python blogging software
Harald Massa
Harald
Greg Krohn
news:Xns93B0F0F1A58A2cp...@62.153.159.134...
> which one should I use? Recommendations?
>
> Harald
Are you running Zope[1]? If so, it's fairly trivial to brew your own[2]. If
your homebrew always turns out stale, be cool like me and try Squishdot[3]
(this _might_ be overkill if you strictly want a blog.) Hope this helps.
[1] http://www.zope.org
[2] http://www.evolt.org/article/A_Quick_and_Dirty_Blog_using_Zope/20/23583/
[3] http://squishdot.org/
Greg (who obviously has brew on his mind)
Van Gale
weblogging:
http://plone.org
http://zwiki.org
Outside of Zope here are a few choices:
http://pyds.muensterland.org/
http://www.pycs.net/
http://www.myelin.co.nz/bzero/
http://roughingit.subtlehints.net/pyblosxom
http://pyrite.org/byline/
Paul Rubin
enough that I thought I'd post it. It's like xrange, except for floating
point values.
def frange(start, stop, step=1.0):
sign = cmp(0, step)
while cmp(start, stop) == sign:
yield start
start += step
>>> for i in frange(3., 4., 0.2): print i
...
3.0
3.2
3.4
3.6
3.8
>>>
Carl Banks
> I needed this for something I was doing just now. It came out pretty
> enough that I thought I'd post it. It's like xrange, except for floating
> point values.
>
> def frange(start, stop, step=1.0):
> sign = cmp(0, step)
> while cmp(start, stop) == sign:
> yield start
> start += step
It's a nice little iterator. Three problems, though:
1. It can accumulate rounding errors
2. It can return an integer on the first iteration (minor)
3. The last iteration might not happen because of floating point
uncertainties, even if rounding errors are minimized
A more robust version would look something like this (but it still has
problem #3):
def frange(start, stop, step=1.0):
sign = cmp(0,step)
for i in xrange(sys.maxint):
v = start+i*step
if cmp(v,stop) != sign:
break
yield v
The most robust way to handle this is to iterpolate, i.e., instead of
passing start, stop, and step, pass start, stop, and n_intervals:
def interiter(start, stop, n_intervals):
diff = stop - start
for i in xrange(n_intervals+1):
yield start + (i*diff)/n_intervals
An IEEE 754 (whatever) geek can probably point out even more accurate
ways to do these.
--
CARL BANKS
Carl Banks
> 3. The last iteration might not happen because of floating point
> uncertainties, even if rounding errors are minimized
Just to be sure: what I have should have said was there might be more
extra iteration. The Python convention, of course, it not to include
the final point. However, with floating point, it could happen to
slightly undershoot the stop value, which would cause an evident extra
iteration.
--
CARL BANKS
Neal Holtz
> ...
> A more robust version would look something like this (but it still has
> problem #3):
>
> def frange(start, stop, step=1.0):
> sign = cmp(0,step)
> for i in xrange(sys.maxint):
> v = start+i*step
> if cmp(v,stop) != sign:
> break
> yield v
>
> The most robust way to handle this is to iterpolate, i.e., instead of
> passing start, stop, and step, pass start, stop, and n_intervals:
>
> def interiter(start, stop, n_intervals):
> diff = stop - start
> for i in xrange(n_intervals+1):
> yield start + (i*diff)/n_intervals
>
> An IEEE 754 (whatever) geek can probably point out even more accurate
> ways to do these.
You can easily invent examples where the number of elements generated
may (or may not) suprise you. Here is an implementation that attempts
to account for some of the problems of floating point arithmetic:
# (See http://www.python.org/doc/current/tut/node14.html)
def epsilon():
"""Compute machine epsilon: smallest number, e, s.t. 1 + e > 1"""
one = 1.0
eps = 1.0
while (one + eps) > one:
eps = eps / 2.0
return eps*2.0
_Epsilon = epsilon()
def frange2( start, stop, step=1.0, fuzz=_Epsilon ):
sign = cmp( step, 0.0 )
_fuzz = sign*fuzz*abs(stop)
for i in xrange(sys.maxint):
value = start + i*step
if cmp( stop, value + _fuzz ) != sign:
break
yield value
And we can generate a bunch of test cases to compare this
implementation with the first one above:
for step in frange( 0.0, 1.0, 0.01 ):
for stop in frange( 0.01, 10.0, 0.01 ):
l1 = [x for x in frange( 0.0, stop, step )]
l2 = [x for x in frange2( 0.0, stop, step )]
if l1 != l2:
print stop, step, len(l1), len(l2)
print repr(stop), repr(step)
print l1
print l2
print
The first of many the differences noted is this:
0.06 0.01 7 6
0.060000000000000005 0.01
[0.0, 0.01, 0.02, 0.029999999999999999, 0.040000000000000001,
0.050000000000000003, 0.059999999999999998]
[0.0, 0.01, 0.02, 0.029999999999999999, 0.040000000000000001,
0.050000000000000003]
Under most normal output conditions, you would think that the stop
value was 0.06 and the step was 0.01, so you would expect to get 6
elements in the sequence. Yet the first implementation gives you 7.
