Re: [Tutor] sorting objects on two attributes
On Mar 4, 2008, at 11:04 AM, Kent Johnson wrote: > Eric Abrahamsen wrote: >> Itertools.groupby is totally impenetrable to me > > Maybe this will help: > http://personalpages.tds.net/~kent37/blog/arch_m1_2005_12.html#e69 > > Kent > It did! Thanks very much. I think I understand now what's going on in the groupby line. And then this line: t.append((sorted_part[-1].submit_date, key, sorted_part)) is basically a Decorate-Sort-Undecorate operation, sorting on sorted_part[-1].submit_date because that's guaranteed to be the latest date in each group? It's starting to come clear... ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sorting objects on two attributes
Eric Abrahamsen wrote: > Itertools.groupby is totally impenetrable to me Maybe this will help: http://personalpages.tds.net/~kent37/blog/arch_m1_2005_12.html#e69 Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sorting objects on two attributes
Well I expected to learn a thing or two, but I didn't expect not to
understand the suggestions at all! :) Thanks to everyone who
responded, and sorry for the typo (it was meant to be object_id
throughout, not content_type).
So far Michael's solution works and is most comprehensible to me – ie
I stand a fighting chance of figuring it out.
I didn't realize you could make direct use of SortedDicts in django,
so that's worth a try, too.
Itertools.groupby is totally impenetrable to me, but works as well! Is
there any consensus about which might go faster? I will go right now
and google until my brain is full.
Thanks again,
Eric
On Mar 4, 2008, at 12:56 AM, Michael H. Goldwasser wrote:
>
>
> Hello Eric,
>
> Your basic outlook is fine, but you can do it much more efficiently
> with a single sort. Here's the way I'd approach the task
> (untested):
>
> # --
> # first compute the latest date for each id group; uses O(n) time
> newest = {}
> for q in queryset:
> id,date = q.object_id, q.submit_date
> if id not in newest or date > newest[id]:
> newest[id] = date
>
> # now sort based on the following decorator as a key
> data.sort(reverse=True, key=lambda x: (newest[x.object_id],
> x.object_id, x.submit_date))
> # --
>
> In essence, you compute the max date within each group, but your
> approach (i.e., building explicit sublists and then repeatedly
> calling max on those sublists) is far more time-consuming than the
> above dictionary based approach.
>
> Note well that using a tuple as a decorator key is more efficient
> than calling sort separately for each subgroup. The
> lexicographical order of the following
>
>(newest[x.object_id], x.object_id, x.submit_date)
>
> should produce the order that you desire. The reason for the middle
> entry is to ensure that items groups by object_id in the case that
> two different groups achieve the same maximum date. It wasn't clear
> to me whether you wanted elements within groups from oldest to
> newest or newest to oldest. I believe that the code I give produces
> the ordering that you intend, but you may adjust the sign of the
> decorate elements if necessary.
>
> With regard,
> Michael
>
> +---
> | Michael Goldwasser
> | Associate Professor
> | Dept. Mathematics and Computer Science
> | Saint Louis University
> | 220 North Grand Blvd.
> | St. Louis, MO 63103-2007
>
>
> On Monday March 3, 2008, Eric Abrahamsen wrote:
>
>> I have a grisly little sorting problem to which I've hacked
>> together a
>> solution, but I'm hoping someone here might have a better
>> suggestion.
>>
>> I have a list of objects, each of which has two attributes,
>> object_id
>> and submit_date. What I want is to sort them by content_type,
>> then by
>> submit_date within content_type, and then sort each content_type
>> block
>> according to which block has the newest object by submit_date.
>> (This
>> sequence of sorting might not be optimal, I'm not sure). I'm
>> actually
>> creating a list of recent comments on blog entries for a python-
>> based
>> web framework, and want to arrange the comments according to blog
>> entry (content_type), by submit_date within that entry, with the
>> entries with the newest comments showing up on top.
>>
>> I don't believe a single cmp function fed to list.sort() can do
>> this,
>> because you can't know how two objects should be compared until you
>> know all the values for all the objects. I'd be happy to be proven
>> wrong here.
>>
>> After some false starts with dictionaries, here's what I've got.
>> Queryset is the original list of comments (I'm doing this in
>> django),
>> and it returns a list of lists, though I might flatten it
>> afterwards.
