Re: complaints about no replies last week
On Mar 31, 4:34 pm, Arnaud Delobelle arno...@googlemail.com wrote: a...@pythoncraft.com (Aahz) writes: Arnaud Delobelle arno...@googlemail.com wrote: There are no comments - I don't have the time to add any, sorry! The margin is too small to contain the proof? I wish I could come up with such a resilient conjecture! Ah but in this case, the proof is in the program, as Curry-Howard may well have said. Is it simple, complex, or complicated? IS IT FLAT OR NESTED! / shouting -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
On Mar 31, 4:07 pm, Arnaud Delobelle arno...@googlemail.com wrote:
prueba...@latinmail.com writes:
[...]
Well since I attracted a couple people's attention I will describe the
problem in more detail. Describing the problem properly is probably as
hard as solving it, so excuse me if I struggle a bit.
The problem is for a health insurance company and involves the period
of time a person is covered. Most insurance companies allow not only
for the main member to be insured but his family: the spouse and the
dependents (children). This additional coverage costs extra but less
than a full new insurance. So for example if Alice buys an insurance
worth at 100 dollars a month, she can insure her husband Bob for an
additional 50 dollars. Under certain circumstances Alice may go off
the insurance and only Bob stays. In that case the price goes back to
100 dollars or maybe there is a deal for 80 or something like that. In
other words the cost of the insurance is dependent on the combination
of family members that participate in it. Additionally not only do we
have different family compositions but also different insurance
products. So you can get medical, dental and vision insurance.
All that data is stored in a database that is not very tidy and looks
something like this:
First Day of Coverage, Last Day of Coverage, Relationship, Product
5/3/2005, 5/3/2005, D, M
9/10/2005, 10/10/2005, S, V
3/15/2005, 7/15/2005, M, M
3/1/2005, 6/1/2005, S, M
5/15/2005, 7/20/2005, D, D
9/10/2005, 1/1/2140, M, V
2/1/2005, 5/3/2005, M, M
Where
Relationship: M=Main Member, S=Spouse, D=Dependent
Product: M=Medical, D=Dental, V=Vision
As far as I know at the present time there are no deals based on
combination of products purchased so we will handle each product
independently. What I want is a simple algorithm that allows me to
calculate something like this out of it (hopefully I didn’t make a
mistake):
Medical:
2/1/2005, 2/28/2005, M
3/1/2005, 5/2/2005, MS
5/3/2005, 5/3/2005, MSD
5/4/2005, 6/1/2005, MS
6/2/2005, 7/15/2005, M
Dental:
5/15/2005, 7/20/2005, D
Vision:
9/10/2005, 10/10/2005, MS
10/11/2005, 1/1/2140, M
OK the approach I describe in my previous email works fine for this
particular problem. I implement it below - the function walk_ivals is at
the heart of it, I've made it as simple as possible and it's pretty
clear that it is O(nlogn).
The function that actually sorts the data is union(), and it's just a
call to walk_ivals with callback the function acc() which is constant
time, therefore union() itself has the same complexity as walk_ivals.
There are no comments - I don't have the time to add any, sorry!
