Numerical Python

I'm still here working away at the code cleanup of Numerical Python.  I
know some of you may be interested in the results of a survey related to
that effort. 

Here are the results so far:  

1: How long have your been using NumPy:

19 response:  
<= 1 year:  8
 1-3 years:  6
 4-5 years:  5

2: How important is NumPy to your daily work now? (1-Depend on it, 
      5-Dabble with it)

 1 - 17
 2 -  5
 3 -  5
 4 -  3

Avg: 1.8


3: How would you rate the current functionality of NumPy? (1-Love it,
     5-Hate it)

1 - 2
2 - 19
3 - 8
4 - 1

Avg: 2.27


4: How important is it to you for NumPy to get into the Python core?
     (1-Very Important, 5-Not Important)

1 - 8
2 - 14
3 - 4
3 - 4

Avg: 2.1


5: How much interest do you have in improving NumPy? (1-Unlimited, 5-None)

1 - 10
2 - 13
3 - 7

Avg: 1.9


6: How concerned are you about alterations to the underlying C-structure 
    of NumPy? (1-Very concerned, 5-Don't care))

1 - 3
2 - 2
3 - 8
4 - 5
5 - 12

Avg: 3.7


7:  How important is memory conservation to you? (1-Very important, 5-Not
     important)

1 - 11
2 - 10
3 - 6
4 - 1
5 - 2

Avg: 2.1


8: How important is it to you that the underlying code to NumPy be easy to
   understand? (1-Very important, 5-Not important)

1 - 7
2 - 13
3 - 5
4 - 5

Avg: 2.3


9: How important is it to you that NumPy be fast? (1-Very important, 5-Not
important)

1 - 22
2 - 7
3 - 1

Avg: 1.3


10: How happy are you with the current coercion rules (including 
    spacesaving arrays)? (1-Happy, 5-Miserable)

1 - 3
2 - 13
3 - 10 
4 - 2

Avg: 2.4


11: Should mixed-precision arithmetic cast to the largest memory type
    (yes) or the least memory type (no)?

Yes - 23
No  - 6


12: Should object arrays (typecode='O') remain a part of NumPy? (1-Agree, 
     5-Disagree)

1 - 13
2 - 3
3 - 10
4 - 2

Avg: 2.1


13: Should slices (e.g. x[3:10]) be changed to be copies? 

Yes - 12 
No  - 16


So, as you can see the results were pretty clear in some areas and quite
controversial in other areas.  

Have fun interpreting them...

Travis

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Hassan,
	This seemed relevant to pass back to the numeric list - hope you don't mind. 
My comments follow.

- ------- Forwarded Message

Date:    Wed, 01 Mar 2000 19:46:42 +0000
From:    Hassan Aurag <aurag@...>
To:      zow@..., Gmath-devel@...
Subject: Re: [Gmath-devel] Re: [Numpy-discussion] Derivatives



 I have to agree with you on most counts that Numerical Recipes in C=20
is not a full-blown encyclopedia on all subtleties of doing numerical=20
computations.

 However it does its job greatly for a big class of stuff I am=20
interested in: minimization, non-linear system of equations solving=20
(The newton routine given there is good, accurate and fast.)

 There are errors and problems as in most other numerical books. In=20
truth, I don't think there is anything fully correct out there.

 When trying to make computations you have to do a lot of testing and=20
a lot of thought even if the algorithm seems perfect. That applies to=20
all books, recipes et al.

 I have just discovered that tan(pi/2) =3D 1.63317787284e+16 in=20
numerical python. And we all know this is bad. It should be infinity=20
period. We should define a symbol called infinity and put the correct=20
definition of tan, sin .... for all known angles then interpolate if=20
needed for the rest, or whatever is used to actually compute those thing=
s.

 Peace

<snip>
- ------- End of Forwarded Message

I believe you're actually talking about the math module, not the numeric 
module (I'm not aware of tan or pi definitions in numeric, but I haven't 
bothered to double check that). Never the less, I think it has relevance here 
as Numeric is all about doing serious number crunching. This problem is caused 
by the lack of infinite precision in python. Of course, how is it even 
possible to perform infinite precision on non-rational numbers?

The obvious solution is to allow the routine (tan() in this case) to recognize 
named constants that have relevance in their domain (pi in this case). This 
would fix the problem:

math.tan(math.pi) = -1.22460635382e-16

but it still doesn't solve your problem because the named constant would have 
the mathematical operation performed on it before it's passed into the 
function, ruining whatever intimate knowledge of the given named constant that 
routine has.