>> It works, but it's ghastly unreadable and if there were a more
>> graceful solution I'd feel better about life in general:
>>
>>
>> def make_com_list(queryset):
>> ids = set([com.object_id for com in queryset])
>> xlist = [[com for com in queryset if com.object_id == i] for
>> i in
>> ids]
>> for ls in xlist:
>> ls.sort(key=lambda x: x.submit_date)
>> xlist.sort(key=lambda x: max([com.submit_date for com in x]),
>> reverse=True)
>> return xlist
>>
>> I'd appreciate any hints!
>>
>> Thanks,
>> Eric
>>
>
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Re: [Tutor] Python and displaying mathematical equations?
On 04/03/2008, Shuai Jiang (Runiteking1) <[EMAIL PROTECTED]> wrote: > Hello, I'm trying to create an application that retrieves and displays > (probably in HTML or PDF format) math problems from a database. > The problem is that I need some sort of mechanism to display mathematical > equations. If you can describe your equations in MathML then there may be options for you -- a quick google for "python mathml" turned up a few hits -- e.g. http://sourceforge.net/projects/pymathml/ or http://www.grigoriev.ru/svgmath/ (if you accept SVG as an output). -- John. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] Python and displaying mathematical equations?
Hello, I'm trying to create an application that retrieves and displays (probably in HTML or PDF format) math problems from a database. The problem is that I need some sort of mechanism to display mathematical equations. I'm trying to not use Latex as it would cause the users to install Latex (and most likely dvipng) on their own computers. Is there a lighter version of Latex for Python or do I have to just use Latex? Thanks! Marshall Jiang -- Visit my blog at runiteking1.blogspot.com ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] [tutor] Finding image statistics
Varsha Purohit wrote: > Yeahh so by doing this i am counting only the difference part since we > have grayscaled the image and assuming it will count only the pixels > that evolve as difference Yes > if i use sum2 instead of sum i think it > will give squared sum which is area... and if i just use count it would > count the number of pixels developed like that... sounds interesting .. > thanks for throwing light for me in right direction No. First, pixels are already a measure of area. Second, if each pixel value is 0 or 1, squaring the values won't make any difference. Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] [tutor] Finding image statistics
Yeahh so by doing this i am counting only the difference part since we have
grayscaled the image and assuming it will count only the pixels that evolve
as difference if i use sum2 instead of sum i think it will give squared
sum which is area... and if i just use count it would count the number of
pixels developed like that... sounds interesting .. thanks for throwing
light for me in right direction
On Sun, Mar 2, 2008 at 4:45 PM, Kent Johnson <[EMAIL PROTECTED]> wrote:
> Varsha Purohit wrote:
> > I am getting this list as an output
> >
> > [268541.0, 264014.0, 324155.0]
>
> This is the sum of the values of the red pixels, etc. They are not
> counts, but sums. So if you have an image with the two pixels
> (1, 2, 3), (4, 5, 6) you would get a sum of (5, 7, 9).
> >
> > Actually in my code i am finding difference between two images and i
> > need to count the number of pixels which appear as a difference of these
> > two images. These values i m getting as output are little large.
>
> So you want a count of the number of pixels that differ between the two
> images? Maybe this:
>
> diff = ImageChops.difference(file1, file2)
>
> # Convert to grayscale so differences are counted just once per pixel
> diff = ImageOps.grayscale(diff)
>
> # Convert each difference to 0 or 1 so we can count them
> # Clipping function
> def clip(x):
> return 1 if x >= 1 else 0
>
> # Apply the clipping function
> diff = Image.eval(diff, clip)
>
> print ImageStat.Stat(diff).sum
>
> Kent
>
> >
> > i m pasting my code again
> >
> > file1=Image.open("./pics/original.jpg")
> > file2=Image.open(val)
> > diff = ImageChops.subtract(file1,file2,0.3)
> > stat1 = ImageStat.Stat(diff)
> >
> > count1=stat1.sum
> > print count1
> > diff.save("./pics/diff"+".jpg")
> >
> > diff.show()
> >
> > AS you can see i am finding difference of two images which are nearly
> > identical. But they have some pixels that are different and i can see
> > that in the output image diff.jpg. So i am trying to put this difference
> > image in the imagestat and trying to see if i can get to count the
> > number of pixels ...
