Re: [Tutor] Ptyhon GUI doubt
2009/8/24 Reddy Etikela, Rajasekhar : > Hi, > > I am new to the Python. I have installed Python 2.6.2 version in windows XP. > > When I try to open the IDLE(Python GUI), getting the below message. Any > configuration settings required for this? > > > Please let me know the details. > > Thanks, > Raj > ___ > Tutor maillist - tu...@python.org > http://mail.python.org/mailman/listinfo/tutor > > This is usually a firewall issue. What firewall software are you using, and do you have access to modify its settings? -- Rich "Roadie Rich" Lovely There are 10 types of people in the world: those who know binary, those who do not, and those who are off by one. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Algorithm
On Sun, Aug 23, 2009 at 6:06 PM, kreglet wrote: > > Hello, > > The problem that I am having is writing an algorithm for finding all the > possible words from a given word. For example: python > > from "python" you can make the words pot, top, hop, not etc. There are few > examples for making anagrams for a word, but only the whole word. I have > tried unsuccessfully to modify some of these to suit my purpose. Also on the > web there are numerous sites where you can type in a word and it will give > you a list of all the words that could be made from it. How would I go about > this in python? Another way to do this: - sort the letters in the target word; for example 'python' becomes 'hnopty' - sort the letters in the word you want to test, for example 'pot' becomes 'opt' - walk through test letters looking for them in the target list. For each test letter, - if it matches the current target letter, go to the next letter in each list - if it is less than the current target letter, go to the next target letter - if the test letter is greater than the target letter, or you run out of target letters, there is no match - if you get to the end of the test letters, you have a match The code for this is pretty simple and doesn't require much in the way of data, just the two lists of letters. If you like I can share my implementation. Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Ptyhon GUI doubt
Hi Rich, I am using my organization machine. I am not aware of the firewall which we are using and don't have the access to it. Thanks, Raj -Original Message- From: Rich Lovely [mailto:roadier...@googlemail.com] Sent: Tuesday, August 25, 2009 4:15 PM To: Reddy Etikela, Rajasekhar Cc: tutor@python.org Subject: Re: [Tutor] Ptyhon GUI doubt 2009/8/24 Reddy Etikela, Rajasekhar : > Hi, > > I am new to the Python. I have installed Python 2.6.2 version in windows XP. > > When I try to open the IDLE(Python GUI), getting the below message. > Any configuration settings required for this? > > > Please let me know the details. > > Thanks, > Raj > ___ > Tutor maillist - tu...@python.org > http://mail.python.org/mailman/listinfo/tutor > > This is usually a firewall issue. What firewall software are you using, and do you have access to modify its settings? -- Rich "Roadie Rich" Lovely There are 10 types of people in the world: those who know binary, those who do not, and those who are off by one. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Algorithm
On Mon, Aug 24, 2009 at 8:58 PM, kreglet wrote:
>
> Wayne,
>
> > def myfunc(cmpword, mainword):
> > for letter in cmpword:
> > if mainword.gets(letter):
> > if cmpword[letter] >mainword[letter]:
> > return False
> > else:
> > return False
>
> I tried your function and couldn't get it to work. It threw an error in
> the
> line "if mainword.gets(letter):" saying that "gets" was not an attribute of
> dictionary. I tried it with "if mainword.get(letter):" -- no s but that
> would't work either.
sorry, 'get' is what I meant. You also need to add "return True" on the same
level as the else.
In [5]: word1 = {'d':1, 'o':1, 'g':1}
In [6]: word2 = {'g':1, 'o':1}
In [7]: in_word(word2, word1)
Out[7]: True
In [24]: word2 = {'b':1, 'a':1, 'r':1}
In [25]: in_word(word2, word1)
Out[25]: False
HTH,
Wayne
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[Tutor] Trouble with a Recipe ...
Hi,
Hope this email finds everyone well - roll on the weekend.
I'm trying to run http://code.activestate.com/recipes/576824/
I'm in IDLE and I try:
>>> email_it_via_gmail( {"To": "garry.bet...@gmail.com", "Subject": "Testing",
>>> "From": "garry.bet...@gmail.com"}, text="Testing")
but I get the following error:
Traceback (most recent call last):
File "", line 1, in
email_it_via_gmail( {"To": "garry.bet...@gmail.com", "Subject":
"Testing", "From": "garry.bet...@gmail.com"}, text="Testing")
File "C:\Documents and Settings\Garry\Desktop\recipe-576824-1.py",
line 50, in email_it_via_gmail
+ headers.get("Bcc", [])
TypeError: cannot concatenate 'str' and 'list' objects
Sorry, but I according to the recipe I don't need a Bcc.
Sorry, again, for such a simple question!
Cheers,
Garry
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Re: [Tutor] Trouble with a Recipe ...
