Re: [Tutor] Python command calls the wrong version!
Dear Python Tutors, I am new to Python, having perviously used IDL for all my scripts. I was hoping to use Python and so I have just downloaded and installed version 2.6 using the mac installer. That all went fine. I then opened up X11, all fine. Then I typed in python and got this message python: execv: /Applications/scisoft/i386/Packages/Python-2.5.4/Python.framework/Versions/2.5/Resources/Python.app/Contents/MacOS/Python: No such file or directory W!!! why is my python command trying to call a version which does not exist? I appear to have broken Python? I am sire this is a really easy thing to fix but I do not know enough about hows these programmes work to figure it out for myself. Most likely your path has been screwed up by the installation of scisoft. From the fact that it now can't find this Python 2.5, it appears you've later removed scisoft, but never changed your path back (check your .bashrc (or .cshrc, depending on your shell). Also check your PATH: $ echo $PATH and perhaps even $ which python to see what you get back. If that points to some awkwardly located python, there's your problem. Btw, if you trying to replace IDL with Python (I assume RSI IDL), you may be better off using macports, because you'll want various Python libraries modules on your Mac (eg numpy, scipy, matplotlib). And, in fact, you don't really need X11, although that may depend on the GUI toolkit you're using (TKinter works without, pygtk needs X11 GTK, the latter also available through macports); you can use start Python from the Terminal. Evert ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Python command calls the wrong version!
I am new to Python, having perviously used IDL for all my scripts. I was hoping to use Python and so I have just downloaded and installed version 2.6 using the mac installer. That all went fine. I then opened up X11, all fine. Why did you open X11? If you used the Mac installer(*) it should run as a standard Mac program, no need for X? What happens if you type Python into the standard MacOS Terminal application? Running X adds a whole bunch of extra path and environment issues to resolve, it might be easier to get it running in the vanilla MacOS first. (*) BTW Which Mac installer? - there are at least 3 that I know of! Just a thought. Alan G. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] pickling codecs
I use codecs to retain consistent unicode/utf-8 encoding and decoding for reading/writing to files. Should the codecs be applied when using the pickle/unpickle function? For example, the standard syntax is: # pickle object f = open(object, 'wb') pickle.dump(object, f, 2) # unpickle object f = open(object, 'rb') object= pickle.load(f) or should it be: # pickle object f = codecs.open(object, 'wb', 'utf-8') pickle.dump(object, f, 2) # unpickle object f = codecs.open(object, 'rb', 'utf-8') object= pickle.load(f) ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Code review, plase
On Wed, Sep 8, 2010 at 12:20 AM, Steven D'Aprano st...@pearwood.info wrote:
On Wed, 8 Sep 2010 06:39:27 am Alex wrote:
Hi all.
Could someone review my code? It's the first time I develop a
reusable module and I would like to have some feedback.
If you think it's good enough I will package it for pypi.
I put the code on pastebin: http://pastebin.com/Tz367gAM
Let's start with some little things... you talk about the Unix CronTab
command, but there's no such thing, at least on Linux:
[st...@sylar ~]$ CronTab
bash: CronTab: command not found
Absolutely right I'll update the docstring
In the _remove_crontab method, you say *All* the information contained
in the crontab file are permanently lost, but that's grammatically
incorrect -- information is a mass noun, i.e. neither singular nor
plural. So in standard English, you would say all the information in
the file is permanently lost, not are lost, in the same way that you
would say the sugar is on the table or the grass is green.
Argh! I should go back to grammar school :-) English is not my first langage
ct1 = micron.CronTab()
ct2 = micron.CronTab()
ct1.add_job('daily', 'echo BOOM!')
ct2.add_job('daily', 'echo No BOOM today')
Do they fight? What happens?
They will not be duplicated because the class will generate a unique
job_id incrementing the last id used. Since the module is just a
wrapper for the crontab command any other clashing will be managed by
the cron daemon. I will add an example to document this behaviour.
Thanks for your helpful comments.