>
>
--
Varsha Purohit,
Graduate Student
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[Tutor] sorting objects on two attributes
Hello Eric,
Your basic outlook is fine, but you can do it much more efficiently
with a single sort. Here's the way I'd approach the task (untested):
# --
# first compute the latest date for each id group; uses O(n) time
newest = {}
for q in queryset:
id,date = q.object_id, q.submit_date
if id not in newest or date > newest[id]:
newest[id] = date
# now sort based on the following decorator as a key
data.sort(reverse=True, key=lambda x: (newest[x.object_id], x.object_id,
x.submit_date))
# --
In essence, you compute the max date within each group, but your
approach (i.e., building explicit sublists and then repeatedly
calling max on those sublists) is far more time-consuming than the
above dictionary based approach.
Note well that using a tuple as a decorator key is more efficient
than calling sort separately for each subgroup. The
lexicographical order of the following
(newest[x.object_id], x.object_id, x.submit_date)
should produce the order that you desire. The reason for the middle
entry is to ensure that items groups by object_id in the case that
two different groups achieve the same maximum date. It wasn't clear
to me whether you wanted elements within groups from oldest to
newest or newest to oldest. I believe that the code I give produces
the ordering that you intend, but you may adjust the sign of the
decorate elements if necessary.
With regard,
Michael
+---
| Michael Goldwasser
| Associate Professor
| Dept. Mathematics and Computer Science
| Saint Louis University
| 220 North Grand Blvd.
| St. Louis, MO 63103-2007
On Monday March 3, 2008, Eric Abrahamsen wrote:
>I have a grisly little sorting problem to which I've hacked together a
>solution, but I'm hoping someone here might have a better suggestion.
>
>I have a list of objects, each of which has two attributes, object_id
>and submit_date. What I want is to sort them by content_type, then by
>submit_date within content_type, and then sort each content_type block
>according to which block has the newest object by submit_date. (This
>sequence of sorting might not be optimal, I'm not sure). I'm actually
>creating a list of recent comments on blog entries for a python-based
>web framework, and want to arrange the comments according to blog
>entry (content_type), by submit_date within that entry, with the
>entries with the newest comments showing up on top.
>
>I don't believe a single cmp function fed to list.sort() can do this,
>because you can't know how two objects should be compared until you
>know all the values for all the objects. I'd be happy to be proven
>wrong here.
>
>After some false starts with dictionaries, here's what I've got.
>Queryset is the original list of comments (I'm doing this in django),
>and it returns a list of lists, though I might flatten it afterwards.
>It works, but it's ghastly unreadable and if there were a more
>graceful solution I'd feel better about life in general:
>
>
>def make_com_list(queryset):
>ids = set([com.object_id for com in queryset])
>xlist = [[com for com in queryset if com.object_id == i] for i in
>ids]
>for ls in xlist:
>ls.sort(key=lambda x: x.submit_date)
>xlist.sort(key=lambda x: max([com.submit_date for com in x]),
>reverse=True)
>return xlist
>
>I'd appreciate any hints!
>
>Thanks,
>Eric
>
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Re: [Tutor] sorting objects on two attributes
Chris Fuller wrote:
> You could have a hierarchical sort
> function:
>
> def hiersort(a,b):
>if a.attr1 != b.attr1:
> return cmp(a.attr1, b.attr1)
>else:
> if a.attr2 != b.attr2:
> return cmp(a.attr2, b.attr2)
> else:
> return cmp(a.attr3, b.att3)
>
>
> l.sort(hiersort)
That is exactly what l.sort(key=lambda x: (x.attr1, x.attr2, x.attr3))
does, except the key= version is simpler and most likely faster.
You can also use
l.sort(key=operator.attrgetter('attr1', 'attr2', 'attr3'))
> You can keep nesting for more than three attributes, or you could make it
> arbitrary by setting it up recursively and setting the attribute hierarchy as
> a parameter somewhere. But that's probably unnecessarily fancy.
l.sort(key=operator.attrgetter(*list_of_attribute_names))
should work...