At the top in the docstring it says, '"To", "Cc" and "Bcc" values must
be *lists*'.
That means instead of "To": "garry.bet...@gmail.com", you need "To":
["garry.bet...@gmail.com"] i.e. a list containing the destination
address. That's so that you could send to, cc and bcc more than one address.
HTH,
Vern
Garry Bettle wrote:
Hi,
Hope this email finds everyone well - roll on the weekend.
I'm trying to run http://code.activestate.com/recipes/576824/
I'm in IDLE and I try:
email_it_via_gmail( {"To": "garry.bet...@gmail.com", "Subject": "Testing", "From":
"garry.bet...@gmail.com"}, text="Testing")
but I get the following error:
Traceback (most recent call last):
File "", line 1, in
email_it_via_gmail( {"To": "garry.bet...@gmail.com", "Subject":
"Testing", "From": "garry.bet...@gmail.com"}, text="Testing")
File "C:\Documents and Settings\Garry\Desktop\recipe-576824-1.py",
line 50, in email_it_via_gmail
+ headers.get("Bcc", [])
TypeError: cannot concatenate 'str' and 'list' objects
Sorry, but I according to the recipe I don't need a Bcc.
Sorry, again, for such a simple question!
Cheers,
Garry
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--
This time for sure!
-Bullwinkle J. Moose
-
Vern Ceder, Director of Technology
Canterbury School, 3210 Smith Road, Ft Wayne, IN 46804
vce...@canterburyschool.org; 260-436-0746; FAX: 260-436-5137
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Re: [Tutor] Trouble with a Recipe ...
Many, many thanks Vern.
For the life of me, I couldn't remember [] are for lists: I was trying with ().
Guess I picked a bad day to stop sniffing glue!
Cheers,
Garry
On Tue, Aug 25, 2009 at 15:55, Vern Ceder wrote:
> At the top in the docstring it says, '"To", "Cc" and "Bcc" values must be
> *lists*'.
>
> That means instead of "To": "garry.bet...@gmail.com", you need "To":
> ["garry.bet...@gmail.com"] i.e. a list containing the destination address.
> That's so that you could send to, cc and bcc more than one address.
>
> HTH,
>
> Vern
>
>
> Garry Bettle wrote:
>>
>> Hi,
>>
>> Hope this email finds everyone well - roll on the weekend.
>>
>> I'm trying to run http://code.activestate.com/recipes/576824/
>>
>> I'm in IDLE and I try:
>>
> email_it_via_gmail( {"To": "garry.bet...@gmail.com", "Subject":
> "Testing", "From": "garry.bet...@gmail.com"}, text="Testing")
>>
>> but I get the following error:
>>
>> Traceback (most recent call last):
>> File "", line 1, in
>> email_it_via_gmail( {"To": "garry.bet...@gmail.com", "Subject":
>> "Testing", "From": "garry.bet...@gmail.com"}, text="Testing")
>> File "C:\Documents and Settings\Garry\Desktop\recipe-576824-1.py",
>> line 50, in email_it_via_gmail
>> + headers.get("Bcc", [])
>> TypeError: cannot concatenate 'str' and 'list' objects
>>
>> Sorry, but I according to the recipe I don't need a Bcc.
>>
>> Sorry, again, for such a simple question!
>>
>> Cheers,
>>
>> Garry
>> ___
>> Tutor maillist - tu...@python.org
>> http://mail.python.org/mailman/listinfo/tutor
>
> --
> This time for sure!