--
import datetime
from collections import defaultdict
def walk_ivals(ivals, callback, endvals=(-1, 1)):
endpoints = [(x, data) for ival, data in ivals for x in zip(ival,
endvals)]
endpoints.sort()
for (endpoint, endval), data in endpoints:
callback(endpoint, endval, data)
def union(ivals):
timelines = defaultdict(list)
mult = defaultdict(lambda: defaultdict(int))
def acc(date, step, data):
rel, prod = data
old_mult = mult[prod][rel]
mult[prod][rel] -= step
if not(old_mult and mult[prod][rel]):
rels = [rel for rel, m in mult[prod].iteritems() if m]
if timelines[prod] and timelines[prod][-1][0] == date:
timelines[prod][-1][1] = rels
else:
timelines[prod].append([date, rels])
walk_ivals(ivals, acc)
return timelines
test_data = 5/3/2005, 5/3/2005, D, M
9/10/2005, 10/10/2005, S, V
3/15/2005, 7/15/2005, M, M
3/1/2005, 6/1/2005, S, M
5/15/2005, 7/20/2005, D, D
9/10/2005, 1/1/2140, M, V
2/1/2005, 5/3/2005, M, M
def parse_date(date_string):
month, day, year = map(int, date_string.split('/'))
return datetime.date(year, month, day)
def parse_data(data_string):
data = []
for line in data_string.split(\n):
start, end, rel, prod = line.split(,)
start, end = map(parse_date, (start + datetime.timedelta(1), end))
rel, prod = rel.strip(), prod.strip()
data.append([(start, end), (rel, prod)])
return data
def test():
ivals = parse_data(test_data)
timelines = union(ivals)
for prod, timeline in timelines.iteritems():
print -*20
print Product, prod
for date, covers in timeline:
print date, ' '.join(covers)
if __name__ == '__main__':
test()
--
Here is what it outputs:
marigold:junk arno$ python intervals2.py
Product M
2005-02-01 M
2005-03-01 S M
2005-05-03 S M D
2005-05-04 S M
2005-06-02 M
2005-07-16
Product D
2005-05-15 D
2005-07-21
Product V
2005-09-10
Re: complaints about no replies last week
Arnaud Delobelle wrote: prueba...@latinmail.com writes: [...] I myself asked about how to write a library to efficiently do union and intersection of sets containing time intervals some time ago on this list and got little to no answers. It is a tricky problem. Since I was getting paid I got an O(n*n) solution working. People on this list on the other hand do not get paid and answer whatever strikes their fancy. Sometimes the question is hard or confusing and nobody is motivated enough to answer. I wasn't around when you posted this I guess. Do you mean intervals sets on the (real) number line such as: 1-3, 6-10 meaning all numbers between 1 and 3 and all numbers between 6 and 10. In this case I think you can achieve union and intersection in O(nlogn) where n is the total number of intervals in the interval sets to unify or intersect. There is an implementation below. I have chosen a very simple data structure for interval sets: an interval set is the list of its endpoints. E.g. 1-3, 6-10 is the list [1, 3, 6, 10] This means that I can't specify whether an interval is closed or open. So in the implementation below all intervals are assumed to be open. The method could be made to work for any kind of intervals with the same complexity, there would just be a few more LOC. I'm focusing on the principle - here it is: -- # Implementation of union and intersection of interval sets. from itertools import * def combine(threshold, intsets): endpoints = sorted(chain(*imap(izip, intsets, repeat(cycle([1,-1]) height = 0 compound = [] for x, step in endpoints: old_height = height height += step if max(height, old_height) == threshold: compound.append(x) return compound def union(*intsets): return combine(1, intsets) def intersection(*intsets): return combine(len(intsets), intsets) # tests def pretty(a): a = iter(a) return ', '.join(%s-%s % (a, b) for a, b in izip(a, a)) tests = [ ([1, 5, 10, 15], [3, 11, 13, 20]), ([2, 4, 6, 8], [4, 7, 10, 11]), ([0, 11], [5, 10, 15, 25], [7, 12, 13, 15]), ] for intsets in tests: print sets: , ; .join(imap(pretty, intsets)) print union: , pretty(union(*intsets)) print intersection: , pretty(intersection(*intsets)) print -*20 -- Is this what you were looking for? -- Arnaud I realised after posting last night that I must be (1) solving the wrong problem (2) solving it badly - My implementation of the combine() function above is O(nlogn) (because of the sorted() call) whereas it could be O(n) by iterating over the interval in the parallel manner, hence (2). This would make union() and intersection() O(n). - As the problem was solved by the OP in O(n^2) I must be solving the wrong problem (1). I apologise for this. However it was a nice and compact implementation IMHO :) -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