Perhaps you could get the routine to recognize rational math on named 
constants (the problem with that solution is how do you not burden other 
routines with the knowledge of how to process that expression). Assuming you 
had that, even for your example, should the answer be positive or negative 
infinity?

Another obvious solution is just to assume that any floating point overflow is 
infinity and any underflow is zero. This obviously won't work because some 
asymptotic functions (say 1/x^3) will overflow or underflow at values for x 
for which the correct answer is not properly infinity or zero.

It is interesting to note that Matlab's behaviour is the same as Python's, 
which would indicate to me that there's not some easy solution to this problem 
that Guido et. al. overlooked. I haven't really researched the problem at all 
(although now I'm interested), but I'd be interested if anyone has a proposal 
for (or reference to) how this problem can be solved in a general purpose 
programming language at all (as there exists the distinct possibility that it 
can not be done in Python without breaking backwards compatibility).

Terry



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Pyfort,  a Python-Fortran connection tool, is now available at
pyfortran.sourceforge.net, instead of xfiles.llnl.gov. Documentation is
available at the site.

Version 3.1 uses distutils for easy extension building. A sample package is in
the test subdirectory.

Support for the pgf90 compiler on Linux is now available. Note that the
limitation to non-explicitly interfaced routines has not been removed yet, but
it will handle using pgf90-compiled F90 files in your Fortran.

The answer is yes!
 It is an 1e10 and not 1e-10. At 0.0 you got to pick the interval yourself. You 
can't just use the starting point x
 
 
> Date: Wed, 1 Mar 2000 16:06:28 +0100 (MET)
> From: Fredrik Stenberg <su96-fst@...>
> To: Hassan Aurag <aurag@...>
> Subject: Re: [Numpy-discussion] Derivatives
> MIME-Version: 1.0
> 
> > Hi,
> > 
> > attached is a file called Derivative.py.
> > 
> > It computes derivatives and is based on an algorithm found in 
> > Numerical Recipes in C.
> > 
> > What to do you think about it and has anyone started a "serious" 
> > calculus oriented subpackage for Numerical Python in general?
> > 
> > I mean: derivatives, partial derivatives, jacobian, hessian 
> > implemented fast and precise.
> > 
> > On another note, why isn't infinity defined in NumPy?
> > 
> > Why is tan(pi/2) a number even if big? Shouldn't it be infinity?
> > 
> >  
> > 
> 
> I tried your algoritm on sin(x) and i got some rather interesting
> results.
> 
> ######### EXAMPLE#############
> from math import sin
> 
> def f(x):
>     return sin(x)
> 
> import Derivative
> 
> print Derivative.Diff(f,0.0)
> 
> ########RESULT################
> -2.03844228853e-10
> 
> It should be approx 1.0 
> 
> 
> 
> I found the error i think..
> check row 28 in Derivative
>         h = random()/1e-10
> should that be h = random()/1e+10??
> 
> 
> Fredrik
>

Forgot to attach file. Here it is

Hi,

attached is a file called Derivative.py.

It computes derivatives and is based on an algorithm found in 
Numerical Recipes in C.

What to do you think about it and has anyone started a "serious" 
calculus oriented subpackage for Numerical Python in general?

I mean: derivatives, partial derivatives, jacobian, hessian 
implemented fast and precise.

On another note, why isn't infinity defined in NumPy?

Why is tan(pi/2) a number even if big? Shouldn't it be infinity?

I installed the Numpy 15.2 and got the following error during the
import of Matrix.  Apparently, the version number is no longer
embedded in the module doc string following the # sign.

>>> import Matrix
Traceback (innermost last):
  File "<stdin>", line 1, in ?
  File "/usr/local/lib/python1.5/site-packages/Numeric/Matrix.py", line 5, in ?
    __version__ = int(__id__[string.index(__id__, '#')+1:-1])
  File "/usr/local/lib/python1.5/string.py", line 138, in index
    return _apply(s.index, args)
ValueError: substring not found in string.index

Jeff

The code sent by Herbert Roitblat pointed out some inconsistencies in the
current NumPy, that I've fixed with two small changes:

1) Long's can no longer be safely cast to Int's (this is not safe on
64-bit machines anyway) -- this makes Numeric more consistent with how it
interprets Python integers.

2) Automatic casting will be done when using rank-0 arrays to set elements
of a Numeric array to be consistent with the behavior for Python scalars.