Alex
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Re: [Tutor] Pastebin.com fork based completely on Python (Django)
Does not *look* Pythonic - have a look at this : http://paste.pocoo.org/ On Tue, Sep 7, 2010 at 6:00 PM, Carlos Guerrero guerrerocar...@gmail.com wrote: I did a pastebin.com fork in Django ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] sort problem
Hello, I have this : def sort_sequence(seq): sort_sequence([3, 4, 6, 7, 8, 2]) [2, 3, 4, 6, 7, 8] sort_sequence((3, 4, 6, 7, 8, 2)) (2, 3, 4, 6, 7, 8) sort_sequence(nothappy) 'ahnoppty' if type(seq) == type([]): seq.sort() elif type(seq)== type(()): seq = tuple(sorted(seq)) else: seq2 = list(seq) seq2.sort() print seq2 seq.join(seq2) return seq The problem is that if I want to sort the characters in a string, the list exist of the sorted characters but as soon as I convert them to a string I get the old string. What went wrong ? Roelof ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sort problem
I have this : def sort_sequence(seq): sort_sequence([3, 4, 6, 7, 8, 2]) [2, 3, 4, 6, 7, 8] sort_sequence((3, 4, 6, 7, 8, 2)) (2, 3, 4, 6, 7, 8) sort_sequence(nothappy) 'ahnoppty' if type(seq) == type([]): seq.sort() elif type(seq)== type(()): seq = tuple(sorted(seq)) else: seq2 = list(seq) seq2.sort() print seq2 seq.join(seq2) return seq The problem is that if I want to sort the characters in a string, the list exist of the sorted characters but as soon as I convert them to a string I get the old string. Carefully read the documentation for str.join: http://docs.python.org/library/stdtypes.html#str.join How does it work, what does it return, etc. Then fix the corresponding line in your code. As a hint: str.join does work quite different than list.sort; I assume you're confusing their syntaxes. Good luck, Evert What went wrong ? Roelof ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sort problem
Subject: Re: [Tutor] sort problem From: evert@gmail.com Date: Wed, 8 Sep 2010 17:26:58 +0200 CC: tutor@python.org To: rwob...@hotmail.com I have this : def sort_sequence(seq): sort_sequence([3, 4, 6, 7, 8, 2]) [2, 3, 4, 6, 7, 8] sort_sequence((3, 4, 6, 7, 8, 2)) (2, 3, 4, 6, 7, 8) sort_sequence(nothappy) 'ahnoppty' if type(seq) == type([]): seq.sort() elif type(seq)== type(()): seq = tuple(sorted(seq)) else: seq2 = list(seq) seq2.sort() print seq2 seq.join(seq2) return seq The problem is that if I want to sort the characters in a string, the list exist of the sorted characters but as soon as I convert them to a string I get the old string. Carefully read the documentation for str.join: http://docs.python.org/library/stdtypes.html#str.join How does it work, what does it return, etc. Then fix the corresponding line in your code. As a hint: str.join does work quite different than list.sort; I assume you're confusing their syntaxes. Good luck, Evert str.join(iterable)¶ How it works. It puts all the elements of iterable into one string named str. So it returns a string. Str is here seq and the iterable is the list made by list.sort so seq2 So I don't see the error in that line. Roelof ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sort problem
On Wed, Sep 8, 2010 at 11:50 AM, Roelof Wobben rwob...@hotmail.com wrote: Subject: Re: [Tutor] sort problem From: evert@gmail.com Date: Wed, 8 Sep 2010 17:26:58 +0200 CC: tutor@python.org To: rwob...@hotmail.com I have this : def sort_sequence(seq): sort_sequence([3, 4, 6, 7, 8, 2]) [2, 3, 4, 6, 7, 8] sort_sequence((3, 4, 6, 7, 8, 2)) (2, 3, 4, 6, 7, 8) sort_sequence(nothappy) 'ahnoppty' if type(seq) == type([]): seq.sort() elif type(seq)== type(()): seq = tuple(sorted(seq)) else: seq2 = list(seq) seq2.sort() print seq2 seq.join(seq2) return seq The problem is that if I want to sort the characters in a string, the list exist of the sorted characters but as soon as I convert them to a string I get the old string. Carefully read the documentation for str.join: http://docs.python.org/library/stdtypes.html#str.join How does it work, what does it return, etc. Then fix the corresponding line in your code. As a hint: str.join does work quite different than list.sort; I assume you're confusing their syntaxes. Good luck, Evert str.join(*iterable*)¶ #12af20c2e150d2eb_str.join How it works. It puts all the elements of iterable into one string named str. So it returns a string. Str is here seq and the iterable is the list made by list.sort so seq2 So I don't see the error in that line. Roelof The error is that you misunderstand the usage of str.join. It doesn't do it in place, i.e. it doesn't change the actual string, so you have to have a variable to capture the response. The biggest thing, though, is that in str.join, str is not the string to store the joined iterator in, it's the separator for the string. so, in your case, where you have seq.join(seq2) You really want seq = .join(seq2) where is the separator to join seq2 on (an empty string in this case) HTH. -- Greg Bair gregb...@gmail.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sort problem