Kent
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Re: [Tutor] sorting objects on two attributes
Almost. And better than my original idea. You could have a hierarchical sort function: def hiersort(a,b): if a.attr1 != b.attr1: return cmp(a.attr1, b.attr1) else: if a.attr2 != b.attr2: return cmp(a.attr2, b.attr2) else: return cmp(a.attr3, b.att3) l.sort(hiersort) You can keep nesting for more than three attributes, or you could make it arbitrary by setting it up recursively and setting the attribute hierarchy as a parameter somewhere. But that's probably unnecessarily fancy. Cheers On Monday 03 March 2008 08:42, Andreas Kostyrka wrote: > Well, this assumes that all named attributes do exist. If not, you need > to replace x.attr with getattr(x, "attr", defaultvalue) ;) > > > l.sort(key=lambda x: (x.content_type, x.submit_date)) > > Now, you can construct a sorted list "t": > > t = [] > for key, item_iterator in itertools.groupby(l, key=lambda x: > (x.content_type, x.submit_date)): sorted_part = sorted(item_iterator, > key=lambda x: x.submit_date) t.append((sorted_part[-1].submit_date, key, > sorted_part)) > > t.sort() > > t = sum([x[2] for x in t], []) > > Totally untested, as written in the MTA :) > > Andreas > > Am Montag, den 03.03.2008, 22:19 +0800 schrieb Eric Abrahamsen: > > I have a grisly little sorting problem to which I've hacked together a > > solution, but I'm hoping someone here might have a better suggestion. > > > > I have a list of objects, each of which has two attributes, object_id > > and submit_date. What I want is to sort them by content_type, then by > > submit_date within content_type, and then sort each content_type block > > according to which block has the newest object by submit_date. (This > > sequence of sorting might not be optimal, I'm not sure). I'm actually > > creating a list of recent comments on blog entries for a python-based > > web framework, and want to arrange the comments according to blog > > entry (content_type), by submit_date within that entry, with the > > entries with the newest comments showing up on top. > > > > I don't believe a single cmp function fed to list.sort() can do this, > > because you can't know how two objects should be compared until you > > know all the values for all the objects. I'd be happy to be proven > > wrong here. > > > > After some false starts with dictionaries, here's what I've got. > > Queryset is the original list of comments (I'm doing this in django), > > and it returns a list of lists, though I might flatten it afterwards. > > It works, but it's ghastly unreadable and if there were a more > > graceful solution I'd feel better about life in general: > > > > > > def make_com_list(queryset): > > ids = set([com.object_id for com in queryset]) > > xlist = [[com for com in queryset if com.object_id == i] for i in > > ids] > > for ls in xlist: > > ls.sort(key=lambda x: x.submit_date) > > xlist.sort(key=lambda x: max([com.submit_date for com in x]), > > reverse=True) > > return xlist > > > > I'd appreciate any hints! > > > > Thanks, > > Eric > > ___ > > Tutor maillist - Tutor@python.org > > http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sorting objects on two attributes
Andreas Kostyrka wrote: > l.sort(key=lambda x: (x.content_type, x.submit_date)) > > Now, you can construct a sorted list "t": > > t = [] > for key, item_iterator in itertools.groupby(l, key=lambda x: (x.content_type, > x.submit_date)): > sorted_part = sorted(item_iterator, key=lambda x: x.submit_date) > t.append((sorted_part[-1].submit_date, key, sorted_part)) I think you mean for key, item_iterator in itertools.groupby(l, key=lambda x: (x.content_type)): # Group by content type only sorted_part = list(item_iterator) # No need to sort again t.append((sorted_part[-1].submit_date, key, sorted_part)) Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sorting objects on two attributes
Eric Abrahamsen wrote: > I have a list of objects, each of which has two attributes, object_id > and submit_date. What I want is to sort them by content_type, then by > submit_date within content_type, and then sort each content_type block > according to which block has the newest object by submit_date. (This > sequence of sorting might not be optimal, I'm not sure). I'm actually > creating a list of recent comments on blog entries for a python-based > web framework, and want to arrange the comments according to blog > entry (content_type), by submit_date within that entry, with the > entries with the newest comments showing up on top. This description doesn't match your code. There is no content_type in the code. I think what you want to do is group the comments by object_id, sort within each object_id group by submit_date, then sort the groups by most recent submit date. > I don't believe a single cmp function fed to list.sort() can do this, > because you can't know how two objects should be compared until you > know all the values for all the objects. I'd be happy to be proven > wrong here. Django's SortedDict might help. Perhaps this: from operator import