > > >t = [1.0*i for i in range(1400)] > > > >start = time.clock() > > > >for x in range(100): > > > > reg = nmean(t,50) > > > >print "regular python took: %f sec."%(time.clock() - start) > > > > > > > >def numpy_nmean(list,n): > > > > a = numpy.empty(len(list),dtype=float) > > > > for i in range(1,len(list)+1): > > > > start = i-n > > > > if start < 0: > > > > start = 0 > > > > a[i-1] = list[start:i].mean(0) > > > > return a > > > > > > > >t = numpy.arange (0,1400,dtype=float) > > > >start = time.clock() > > > >for x in range(100): > > > > npm = numpy_nmean(t,50) > > > >print "numpy mean slice took: %f sec."%(time.clock() - start) > > > > > > > >def numpy_nmean_conv(list,n): > > > > b = numpy.ones(n,dtype=float) > > > > a = numpy.convolve(list,b,mode="full") > > > > for i in range(0,len(list)): > > > > if i < n : > > > > a[i] /= i + 1 > > > > else : > > > > a[i] /= n > > > > return a[:len(list)] > > > > > > > >t = numpy.arange(0,1400,dtype=float) > > > >start = time.clock () > > > >for x in range(100): > > > > npc = numpy_nmean_conv(t,50) > > > >print "numpy convolve took: %f sec."%(time.clock() - start) > > > > > > > >def numpy_nmean_conv_nl(list,n): > > > > b = numpy.ones(n,dtype=float) > > > > a = numpy.convolve(list,b,mode="full") > > > > for i in range(n): > > > > a[i] /= i + 1 > > > > a[n:] /= n > > > > return a[:len(list)] > > > > > > > >t = numpy.arange(0,1400,dtype=float) > > > >start = time.clock() > > > >for x in range(100): > > > > npn = numpy_nmean_conv_nl(t,50) > > > >print "numpy convolve noloop took: %f sec."%( time.clock() - start) > > > > > > > >numpy.testing.assert_equal(reg,npm) > > > >numpy.testing.assert_equal(reg,npc) > > > >numpy.testing.assert_equal(reg,npn) > > > > > > > >On 7/29/06, David Grant < davidgrant(a)gmail.com> wrote: > > > >> > > > >> > > > >> > > > >> On 7/29/06, Charles R Harris <charlesr.harris(a)gmail.com > wrote: > > > >> > > > > >> > Hmmm, > > > >> > > > > >> > I rewrote the subroutine a bit. > > > >> > > > > >> > > > > >> > def numpy_nmean(list,n): > > > >> > a = numpy.empty(len(list),dtype=float) > > > >> > > > > >> > b = numpy.cumsum(list) > > > >> > for i in range(0,len(list)): > > > >> > if i < n : > > > >> > a[i] = b[i]/(i+1) > > > >> > else : > > > >> > a[i] = (b[i] - b[i-n])/(i+1) > > > >> > return a > > > >> > > > > >> > and got > > > >> > > > > >> > regular python took: 0.750000 sec. > > > >> > numpy took: 0.380000 sec. > > > >> > > > >> > > > >> I got rid of the for loop entirely. Usually this is the thing to do, > at > > > >> least this will always give speedups in Matlab and also in my limited > > > >> experience with Numpy/Numeric: > > > >> > > > >> def numpy_nmean2(list,n): > > > >> > > > >> a = numpy.empty(len(list),dtype=float) > > > >> b = numpy.cumsum(list) > > > >> c = concatenate((b[n:],b[:n])) > > > >> a[:n] = b[:n]/(i+1) > > > >> a[n:] = (b[n:] - c[n:])/(i+1) > > > >> return a > > > >> > > > >> I got no noticeable speedup from doing this which I thought was > pretty > > > >> amazing. I even profiled all the functions, the original, the one > > > >written by > > > >> Charles, and mine, using hotspot just to make sure nothing funny was > > > >going > > > >> on. I guess plain old Python can be better than you'd expect in > certain > > > >> situtations. > > > >> > > > >> -- > > > >> David Grant > > > > > > > > >------------------------------------------------------------------------- > > > >Take Surveys. Earn Cash. Influence the Future of IT > > > >Join SourceForge.net's Techsay panel and you'll get the chance to share > > > >your > > > >opinions on IT & business topics through brief surveys -- and earn cash > > > > > http://www.techsay.com/default.php?page=join.php&p=sourceforge&CID=DEVDEV > > > >_______________________________________________ > > > >Numpy-discussion mailing list > > > > Numpy-discussion(a)lists.sourceforge.net > > > > >https://lists.sourceforge.net/lists/listinfo/numpy-discussion > > > > > > > > > > ------------------------------------------------------------------------- > > > Take Surveys. Earn Cash. Influence the Future of IT > > > Join SourceForge.net's Techsay panel and you'll get the chance to share > your > > > opinions on IT & business topics through brief surveys -- and earn cash > > > > http://www.techsay.com/default.php?page=join.php&p=sourceforge&CID=DEVDEV > > > _______________________________________________ > > > Numpy-discussion mailing list > > > Numpy-discussion(a)lists.sourceforge.net > > > > https://lists.sourceforge.net/lists/listinfo/numpy-discussion > > > > > > -- > > -- > > Scott M. Ransom Address: NRAO > > Phone: (434) 296-0320 520 Edgemont Rd. > > email: sransom(a)nrao.edu Charlottesville, VA 22903 USA > > GPG Fingerprint: 06A9 9553 78BE 16DB 407B FFCA 9BFA B6FF FFD3 2989

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bug tracker to cc email address by default
by Sebastian Haase Aug. 4, 2006

Aug. 4, 2006

Hi, Is it possible to have 'cc'-ing the poster of a bug ticket be the default !? Or is/can this be set in a per user preference somehow ? Thanks, Sebastian Haase