On Mar 31, 2:56 am, Arnaud Delobelle arno...@googlemail.com wrote: Arnaud Delobelle wrote: prueba...@latinmail.com writes: [...] I myself asked about how to write a library to efficiently do union and intersection of sets containing time intervals some time ago on this list and got little to no answers. It is a tricky problem. Since I was getting paid I got an O(n*n) solution working. People on this list on the other hand do not get paid and answer whatever strikes their fancy. Sometimes the question is hard or confusing and nobody is motivated enough to answer. I wasn't around when you posted this I guess. Do you mean intervals sets on the (real) number line such as: 1-3, 6-10 meaning all numbers between 1 and 3 and all numbers between 6 and 10. In this case I think you can achieve union and intersection in O(nlogn) where n is the total number of intervals in the interval sets to unify or intersect. There is an implementation below. I have chosen a very simple data structure for interval sets: an interval set is the list of its endpoints. E.g. 1-3, 6-10 is the list [1, 3, 6, 10] This means that I can't specify whether an interval is closed or open. So in the implementation below all intervals are assumed to be open. The method could be made to work for any kind of intervals with the same complexity, there would just be a few more LOC. I'm focusing on the principle - here it is: -- # Implementation of union and intersection of interval sets. from itertools import * def combine(threshold, intsets): endpoints = sorted(chain(*imap(izip, intsets, repeat(cycle([1,-1]) height = 0 compound = [] for x, step in endpoints: old_height = height height += step if max(height, old_height) == threshold: compound.append(x) return compound def union(*intsets): return combine(1, intsets) def intersection(*intsets): return combine(len(intsets), intsets) # tests def pretty(a): a = iter(a) return ', '.join(%s-%s % (a, b) for a, b in izip(a, a)) tests = [ ([1, 5, 10, 15], [3, 11, 13, 20]), ([2, 4, 6, 8], [4, 7, 10, 11]), ([0, 11], [5, 10, 15, 25], [7, 12, 13, 15]), ] for intsets in tests: print sets: , ; .join(imap(pretty, intsets)) print union: , pretty(union(*intsets)) print intersection: , pretty(intersection(*intsets)) print -*20 -- Is this what you were looking for? -- Arnaud I realised after posting last night that I must be (1) solving the wrong problem (2) solving it badly - My implementation of the combine() function above is O(nlogn) (because of the sorted() call) whereas it could be O(n) by iterating over the interval in the parallel manner, hence (2). This would make union() and intersection() O(n). - As the problem was solved by the OP in O(n^2) I must be solving the wrong problem (1). I apologise for this. However it was a nice and compact implementation IMHO :) -- Arnaud I am pretty sure the problem can be solved in O(n log n). I just wasn't feeling overly smart when I was writing the algorithm. N is on average 4 and it had eventually to be implemented inside a framework using C++ anyway, so it is pretty fast. I can’t believe that no programmer has come over the same kind of problem before, yet my Google fu didn’t do anything for me. Well since I attracted a couple people's attention I will describe the problem in more detail. Describing the problem properly is probably as hard as solving it, so excuse me if I struggle a bit. The problem is for a health insurance company and involves the period of time a person is covered. Most insurance companies allow not only for the main member to be insured but his family: the spouse and the dependents (children). This additional coverage costs extra but less than a full new insurance. So for example if Alice buys an insurance worth at 100 dollars a month, she can insure her husband Bob for an additional 50 dollars. Under certain circumstances Alice may go off the insurance and only Bob stays. In that case the price goes back to 100 dollars or maybe there is a deal for 80 or something like that. In other words the cost of the insurance is dependent on the combination of family members that participate in it. Additionally not only do we have different family compositions but also different insurance products. So you can get medical, dental and vision insurance. All that data is stored in a database that is not very tidy and looks something like this: First Day of Coverage, Last Day of Coverage, Relationship, Product 5/3/2005, 5/3/2005, D, M 9/10/2005, 10/10/2005, S, V 3/15/2005, 7/15/2005, M, M 3/1/2005, 6/1/2005, S, M 5/15/2005, 7/20/2005, D, D 9/10/2005, 1/1/2140, M, V 2/1/2005, 5/3/2005, M,
Re: complaints about no replies last week
prueba...@latinmail.com writes:
[...]