This might be surprising, as under most output conditions the last
value would appear to be equal to the stop value. Of course, the
actual stop value is just slightly greater than 0.06. The second
implementation gives you 6 elements in the sequence; though it can be
argued which is "correct", the second implementation is probably less
surprising in most cases.
This is just another example, of course, of the idea that if it
really, really matters how many elements are in the sequence, don't
use floating point arithmetic to control the # of iterations.
Bengt Richter
[...]
>
>The most robust way to handle this is to iterpolate, i.e., instead of
>passing start, stop, and step, pass start, stop, and n_intervals:
>
> def interiter(start, stop, n_intervals):
> diff = stop - start
> for i in xrange(n_intervals+1):
> yield start + (i*diff)/n_intervals
>
To guarantee the exact end points, maybe:
def interiter(start, stop, n_intervals):
fn=float(n_intervals)
for i in xrange(n_intervals+1):
yield ((n_intervals-i)/fn)*start + (i/fn)*stop
but shouldn't that be xrange(n_intervals) and leave out
the final point (thus exact but invisible ;-).
Regards,
Bengt Richter
Carl Banks
> On Mon, 28 Jul 2003 02:22:59 GMT, Carl Banks <imb...@aerojockey.com> wrote:
> [...]
>>
>>The most robust way to handle this is to iterpolate, i.e., instead of
>>passing start, stop, and step, pass start, stop, and n_intervals:
>>
>> def interiter(start, stop, n_intervals):
>> diff = stop - start
>> for i in xrange(n_intervals+1):
>> yield start + (i*diff)/n_intervals
>>
>
> To guarantee the exact end points, maybe:
>
> def interiter(start, stop, n_intervals):
> fn=float(n_intervals)
> for i in xrange(n_intervals+1):
> yield ((n_intervals-i)/fn)*start + (i/fn)*stop
Good.
> but shouldn't that be xrange(n_intervals) and leave out
> the final point (thus exact but invisible ;-).
Well, I thought about that.
In the interests of practicality beats purity, when you're
interpolating (instead of stepping as in range) you should include the
final endpoint. In my experience, when when you divide a segment into
intervals, you almost always want to include both endpoints.
If you use the Python convention of not including the final point in
the iteration, and someone decides they do want the final endpoint,
then it's kind of a pain to calculate what point you should use to get
the final "endpoint", and you lose accuracy on top of it.
So, my suggestion when iterating over an interval is this: If you pass
in a stepsize, then don't include the final point. If you pass in a
number of intervals, then do include the final point. I think the
practicality of it justifies the different conventions.
--
CARL BANKS
Paul Rubin
> In the interests of practicality beats purity, when you're
> interpolating (instead of stepping as in range) you should include the
> final endpoint. In my experience, when when you divide a segment into
> intervals, you almost always want to include both endpoints.
I think you're right about this, but it goes against the Python notion
of a "range" function, so an frange that includes both endpoints
should be called something different, maybe "fstep" or something like
that. For what I was doing, the frange that I posted did the job.
I see now that that the problem is subtle enough that IMO, a fully
worked out solution should be written and included in the Python
library or at least the Python cookbook.
Bengt Richter
How about if we indicate the kind of interval, e.g.,
>>> def interiter(start, stop, n_intervals, ends='[]'):
... fn=float(n_intervals)
... for i in xrange(ends[0]=='(', n_intervals+(ends[1]==']')):
... yield ((n_intervals-i)/fn)*start + (i/fn)*stop
...
>>>
>>> for f in interiter(0, 8, 8): print f,
...
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
>>> for f in interiter(0, 8, 8, ends='[)'): print f,
...
1.0 2.0 3.0 4.0 5.0 6.0 7.0
>>> for f in interiter(0, 8, 8, ends='(]'): print f,
...
1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
>>> for f in interiter(0, 8, 8, ends='()'): print f,
...
1.0 2.0 3.0 4.0 5.0 6.0 7.0
>>> for f in interiter(0, 8, 8, ends='[]'): print f,
...
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
Regards,
Bengt Richter
Michael Hudson
> I see now that that the problem is subtle enough that IMO, a fully
> worked out solution should be written and included in the Python
> library or at least the Python cookbook.
I also think that this is a problem without One True Answer
wrt. endpoint handling which makes a library solution a bit of a
challenge. Maybe I'm wrong, though (haven't done numeric stuff for a
while now).
Cheers,
mwh
--
6. Symmetry is a complexity-reducing concept (co-routines include
subroutines); seek it everywhere.
-- Alan Perlis, http://www.cs.yale.edu/homes/perlis-alan/quotes.html