Date: Wed, 8 Sep 2010 12:38:03 -0400 From: gregb...@gmail.com To: tutor@python.org Subject: Re: [Tutor] sort problem On Wed, Sep 8, 2010 at 11:50 AM, Roelof Wobben rwob...@hotmail.com wrote: Subject: Re: [Tutor] sort problem From: evert@gmail.com Date: Wed, 8 Sep 2010 17:26:58 +0200 CC: tutor@python.org To: rwob...@hotmail.com I have this : def sort_sequence(seq): sort_sequence([3, 4, 6, 7, 8, 2]) [2, 3, 4, 6, 7, 8] sort_sequence((3, 4, 6, 7, 8, 2)) (2, 3, 4, 6, 7, 8) sort_sequence(nothappy) 'ahnoppty' if type(seq) == type([]): seq.sort() elif type(seq)== type(()): seq = tuple(sorted(seq)) else: seq2 = list(seq) seq2.sort() print seq2 seq.join(seq2) return seq The problem is that if I want to sort the characters in a string, the list exist of the sorted characters but as soon as I convert them to a string I get the old string. Carefully read the documentation for str.join: http://docs.python.org/library/stdtypes.html#str.join How does it work, what does it return, etc. Then fix the corresponding line in your code. As a hint: str.join does work quite different than list.sort; I assume you're confusing their syntaxes. Good luck, Evert str.join(iterable)¶ How it works. It puts all the elements of iterable into one string named str. So it returns a string. Str is here seq and the iterable is the list made by list.sort so seq2 So I don't see the error in that line. Roelof The error is that you misunderstand the usage of str.join. It doesn't do it in place, i.e. it doesn't change the actual string, so you have to have a variable to capture the response. The biggest thing, though, is that in str.join, str is not the string to store the joined iterator in, it's the separator for the string. so, in your case, where you have seq.join(seq2) You really want seq = .join(seq2) where is the separator to join seq2 on (an empty string in this case) HTH. -- Greg Bair gregb...@gmail.com Oke, If I understand it right with join I can put two strings into 1 string. Roelof ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sort problem
On 9/8/2010 1:12 PM, Roelof Wobben wrote: If I understand it right You don't. What does put two strings into 1 string mean. Provide an example. What does the documentation say about join? What part of that do you not understand? -- Bob Gailer 919-636-4239 Chapel Hill NC ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sort problem
Roelof Wobben rwob...@hotmail.com wrote Carefully read the documentation for str.join: http://docs.python.org/library/stdtypes.html#str.join How does it work, what does it return, etc. Then fix the corresponding line in your code. str.join(iterable)¶ It puts all the elements of iterable into one string named str. Thats not what the documentation says... So it returns a string. Thats true., When trying to understand how a function works, or debug these kinds of errors use the prompt to experiment. It's the definitive way of seeing what Python will do. For example try: 123.join([5,6,7]) Can you see what Python has done? Use the prompt it is one of the most powerful tools you have. HTH, -- Alan Gauld Author of the Learn to Program web site http://www.alan-g.me.uk/ ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] how to create a persistent dictionary w/ cpickle?
Hi, I'm trying to a create a basic addressbook for practice. I'm using a
dictionary with cpickle, though I'm not sure how to persistently store each
instance in the dictionary. Below is the code I have so far.
If I run it once and add a contact and the details, it's fine. p.load(f)
shows the details next time I run it, but if I add another contact, and run
it again, the previous key:value doesn't show. It gets replaced by the new
one.
How can I fix this so that it adds the new key:value to the dictionary
instead of replacing the existing one? Appreciate any help, thanks.