Well since I attracted a couple people's attention I will describe the
problem in more detail. Describing the problem properly is probably as
hard as solving it, so excuse me if I struggle a bit.
The problem is for a health insurance company and involves the period
of time a person is covered. Most insurance companies allow not only
for the main member to be insured but his family: the spouse and the
dependents (children). This additional coverage costs extra but less
than a full new insurance. So for example if Alice buys an insurance
worth at 100 dollars a month, she can insure her husband Bob for an
additional 50 dollars. Under certain circumstances Alice may go off
the insurance and only Bob stays. In that case the price goes back to
100 dollars or maybe there is a deal for 80 or something like that. In
other words the cost of the insurance is dependent on the combination
of family members that participate in it. Additionally not only do we
have different family compositions but also different insurance
products. So you can get medical, dental and vision insurance.
All that data is stored in a database that is not very tidy and looks
something like this:
First Day of Coverage, Last Day of Coverage, Relationship, Product
5/3/2005, 5/3/2005, D, M
9/10/2005, 10/10/2005, S, V
3/15/2005, 7/15/2005, M, M
3/1/2005, 6/1/2005, S, M
5/15/2005, 7/20/2005, D, D
9/10/2005, 1/1/2140, M, V
2/1/2005, 5/3/2005, M, M
Where
Relationship: M=Main Member, S=Spouse, D=Dependent
Product: M=Medical, D=Dental, V=Vision
As far as I know at the present time there are no deals based on
combination of products purchased so we will handle each product
independently. What I want is a simple algorithm that allows me to
calculate something like this out of it (hopefully I didn’t make a
mistake):
Medical:
2/1/2005, 2/28/2005, M
3/1/2005, 5/2/2005, MS
5/3/2005, 5/3/2005, MSD
5/4/2005, 6/1/2005, MS
6/2/2005, 7/15/2005, M
Dental:
5/15/2005, 7/20/2005, D
Vision:
9/10/2005, 10/10/2005, MS
10/11/2005, 1/1/2140, M
OK the approach I describe in my previous email works fine for this
particular problem. I implement it below - the function walk_ivals is at
the heart of it, I've made it as simple as possible and it's pretty
clear that it is O(nlogn).
The function that actually sorts the data is union(), and it's just a
call to walk_ivals with callback the function acc() which is constant
time, therefore union() itself has the same complexity as walk_ivals.
There are no comments - I don't have the time to add any, sorry!
--
import datetime
from collections import defaultdict
def walk_ivals(ivals, callback, endvals=(-1, 1)):
endpoints = [(x, data) for ival, data in ivals for x in zip(ival, endvals)]
endpoints.sort()
for (endpoint, endval), data in endpoints:
callback(endpoint, endval, data)
def union(ivals):
timelines = defaultdict(list)
mult = defaultdict(lambda: defaultdict(int))
def acc(date, step, data):
rel, prod = data
old_mult = mult[prod][rel]
mult[prod][rel] -= step
if not(old_mult and mult[prod][rel]):
rels = [rel for rel, m in mult[prod].iteritems() if m]
if timelines[prod] and timelines[prod][-1][0] == date:
timelines[prod][-1][1] = rels
else:
timelines[prod].append([date, rels])
walk_ivals(ivals, acc)
return timelines
test_data = 5/3/2005, 5/3/2005, D, M
9/10/2005, 10/10/2005, S, V
3/15/2005, 7/15/2005, M, M
3/1/2005, 6/1/2005, S, M
5/15/2005, 7/20/2005, D, D
9/10/2005, 1/1/2140, M, V
2/1/2005, 5/3/2005, M, M
def parse_date(date_string):
month, day, year = map(int, date_string.split('/'))
return datetime.date(year, month, day)
def parse_data(data_string):
data = []
for line in data_string.split(\n):
start, end, rel, prod = line.split(,)
start, end = map(parse_date, (start + datetime.timedelta(1), end))
rel, prod = rel.strip(), prod.strip()
data.append([(start, end), (rel, prod)])
return data
def test():
ivals = parse_data(test_data)
timelines = union(ivals)
for prod, timeline in timelines.iteritems():
print -*20
print Product, prod
for date, covers in timeline:
print date, ' '.join(covers)
if __name__ == '__main__':
test()
--
Here is what it outputs:
marigold:junk arno$ python intervals2.py
Product M
2005-02-01 M
2005-03-01 S M
2005-05-03 S M D
2005-05-04 S M
2005-06-02 M
2005-07-16