attrgetter from django.utils.datastructures import SortedDict sd = SortedDict() for com in sorted(queryset, key=attrgetter.submit_date, reverse=True): sd.setdefault(com.object.id, []).append(com) Now sd.keys() is a list of object_ids in descending order by most recent comment, and sd[object_id] is a list of comments for object_id, also in descending order by submit_date. If you want the comments in increasing date order (which I think your code below does) then you have to reverse the lists of comments, e.g. for l in sd.values(): l.reverse() or just reverse at the point of use with the reversed() iterator. > def make_com_list(queryset): > ids = set([com.object_id for com in queryset]) > xlist = [[com for com in queryset if com.object_id == i] for i in > ids] > for ls in xlist: > ls.sort(key=lambda x: x.submit_date) > xlist.sort(key=lambda x: max([com.submit_date for com in x]), > reverse=True) No need for max() since the list is sorted; use key=lambda x: x[-1].submit_date Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sorting objects on two attributes
Well, this assumes that all named attributes do exist. If not, you need to replace x.attr with getattr(x, "attr", defaultvalue) ;) l.sort(key=lambda x: (x.content_type, x.submit_date)) Now, you can construct a sorted list "t": t = [] for key, item_iterator in itertools.groupby(l, key=lambda x: (x.content_type, x.submit_date)): sorted_part = sorted(item_iterator, key=lambda x: x.submit_date) t.append((sorted_part[-1].submit_date, key, sorted_part)) t.sort() t = sum([x[2] for x in t], []) Totally untested, as written in the MTA :) Andreas Am Montag, den 03.03.2008, 22:19 +0800 schrieb Eric Abrahamsen: > I have a grisly little sorting problem to which I've hacked together a > solution, but I'm hoping someone here might have a better suggestion. > > I have a list of objects, each of which has two attributes, object_id > and submit_date. What I want is to sort them by content_type, then by > submit_date within content_type, and then sort each content_type block > according to which block has the newest object by submit_date. (This > sequence of sorting might not be optimal, I'm not sure). I'm actually > creating a list of recent comments on blog entries for a python-based > web framework, and want to arrange the comments according to blog > entry (content_type), by submit_date within that entry, with the > entries with the newest comments showing up on top. > > I don't believe a single cmp function fed to list.sort() can do this, > because you can't know how two objects should be compared until you > know all the values for all the objects. I'd be happy to be proven > wrong here. > > After some false starts with dictionaries, here's what I've got. > Queryset is the original list of comments (I'm doing this in django), > and it returns a list of lists, though I might flatten it afterwards. > It works, but it's ghastly unreadable and if there were a more > graceful solution I'd feel better about life in general: > > > def make_com_list(queryset): > ids = set([com.object_id for com in queryset]) > xlist = [[com for com in queryset if com.object_id == i] for i in > ids] > for ls in xlist: > ls.sort(key=lambda x: x.submit_date) > xlist.sort(key=lambda x: max([com.submit_date for com in x]), > reverse=True) > return xlist > > I'd appreciate any hints! > > Thanks, > Eric > ___ > Tutor maillist - Tutor@python.org > http://mail.python.org/mailman/listinfo/tutor signature.asc Description: Dies ist ein digital signierter Nachrichtenteil ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] sorting objects on two attributes
I have a grisly little sorting problem to which I've hacked together a solution, but I'm hoping someone here might have a better suggestion. I have a list of objects, each of which has two attributes, object_id and submit_date. What I want is to sort them by content_type, then by submit_date within content_type, and then sort each content_type block according to which block has the newest object by submit_date. (This sequence of sorting might not be optimal, I'm not sure). I'm actually creating a list of recent comments on blog entries for a python-based web framework, and want to arrange the comments according to blog entry (content_type), by submit_date within that entry, with the entries with the newest comments showing up on top. I don't believe a single cmp function fed to list.sort() can do this, because you can't know how two objects should be compared until you know all the values for all the objects. I'd be happy to be proven wrong here. After some false starts with dictionaries, here's what I've got. Queryset is the original list of comments (I'm doing this in django), and it returns a list of lists, though I might flatten it afterwards. It works, but it's ghastly unreadable and if there were a more graceful solution I'd feel better about life in general: def make_com_list(queryset): ids = set([com.object_id for com in queryset]) xlist = [[com for com in queryset if com.object_id == i] for i in ids] for ls in xlist: ls.sort(key=lambda x: x.submit_date) xlist.sort(key=lambda x: max([com.submit_date for com in x]), reverse=True) return xlist I'd appreciate any hints! Thanks, Eric ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Need help with encoder & decryption keys