import cPickle as p
addressbook = 'addressbook.data'
f = file(addressbook, 'r+')
class address:
def __init__(self, name, tel, email):
self.name = name
self.tel = tel
self.email = email
ab = {self.name : self.tel}
f = file(addressbook, 'r+')
p.dump(ab, f)
print p.load(f)
x = raw_input()
if x == 'add':
name = raw_input('\nName: ')
tel = raw_input('Tel: ')
email = raw_input('Email: ')
contact = address(name, tel, email)
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Re: [Tutor] sort problem
On 08/09/2010 19.12, Roelof Wobben wrote: ... Oke, If I understand it right with join I can put two strings into 1 string. Roelof Not quite. With join you can put together in one string all the elements of a list of strings. While you do so, you can also put another string as a wall between each element of the list. Let's make a little example: separator = Roelof list = [Wobben, Python, Learner] print separator.join(list) ... what you will get? Guess before you try. Francesco Nessun virus nel messaggio in uscita. Controllato da AVG - www.avg.com Versione: 9.0.851 / Database dei virus: 271.1.1/3119 - Data di rilascio: 09/07/10 08:34:00 ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sort problem
On 08/09/2010 19.12, Francesco Loffredo wrote: ... a little example: separator = Roelof list = [Wobben, Python, Learner] print separator.join(list) ... what you will get? Guess before you try. There's more, I forgot to add: print separator This is important, this method *returns* a *NEW* string, it does not modify the providing string (here separator)! This means you must save the result somewhere, if you want to use it later: together = separator.join(list) Francesco Nessun virus nel messaggio in uscita. Controllato da AVG - www.avg.com Versione: 9.0.851 / Database dei virus: 271.1.1/3119 - Data di rilascio: 09/07/10 08:34:00 ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] sort problem
Date: Wed, 8 Sep 2010 20:10:28 +0200 From: f...@libero.it To: tutor@python.org Subject: Re: [Tutor] sort problem On 08/09/2010 19.12, Francesco Loffredo wrote: ... a little example: separator = Roelof list = [Wobben, Python, Learner] print separator.join(list) ... what you will get? Guess before you try. There's more, I forgot to add: print separator This is important, this method *returns* a *NEW* string, it does not modify the providing string (here separator)! This means you must save the result somewhere, if you want to use it later: together = separator.join(list) Francesco ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor Oke, I now see what everyone try to teach me. Roelof ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] Mutable Properties
Dear Tutors, I noticed that when you use a property to represent a mutable value, I you try to use its methods, it will directly change the value returned. I know this happens because I'm not really assigning it to something new, but changing whats already there, which won't fire off the set method. I was wondering if there was a way around this. Sincerely, Me ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Mutable Properties
On Thu, 9 Sep 2010 06:59:57 am Chris King wrote: Dear Tutors, I noticed that when you use a property to represent a mutable value, I you try to use its methods, it will directly change the value returned. I know this happens because I'm not really assigning it to something new, but changing whats already there, which won't fire off the set method. I was wondering if there was a way around this. You're going to need to give an example of: (1) what you do; (2) what you want to happen; and (3) what actually happens. The simplest example I can think of is: class K(object): def __init__(self): self._private = [] def _getter(self): return self._private def _setter(self, value): self._private = list(value) seq = property(_getter, _setter) And in use: k = K() k.seq [] k.seq.append(1) k.seq [1] Works fine. -- Steven D'Aprano ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] how to create a persistent dictionary w/ cpickle?
On Thu, 9 Sep 2010 03:43:42 am Carter Danforth wrote:
Hi, I'm trying to a create a basic addressbook for practice. I'm
using a dictionary with cpickle, though I'm not sure how to
persistently store each instance in the dictionary. Below is the code
I have so far.
If I run it once and add a contact and the details, it's fine.
p.load(f) shows the details next time I run it, but if I add another
contact, and run it again, the previous key:value doesn't show. It
gets replaced by the new one.
Where do you think you are *adding* a new contact? You don't. You
*replace* the existing contact with a brand new one, every time.
The problem has nothing to do with pickle, or storing each instance in
the dictionary. Pickle is already storing each instance in the
dictionary. The problem is that you never *add* anything to the address
book, you *replace* it each time, so there is never more than two
instances in the dictionary (one key, one value).
You don't have an address BOOK, you only have a single address.
How can I fix this so that it adds the new key:value to the
dictionary instead of replacing the existing one? Appreciate any
help, thanks.
I would dump the entire address class for now and just go for something
nice and minimal. Get that working first, and then, *if necessary*,
wrap it in a class. This is Python, not Java -- we use whatever works,
and don't force everything to be a class when it doesn't have to be.
What's the simplest address record you might have? How about a name
linked to a telephone number and email?
address_book = {name: (tel, email), another_name: (tel, email), ...}
So, here's the basic algorithm:
(1) Look for the address-book. If it doesn't exist, create an empty
dictionary, and pickle it as the address-book.
(2) Read the address-book from the pickle file. It will be a dictionary,
possibly empty.
(3) Add an address to the dictionary. Don't create a new dictionary:
addresses = {} # creates a new, empty address book
addresses[Fred] = (1234 5678, f...@example.com)
addresses[Betty] = (2468 1357, be...@nowhere.com)
addresses # not empty any more
{'Betty': ('2468 1357', 'be...@nowhere.com'), 'Fred': ('1234
5678', 'f...@example.com')}
(3) Save the dictionary to the pickle file.
Once you have those steps happening manually, then wrap it into a class
so they happen automatically.
Some more comments on your code:
import cPickle as p
Now that's just lazy. While laziness in a programmer in a good thing,
this is taking it to extremes!!! You use pickle twice, three times if
you count the import. Is it really such a burden on you to
type cPickle (or pickle) two more times, that you need to rename
it p?