Product D
2005-05-15 D
2005-07-21
Product V
2005-09-10 S M
2005-10-11 M
2140-01-02
--
The date output is slightly different from yours - I didn't realise you
had time intervals and now I don't
Re: complaints about no replies last week
In article m263hpa21u@googlemail.com, Arnaud Delobelle arno...@googlemail.com wrote: There are no comments - I don't have the time to add any, sorry! The margin is too small to contain the proof? -- Aahz (a...@pythoncraft.com) * http://www.pythoncraft.com/ Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it. --Brian W. Kernighan -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
a...@pythoncraft.com (Aahz) writes: Arnaud Delobelle arno...@googlemail.com wrote: There are no comments - I don't have the time to add any, sorry! The margin is too small to contain the proof? I wish I could come up with such a resilient conjecture! Ah but in this case, the proof is in the program, as Curry-Howard may well have said. -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
On Mar 28, 11:07 am, Aaron Brady castiro...@gmail.com wrote: Hi, A week ago, I posted a question and an idea about Python's garbage collector. I got a few replies. Some days later, I posted a mock-up implementation of it, and got *NO* replies. Does this mean: a) It works b) It doesn't work c) It's not particularly applicable to Python at that point (particularly) d) It's not news Thanks and sincerely. P.S. Non-plugging links:http://groups.google.com/group/comp.lang.python/browse_thread/thread/...http://groups.google.com/group/comp.lang.python/browse_thread/thread/... e) It is a hard or complex problem that requires significant investment of time on a problem or approach that few people are interested in at the moment. f) The description is confusing or incomplete or the source code is long and difficult to read. I myself asked about how to write a library to efficiently do union and intersection of sets containing time intervals some time ago on this list and got little to no answers. It is a tricky problem. Since I was getting paid I got an O(n*n) solution working. People on this list on the other hand do not get paid and answer whatever strikes their fancy. Sometimes the question is hard or confusing and nobody is motivated enough to answer. -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
On Mon, 30 Mar 2009 07:50:49 -0700, pruebauno wrote: I myself asked about how to write a library to efficiently do union and intersection of sets containing time intervals some time ago on this list and got little to no answers. It is a tricky problem. With all the confidence of somebody who doesn't need to solve it, I can say, no, it's an easy problem, provided you use the correct data structure. What you want is an interval tree: http://en.wikipedia.org/wiki/Interval_tree Since I was getting paid I got an O(n*n) solution working. Just off the top of my head, I *think* you should be able to get that down to O(m * log n) where m is the size of one set and n the size of the other. Don't quote me on that though. People on this list on the other hand do not get paid We don't??? Damn! I was expecting a HUGE cheque at the end of the month! BTW Aaron, I haven't replied to your post about the garbage collector because that's one part of Python programing where I cherish my ignorance. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
prueba...@latinmail.com writes: [...] I myself asked about how to write a library to efficiently do union and intersection of sets containing time intervals some time ago on this list and got little to no answers. It is a tricky problem. Since I was getting paid I got an O(n*n) solution working. People on this list on the other hand do not get paid and answer whatever strikes their fancy. Sometimes the question is hard or confusing and nobody is motivated enough to answer. I wasn't around when you posted this I guess. Do you mean intervals sets on the (real) number line such as: 1-3, 6-10 meaning all numbers between 1 and 3 and all numbers between 6 and 10. In this case I think you can achieve union and intersection in O(nlogn) where n is the total number of intervals in the interval sets to unify or intersect. There is an implementation below. I