The changes are in CVS right now, but are simple to change back if there
is a problem.

-Travis

Hi Herb,

It has taken awhile for me to respond to this, but your problem's here
illustrate exactly the kinds of difficulties one encounters with the
current NumPy coercion rules:

You do not have a bad version of Numeric.  The behavior you describe is
exactly what "should" happen though it needs to be fixed.  I'll trace
for you exactly what is going on as it could be illustrative to others:

>>> a = zeros((5,5),'b')
# You've just created a 5x5 byte array that follows "normal" coercion
#   rules filled with zeros.

>>> a[3,3] = 8
# This line copies the rank-0 array of type 'b' created from the Python
#  Integer 8 (by a direct coercion in C) into element (3,3) of matrix a

>>> temp = a[3,3]
# This selects out the rank-0 array of typecode 'b' at position (3,3).  As
# of 15.2 this is nolonger changed to a scalar.  Note that rank-0 arrays
# act alot like scalars, but because there is not a one-to-one
# correspondence between the Python Scalars and rank-0 arrays, this is not
# automatically converted to a Python scalar (this is a change in 15.2)

>>> temp = temp + 3
# This is the problem line for you right here.  Something is wrong though,
# since it should not be, a problem.
# You are adding a rank-0 array of typecode 'b' to a Python Integer which 
# is interpreted by Numeric as a rank-0 array of typecode 'l'.  The result
# should be a Python Integer.  For some reason this is returning an array 
# of typecode 'i' (which does not get automatically converted to a Python
# scalar).

>>> a[3,3] = temp
# This would work fine if temp were the Python scalar it should be. 
# Right now, assignment doesn't let you assign an array of a "larger" type 
# to elements of a smaller type (except for Python scalars).  Since temp
# is (incorrectly I think) a type 'i' rank-0 array, it does not let you 
# make the assignment.  At any rate it is inconsistent to let you assign
# Python scalars but not rank-0 arrays of arbitrary precision, this should
# be fixed.  It is also a problem that temp + 3 returns an array of
# typecode 'i'.

I will look into fixing the above problems this example points out.  Of 
course, it could also be fixed by having long integers lower in the
coercion tree than byte arrays.

Thanks for the feedback,

Travis Oliphant

> We get the type error from trying to set the matrix element with a matrix
> element (apparently).  In the old version (1.9) on our NT box,
> temp=a[kwd,kwd] results in temp being an int type.  How can we either cast
> the temp to an int or enable what we really want, which is to add an int to
> a[kwd,kwd], as in a[kwd,kwd] = a[kwd,kwd] + jwd ?
> 
> Do we have a bad version of Numeric?

Maybe an experimental version. If you check the archives of this
mailing list, you can find a recent discussion about proposed
modifications. One of them was to eliminate the automatic conversion
of rank-0 arrays to scalars, in order to prevent type promotion.
Perhaps this proposal was implemented in the version you have.


Note to the NumPy maintainers: please announce all new releases on
this list, mentioning changes, especially those that affect backward
compatibility. As a maintainer of code that makes heavy use of NumPy,
I keep getting questions and bug reports caused by some new NumPy
release that I haven't even heard of. A recent example is the change
of location of the header files; C modules using arrays now have to
include Numeric/arrayobject.h instead of simply arrayobject.h. I can
understand this change (although I am not sure it's important enough
to break compatibility), but I'd have preferred to learn about it
directly and as early as possible. It's really no fun working through
a 2 KB bug report sent by someone with zero knowledge of C.

Konrad.
-- 
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Konrad Hinsen                            | E-Mail: hinsen@...
Centre de Biophysique Moleculaire (CNRS) | Tel.: +33-2.38.25.55.69
Rue Charles Sadron                       | Fax:  +33-2.38.63.15.17
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Hi There,

I just tried to add SGI running IRIX6.5 to the collection of Unix boxes I will
support and I ran into the following problem:

If I compile Python -O2 loading the Numeric extensions dumps the core,
if I compile Python -g it works just fine and this regardless if Numeric is
compile -g or -O2.

After I re-compiled Objects/complexobject.o using -g (everything else being
compiled -O2) I got it to work ...

did anyone else out there see this kind of behavior ?

I also post this to psa-members just in case this might be Python related


-Michel


-- 

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>>>>>>>>>> AREA CODE CHANGE <<<<<<<<< we are now 858 !!!!!!!