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link to numpy ticket tracker ontp the wiki
by Sebastian Haase Aug. 4, 2006

Aug. 4, 2006

Hi! I would like to suggest to put a link to the bug/wishlist tracker web site on the scipy.org wiki site. http://projects.scipy.org/scipy/numpy/ticket I did not do it myself because I could not decide what the best place for it would - I think it should be rather exposed ... The only link I could find was somewhere inside an FAQ for the SciPy package and it was only for the scipy-bug tracker. Thanks, Sebastian Haase

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Numpy 1.0b2 for this weekend
by Travis Oliphant Aug. 4, 2006

Aug. 4, 2006

I'd like to release NumPy beta 2.0 on Saturday to get ready for the SciPy 2006 conference. Please post any bugs and commit any fixes by then. I suspect there will be 4 or 5 beta releases and then a couple of release candidates before the final release comes out at the first of October. -Travis

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Re: [Numpy-discussion] (no subject)
by misa Aug. 4, 2006

Aug. 4, 2006

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Re: [Numpy-discussion] help! type 'float64scalar' is not type 'float'
by Travis Oliphant Aug. 3, 2006

Aug. 3, 2006

Sebastian Haase wrote: >On Wednesday 02 August 2006 22:43, Travis Oliphant wrote: >>Sebastian Haase wrote: >>>Thanks, >>>I just found >>>numpy.isscalar() and numpy.issctype() ? >>>These sound like they would do what I need - what is the difference >>>between the two ? >> >>Oh, yeah. >> >>numpy.issctype works with type objects >>numpy.isscalar works with instances >> >>Neither of them distinguish between scalars and "numbers." >> >>If you get errors with isscalar it would be nice to know what they are. > >I'm still trying to reproduce the exception, but here is a first comparison >that - honestly - does not make much sense to me: >(type vs. instance seems to get mostly the same results and why is there a >difference with a string ('12') ) These routines are a little buggy. I've cleaned them up in SVN to reflect what they should do. When the dtype object came into existence a lot of what the scalar types where being used for was no longer needed. Some of these functions weren't updated to deal with the dtype objects correctly either. This is what you get now: >>> import numpy as N >>> N.isscalar(12) True >>> N.issctype(12) False >>> N.isscalar('12') True >>> N.issctype('12') False >>> N.isscalar(N.array([1])) False >>> N.issctype(N.array([1])) False >>> N.isscalar(N.array([1]).dtype) False >>> N.issctype(N.array([1]).dtype) True >>> N.isscalar(N.array([1])[0].dtype) False >>> N.issctype(N.array([1])[0].dtype) True >>> N.isscalar(N.array([1])[0]) True >>> N.issctype(N.array([1])[0]) False -Travis >>>>N.isscalar(12) > >True > >>>>N.issctype(12) > >True > >>>>N.isscalar('12') > >True > >>>>N.issctype('12') > >False > >>>>N.isscalar(N.array([1])) > >False > >>>>N.issctype(N.array([1])) > >True > >>>>N.isscalar(N.array([1]).dtype) > >False > >>>>N.issctype(N.array([1]).dtype) > >False > > # apparently new 'scalars' have a dtype attribute ! > >>>>N.isscalar(N.array([1])[0].dtype) > >False > >>>>N.issctype(N.array([1])[0].dtype) > >False > >>>>N.isscalar(N.array([1])[0]) > >True > >>>>N.issctype(N.array([1])[0]) > >True > >-Sebastian

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Re: [Numpy-discussion] help! type 'float64scalar' is not type 'float'
by Sebastian Haase Aug. 3, 2006

Aug. 3, 2006

On Thursday 03 August 2006 13:19, Travis Oliphant wrote: > Sebastian Haase wrote: > >On Wednesday 02 August 2006 22:43, Travis Oliphant wrote: > >>Sebastian Haase wrote: > >>>Thanks, > >>>I just found > >>>numpy.isscalar() and numpy.issctype() ? > >>>These sound like they would do what I need - what is the difference > >>>between the two ? > >> > >>Oh, yeah. > >> > >>numpy.issctype works with type objects > >>numpy.isscalar works with instances > >> > >>Neither of them distinguish between scalars and "numbers." > >> > >>If you get errors with isscalar it would be nice to know what they are. > > > >I'm still trying to reproduce the exception, but here is a first > > comparison that - honestly - does not make much sense to me: > >(type vs. instance seems to get mostly the same results and why is there > > a difference with a string ('12') ) > > These routines are a little buggy. I've cleaned them up in SVN to > reflect what they should do. When the dtype object came into > existence a lot of what the scalar types where being used for was no > longer needed. Some of these functions weren't updated to deal with > the dtype objects correctly either. > > This is what you get now: > >>> import numpy as N > >>> N.isscalar(12) > > True > > >>> N.issctype(12) > > False > > >>> N.isscalar('12') > > True > > >>> N.issctype('12') > > False > > >>> N.isscalar(N.array([1])) > > False > > >>> N.issctype(N.array([1])) > > False > > >>> N.isscalar(N.array([1]).dtype) > > False > > >>> N.issctype(N.array([1]).dtype) > > True > > >>> N.isscalar(N.array([1])[0].dtype) > > False > > >>> N.issctype(N.array([1])[0].dtype) > > True > > >>> N.isscalar(N.array([1])[0]) > > True > > >>> N.issctype(N.array([1])[0]) > > False > > > -Travis Great! Just wanted to point out that '12' is a scalar - I suppose that's what it is. (To determine if something is a number it seems best to implement a try: ... except: ... something like float(x) - as Chris has suggested ) -S.

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Plenum
by terms inversion Aug. 3, 2006

Aug. 3, 2006

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