have chosen a very simple data structure for interval sets: an interval set is the list of its endpoints. E.g. 1-3, 6-10 is the list [1, 3, 6, 10] This means that I can't specify whether an interval is closed or open. So in the implementation below all intervals are assumed to be open. The method could be made to work for any kind of intervals with the same complexity, there would just be a few more LOC. I'm focusing on the principle - here it is: -- # Implementation of union and intersection of interval sets. from itertools import * def combine(threshold, intsets): endpoints = sorted(chain(*imap(izip, intsets, repeat(cycle([1,-1]) height = 0 compound = [] for x, step in endpoints: old_height = height height += step if max(height, old_height) == threshold: compound.append(x) return compound def union(*intsets): return combine(1, intsets) def intersection(*intsets): return combine(len(intsets), intsets) # tests def pretty(a): a = iter(a) return ', '.join(%s-%s % (a, b) for a, b in izip(a, a)) tests = [ ([1, 5, 10, 15], [3, 11, 13, 20]), ([2, 4, 6, 8], [4, 7, 10, 11]), ([0, 11], [5, 10, 15, 25], [7, 12, 13, 15]), ] for intsets in tests: print sets: , ; .join(imap(pretty, intsets)) print union: , pretty(union(*intsets)) print intersection: , pretty(intersection(*intsets)) print -*20 -- Is this what you were looking for? -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
On Mar 28, 7:31 pm, ajaksu aja...@gmail.com wrote: Hi! Aaron Brady wrote: A week ago, I posted a question and an idea about Python's garbage collector. I got a few replies. Some very nice, too :) Yes. Some days later, I posted a mock-up implementation of it, and got *NO* replies. Does this mean: It's not particularly clear to me what your goal is in that post. Something like I'm trying to model Python's GC or here's an alternative way for the Python GC to work would make it clear, so that feedback could be more objective. Also, a docstring with the executive overview on top of the module helps a lot, as does including your usage examples as doctests. I've heard these suggestion regarding my posts recently ;) However, I think the kind of deep, conceptual feedback I believe you'd like is available here and elsewhere[1]. It's just a matter of bad luck/bad timing IMHO :) Daniel [1]http://groups.google.com/group/unladen-swallow/browse_thread/thread/c... Hi, Daniel, I added an explanation of what I want-- a technical discussion of it. I also included the output in a reply to the code. Unfortunately it's 367 lines long and rather obscure, albeit systematic. If it works, of course Python could adopt it, but my self-esteem kind of interferes with thinking that I could have improved something that is a decade old and hundreds of people have stared at... even though I do have a CS Bachelor's. Besides, Martin Lowis said: There is an easy design pattern around it, so I'm -1 on complicating the GC protocol. That doesn't mean that no one will help me add it to /my/ project though. -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
On Sat, Mar 28, 2009 at 11:07 AM, Aaron Brady castiro...@gmail.com wrote: Hi, A week ago, I posted a question and an idea about Python's garbage collector. I got a few replies. Some days later, I posted a mock-up implementation of it, and got *NO* replies. Does this mean: a) It works b) It doesn't work c) It's not particularly applicable to Python at that point (particularly) d) It's not news e) Most of us don't have the time or inclination to look through and test your code. Thanks and sincerely. P.S. Non-plugging links: http://groups.google.com/group/comp.lang.python/browse_thread/thread/82ebfec89bcc906c/de901f689ae7a962 http://groups.google.com/group/comp.lang.python/browse_thread/thread/d3bb410cc6dcae54/f4b282e545335c30 -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