Michel F. Sanner Ph.D.                   The Scripps Research Institute
Assistant Professor			Department of Molecular Biology
					  10550 North Torrey Pines Road
Tel. (858) 784-2341				     La Jolla, CA 92037
Fax. (858) 784-2860
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Adobe Acrobat has a shrink to fit option in their print menu.  I'm not sure
if it comes with their free-reader.
Try printing as a 1up.  It seems a small adaptation.
HLR
-----Original Message-----
From: Peter Chang <peter@...>
To: Numpy-discussion@...
<Numpy-discussion@...>
Date: Friday, February 18, 2000 7:09 AM
Subject: [Numpy-discussion] numpy documentation - alternative format?


>
>Hi there,
>
>I've just started to use python and numpy and want to print out the numpy
>document but the PDF file has a strange aspect ratio which makes it hard
>to print it as 2up on A4 paper. (I've tried hacking about with the
>postscript generated by xpdf but it seems that there is no global setting
>for page size!)
>
>Could the authors please provide alternative formats for the doc, eg.
>as postscript files sized for A4 and letter so that people can print them
>out easier?
>
>Thanks
> Peter
>
>
>
>_______________________________________________
>Numpy-discussion mailing list
>Numpy-discussion@...
>http://lists.sourceforge.net/mailman/listinfo/numpy-discussion
>

Hi there,

I've just started to use python and numpy and want to print out the numpy
document but the PDF file has a strange aspect ratio which makes it hard
to print it as 2up on A4 paper. (I've tried hacking about with the
postscript generated by xpdf but it seems that there is no global setting
for page size!)

Could the authors please provide alternative formats for the doc, eg.
as postscript files sized for A4 and letter so that people can print them
out easier?

Thanks
 Peter

This message was sent from Geocrawler.com by "andrew x swan" <a.swan@...>
Be sure to reply to that address.

hi - i've only just started using python and
numpy... the program i wrote below runs much more
slowly than a fortran equivalent. ie. on a dataset
where the order of the matrix is (3325,3325),
python took this long:

362.25user 0.74system 6:09.78elapsed 98%CPU 

and fortran took this long:

2.68user 1.12system 0:03.89elapsed 97%CPU

is this because the element by element
calculations involved are contained in python for
loops?

thanks

#!/usr/bin/python

from Numeric import *

def nrm(pedigree):

    n_animals = len(pedigree) + 1
    
    nrm = zeros((n_animals,n_animals),Float)

    for i in xrange(1,n_animals):
        isire = pedigree[i-1][1]
        idam = pedigree[i-1][2]
        nrm[i,i] = 1.0 + 0.5 * nrm[isire,idam]
        for j in xrange(i+1,n_animals):
            jsire = pedigree[j-1][1]
            jdam = pedigree[j-1][2]
            nrm[j,i] = 0.5 * (nrm[jsire,i] +
nrm[jdam,i])
            nrm[i,j] = nrm[j,i]

    return nrm

if __name__ == '__main__':

    test_ped = [(1,0,0),(2,0,0),(3,1,0),(4,1,2),
                (5,3,4),(6,1,4),(7,5,6)]

    a = nrm(test_ped)
    print a
    


Geocrawler.com - The Knowledge Archive

Hi, folks. A former colleague of mine is now editing a magazine devoted to
scientific computing, and is looking for articles.  If you're doing
something scientific with Python, and want to tell the world about it,
please give me a shout, and I'll forward more information.

Greg Wilson
http://www.software-carpentry.com

Konrad Hinsen wrote:
> 
> > silently). Consider a function
> >
> >       k0 = 100
> >       k = 99
> >       while k < k0:
> >               ..
> >               k0 = k
> >               k = ...
> >
> > which refines a calculation until the measure k stops decreasing.
> > This algorithm may terminate when k is a float, but _fail_ when
> > k is a double  -- the extra precision may cause the algorithm
> 
> I'd call this a buggy implementation. Convergence criteria should be
> explicit and not rely on the internal representation of data
> types. 

> If you care at all about portability, you shouldn't even think about
> this.

	But sometimes you DON'T care about portability.
Sometimes, you want the best result the architecture can support,
and so you need to perform a portable computation of an architecture
dependent value.

-- 
John (Max) Skaller, mailto:skaller@...
10/1 Toxteth Rd Glebe NSW 2037 Australia voice: 61-2-9660-0850
homepage: http://www.maxtal.com.au/~skaller
download: ftp://ftp.cs.usyd.edu/au/jskaller