Well, actually, ssh can also protect private keys with a cryptographic pass phrase. But this is often not what is wanted as it implies that the user needs to provide the pass phrase every time it is used. (Well, that's not the complete truth, man ssh-agent, but that's completely different thing ;) ) Andreas Kent Johnson wrote: > Trey Keown wrote: > >> mmm... So, what would be an effective way to hide the data's key? >> I'm kind of new to the whole encryption scene, although I've had some >> experience with it whilst working on homebrew software on gaming >> platforms. > > I don't know, I'm not a crypto expert. I guess it depends partly on what > you are trying to do. > > I do have a little experience with SSH, it stores keys in files in the > filesystem and relies on the OS to protect them from unauthorized access. > > Kent > > PS Please use Reply All to reply to the list. > ___ > Tutor maillist - Tutor@python.org > http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Need help with encoder & decryption keys
And if it's a string constant, in many cases running strings (Unix program) on the pyc file will reveal it too. All this basically turns down to the problem, that it's hard to embed an encryption key in a program, so that it's not possible to extract it. Notice the the current crop of HDDVD/Blueray decrypters, that tend to derive their copy of the key by extracting it from legal software players. (Hint: it's a basically unsolvable problem, notice how the industry "solved" it by making that kind of analysis illegal, DMCA is the hint here.) If you want to have "typical" C program security for embedded key data, the only thing that you can do is to make a Pyrex module (Which gets compiled to C, and is loaded as an .so shared library). As pointed out above, this is not really safe, it just makes it slightly harder to extract the data. Some thoughts: 1.) the data cannot be made accessible to Python, or else you can read it out. That means decryption/encryption needs to be done in Pyrex/C. PLUS Pyrex should not use any Python functions/modules to achieve the goal. (import secretmodule ; print secretmodule.value. OR: import md5 ; md5.md5 = myLoggingMd5Replacement ) 2.) the stuff done in your C module cannot be to trivial, or an attacker that is determinated can easily remove references to the module. So to summarize: Your idea usually makes no sense. It clearly falls into the "know what you are doing and why" ** n category, with n > 1 ;) There might be technical or legal reasons to do that, BUT they are really, really seldom. And this category question is usually not really appropriate for a tutoring mailing list :) The only way to make sure something is not compromised is to avoid giving it out completely, and put it on your own servers. Again, the above thoughts apply. Plus consider that your program is running in an environment that you do not control: E.g. even if you communicate via SSL and check certificates, and embed the CA certificate into your app so that it cannot be easily replaced. Consider loading a small wrapper for SSL_read/SSL_write via LD_PRELOAD. Oops, the user just learned the complete clear text of your communication. Worse if it's to simple he can just replace the data, worst case by loading a wrapper around openssl that mimics the server. It's no fun, and the easiest way is to give your users the access (e.g. the source code). This stops the crowd that has to prove itself for the fun of breaking a security system. (And yes, there are people that do that). Then put your rules into the LICENSE. That lays out the rules what is allowed. And in some cases, add some lines that check "licenses", so a customer cannot claim that he ran your program unauthorized by mistake. Not much more you can do here, sorry. Making it hard to remove the license check means just that the fun crowd might get motivated into breaking it. Leaving out at least some checks means that users can claim that they ran a copy of your program by mistake. That might have legal ramifications, and worse, in many cases (if you are not a member of BSA), you wouldn't want to sue a customer anyway, right? (Nothing surer to piss off a customer than to sue him.) Andreas Kent Johnson wrote: > Trey Keown wrote: >> is it >> possible to decompile things within a .pyc file? > > Yes, it is possible. There is a commercial service that will do this, > for older versions of Python at least. > > To figure out a secret key kept in a .pyc file it might be enough to > disassemble functions in the module; that can be done with the standard > dis module. > > And of course if you stored the secrets as module globals all you have > to do is import the module and print the value... > > Kent > ___ > Tutor maillist - Tutor@python.org > http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