On Mar 28, 10:07 am, Aaron Brady castiro...@gmail.com wrote: Hi, A week ago, I posted a question and an idea about Python's garbage collector. I got a few replies. Some days later, I posted a mock-up implementation of it, and got *NO* replies. Does this mean: a) It works b) It doesn't work c) It's not particularly applicable to Python at that point (particularly) d) It's not news Thanks and sincerely. P.S. Non-plugging links:http://groups.google.com/group/comp.lang.python/browse_thread/thread/...http://groups.google.com/group/comp.lang.python/browse_thread/thread/... It's probably more like e) You are not a member of the Illuminati, so the elite don't give a monkey's toss! Go back and look over my most infamous thread (i will not utter the title here but you know of which i speak) and you will see first hand the wall of resistance put before any non-member who has a good idea. just ideas -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
In article 2d80ec1b-5eb5-4e82-9a4a-36934dd53...@z9g2000yqi.googlegroups.com, Aaron Brady castiro...@gmail.com wrote: A week ago, I posted a question and an idea about Python's garbage collector. I got a few replies. Some days later, I posted a mock-up implementation of it, and got *NO* replies. Does this mean: a) It works b) It doesn't work c) It's not particularly applicable to Python at that point (particularly) d) It's not news e) the best place to start these days with actual implementations of ideas is the python-ideas list -- Aahz (a...@pythoncraft.com) * http://www.pythoncraft.com/ At Resolver we've found it useful to short-circuit any doubt and just refer to comments in code as 'lies'. :-) --Michael Foord paraphrases Christian Muirhead on python-dev, 2009-3-22 -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
On Mar 28, 11:41 am, a...@pythoncraft.com (Aahz) wrote: In article 2d80ec1b-5eb5-4e82-9a4a-36934dd53...@z9g2000yqi.googlegroups.com, Aaron Brady castiro...@gmail.com wrote: A week ago, I posted a question and an idea about Python's garbage collector. I got a few replies. Some days later, I posted a mock-up implementation of it, and got *NO* replies. Does this mean: a) It works b) It doesn't work c) It's not particularly applicable to Python at that point (particularly) d) It's not news e) the best place to start these days with actual implementations of ideas is the python-ideas list I'm not convinced I'm altogether welcome there. I treated that list as a social list for some time, not discovering its intensity until after estranging everybody. Not to mention, I'm not primarily proposing this idea as an improvement to the reference management; it's an idea I would probably implement on the side, for an extension. I just want to know if it's valid and functional. Should I ask about it on comp.programming? Am I in the right place? Or, is my idea sophisticated and relevant enough to test the waters again? At Resolver we've found it useful to short-circuit any doubt and just refer to comments in code as 'lies'. :-) --Michael Foord paraphrases Christian Muirhead on python-dev, 2009-3-22 P.S. No one read your sig or liked it. P.P.S. Ok, not bad. -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
Hi! Aaron Brady wrote: A week ago, I posted a question and an idea about Python's garbage collector. I got a few replies. Some very nice, too :) Some days later, I posted a mock-up implementation of it, and got *NO* replies. Does this mean: It's not particularly clear to me what your goal is in that post. Something like I'm trying to model Python's GC or here's an alternative way for the Python GC to work would make it clear, so that feedback could be more objective. Also, a docstring with the executive overview on top of the module helps a lot, as does including your usage examples as doctests. I've heard these suggestion regarding my posts recently ;) However, I think the kind of deep, conceptual feedback I believe you'd like is available here and elsewhere[1]. It's just a matter of bad luck/bad timing IMHO :) Daniel [1] http://groups.google.com/group/unladen-swallow/browse_thread/thread/c39e75ad6100893f -- http://mail.python.org/mailman/listinfo/python-list
Re: complaints about no replies last week
Aaron Brady wrote: Hi, c) It's not particularly applicable to Python at that point (particularly) BTW, here's some interesting read: http://mail.python.org/pipermail/python-3000/2006-September/003855.html http://mail.python.org/pipermail/python-3000/2007-May/007129.html http://mail.python.org/pipermail/python-dev/2007-October/074820.html HTH, Daniel -- http://mail.python.org/mailman/listinfo/python-list