Excessive use of one-character names is one of the worst programming
habits you can have. Don't do this.
addressbook = 'addressbook.data'
Misleading variable name. addressbook isn't an addressbook at all,
it's a filename.
f = file(addressbook, 'r+')
You shouldn't keep the file open for large periods of time. On Windows,
it may mean that it will be locked from any other program accessing it
during that time, and it risks data corruption if your program crashes
or the power goes out.
Open and close the file just when you need it.
class address:
A minor point: it is the convention in Python that (most) classes start
with a capital letter. So Address would be the class, leaving address
available for an instance of the class:
address = Address()
def __init__(self, name, tel, email):
self.name = name
self.tel = tel
self.email = email
ab = {self.name : self.tel}
f = file(addressbook, 'r+')
p.dump(ab, f)
print p.load(f)
x = raw_input()
if x == 'add':
name = raw_input('\nName: ')
To get a blank line before the prompt, it might be nicer to do this:
print
name = raw_input('Name: ')
That's a matter of personal preference though.
tel = raw_input('Tel: ')
email = raw_input('Email: ')
contact = address(name, tel, email)
Hope this helps,
--
Steven D'Aprano
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Re: [Tutor] how to create a persistent dictionary w/ cpickle?
On 2:59 PM, Carter Danforth wrote:
Hi, I'm trying to a create a basic addressbook for practice. I'm using a
dictionary with cpickle, though I'm not sure how to persistently store each
instance in the dictionary. Below is the code I have so far.
If I run it once and add a contact and the details, it's fine. p.load(f)
shows the details next time I run it, but if I add another contact, and run
it again, the previous key:value doesn't show. It gets replaced by the new
one.
How can I fix this so that it adds the new key:value to the dictionary
instead of replacing the existing one? Appreciate any help, thanks.
import cPickle as p
addressbook = 'addressbook.data'
f = file(addressbook, 'r+')
class address:
def __init__(self, name, tel, email):
self.name = name
self.tel = tel
self.email = email
ab = {self.name : self.tel}
f = file(addressbook, 'r+')
p.dump(ab, f)
print p.load(f)
x = raw_input()
if x == 'add':
name = raw_input('\nName: ')
tel = raw_input('Tel: ')
email = raw_input('Email: ')
contact = address(name, tel, email)
I have no clue what you're trying to do with that address object; it's
confusing initialization with persistence, and you create such an
object, but never actually use it. I would remove any reference to the
address book from that object. I'd also rename it to Address, since
class names are supposed to be uppercase (PEP8)
But anyway, in your flow, you do a p.load(), but never save the result.
Every time you save to the addressbook file, you start over with just
the one entry.
One way to fix this is to save the p.load() result as a variable,
presumably of type dictionary, then add the 'contact' to that
dictionary, and at the end of the script, save that variable with
p.dump(addresses, f)
addresses = p.load(f)
interact with user --
addresses[name] = Address(name, tel, email)
p.dump(addresses, f)
I also don' t know how you get away with running your script the first
time, since it blows up if there's no existing pickle file.
DaveA
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Re: [Tutor] how to create a persistent dictionary w/ cpickle?
Carter Danforth carter.danfo...@gmail.com wrote Hi, I'm trying to a create a basic addressbook for practice. I'm using a dictionary with cpickle, though I'm not sure how to persistently store each instance in the dictionary. Below is the code I have so far. If you use a dictionary it makes more sense to use shelve rather than pickle for storage. Shelve uses pickle under the hood but it makes a file look like a dictionary so you don't need to load all the data and store it all again as you would with pickle. Unfioortunately your code appears to be truncated, however... import cPickle as p addressbook = 'addressbook.data' f = file(addressbook, 'r+') I think you should use a binary file for pickle so the mode should be 'rb' not 'r+' HTH, Alan G ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] slicing a string
On Tue, Sep 7, 2010 at 1:49 PM, Roel Schroeven rschroev_nospam...@fastmail.fm wrote: But remember that you can make it simpler if you simply don't specify the start and end points: 'hello'[::-1] 'olleh' While I know that idiom works, I haven't really found an explanation as to *why* it works that way. For a string S: * Using range, you need range(len(S),-1,-1) to give you the indexes for the string in reverse. * For slices, if you dont specify the start and end indices, they are supposed to be filled in by 0 and len(S) respectively. - So S[::-1] means, S[0:len(S):-1] , so why dont we start with index 0 here, and then go to -1 (last char) and then into an infinite loop? - Especially, when S[0:len(S):1] works that way? - Sandip ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