> -Bullwinkle J. Moose
> -
> Vern Ceder, Director of Technology
> Canterbury School, 3210 Smith Road, Ft Wayne, IN 46804
> vce...@canterburyschool.org; 260-436-0746; FAX: 260-436-5137
>
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[Tutor] Tutor Mailing List
For the time being, I would like to be removed from the Tutor Mailing List for Python Users. Thanks, - John Jenkinson ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Tutor Mailing List
On Tue, Aug 25, 2009 at 10:28 AM, John Jenkinson wrote: > For the time being, I would like to be removed from the Tutor Mailing List > for Python Users. Click on the link below and follow the directions... Kent > Thanks, > - John Jenkinson > ___ > Tutor maillist - tu...@python.org > http://mail.python.org/mailman/listinfo/tutor > > ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Algorithm
Wayne,
I appreciate your patience with me. I still can't get this to work:
from operator import itemgetter
class testwords:
def __init__(self):
self.lettercount={}
self.inword=False
self.mainword=""
self.cmpword=""
def countletters(word):
lc.lettercount = {}
for letter in word:
lc.lettercount[letter] =lc.lettercount.get(letter,0) + 1
print sorted(lc.lettercount.iteritems(), key=itemgetter(1))
def comparewords(cmpword, mainword):
for letter in cmpword:
if mainword.get(letter):
print letter, cmpword[letter], mainword[letter]
if cmpword[letter] >mainword[letter]:
lc.inword=False
else:
if cmpword[letter] <=mainword[letter]:
lc.inword=True
lc=testwords()
lc.mainword="batty"
lc.cmpword="bat"
countletters(lc.mainword)
mainword = lc.lettercount
countletters(lc.cmpword)
cmpword = lc.lettercount
comparewords(cmpword, mainword)
if lc.inword==True:
print lc.cmpword + " IS in: " + lc.mainword
if lc.inword==False:
print lc.cmpword + " IS NOT in: " + lc.mainword
This is confusing me:
lc.mainword="batty"
lc.cmpword="bat"
[('a', 1), ('y', 1), ('b', 1), ('t', 2)]
[('a', 1), ('b', 1), ('t', 1)]
a 1 1
b 1 1
t 1 2
bat IS in: batty
lc.mainword="batty"
lc.cmpword="byyt"
[('a', 1), ('y', 1), ('b', 1), ('t', 2)]
[('b', 1), ('t', 1), ('y', 2)]
y 2 1
b 1 1
t 1 2
byyt IS in: batty
if I put : if cmpword[letter] <=mainword[letter]:
lc.inword=True
on the same level as the else statement:
lc.mainword="batty"
lc.cmpword="bat"
[('a', 1), ('y', 1), ('b', 1), ('t', 2)]
[('a', 1), ('b', 1), ('t', 1)]
a 1 1
b 1 1
t 1 2
bat IS Not in: batty
Neither is: byyt
Isn't what comes after the else statment to catch if a letter is in the
cmpword that is not in the mainword?
lc.mainword="batty"
lc.cmpword="bst"
KeyError: 's'
Wayne-68 wrote:
>
> On Mon, Aug 24, 2009 at 8:58 PM, kreglet wrote:
>
>>
>> Wayne,
>>
>> > def myfunc(cmpword, mainword):
>> > for letter in cmpword:
>> > if mainword.gets(letter):
>> > if cmpword[letter] >mainword[letter]:
>> > return False
>> > else:
>> > return False
>>
>> I tried your function and couldn't get it to work. It threw an error in
>> the
>> line "if mainword.gets(letter):" saying that "gets" was not an attribute
>> of
>> dictionary. I tried it with "if mainword.get(letter):" -- no s but that
>> would't work either.
>
>
> sorry, 'get' is what I meant. You also need to add "return True" on the
> same
> level as the else.
>
> In [5]: word1 = {'d':1, 'o':1, 'g':1}
>
> In [6]: word2 = {'g':1, 'o':1}
>
> In [7]: in_word(word2, word1)
> Out[7]: True
>
> In [24]: word2 = {'b':1, 'a':1, 'r':1}
>
> In [25]: in_word(word2, word1)
> Out[25]: False
>
> HTH,
> Wayne
>
> ___
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>
>
--
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Sent from the Python - tutor mailing list archive at Nabble.com.
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Re: [Tutor] Algorithm
Hello Kent, Yes I would like to see your implementation. It seems to work but I get error at the end. concat=''.join mainword="python" cmpword="pot" # sort the letters in the target word; for example 'python' becomes 'hnopty' tmplst=[] for letters in mainword: tmplst.append(letters) tmplst.sort() mainword=concat(tmplst) # sort the letters in the word you want to test, for example 'pot' becomes 'opt' tmplst=[] for letters in cmpword: tmplst.append(letters) tmplst.sort() cmpword=concat(tmplst) inword=False # walk through test letters looking for them in the target list. print cmpword, mainword for letter in range(len(cmpword)): #Test for let in range(len(mainword)):#Target print cmpword[letter], mainword[let], # if it matches the current target letter, go to the next letter in each list if cmpword[letter]==mainword[let]: print "Match: " let +=1 letter +=1 # if it is less than the current target letter, go to the next target letter if cmpword[letter]mainword[let]: print "No Match:" inword=False inword=True if inword==True: print cmpword + " is IN: " + mainword else: print cmpword + " is NOT in: " + mainrowd opt hnopty o h No Match: o n No Match: o o Match: p p Match: t t Match: Traceback (most recent call last): File "/home/kreglet/bin/knt.py", line 45, in if cmpword[letter] > > Another way to do this: > - sort the letters in the target word; for example 'python' becomes > 'hnopty' > - sort the letters in the word you want to test, for example 'pot' becomes > 'opt' > - walk through test letters looking for them in the target list. For > each test letter, > - if it matches the current target letter, go to the next letter in each > list > - if it is less than the current target letter, go to the next target > letter > - if the test letter is greater than the target letter, or you run > out of target letters, there is no match > - if you get to the end of the test letters, you have a match > > The code for this is pretty simple and doesn't require much in the way > of data, just the two lists of letters. If you like I can share my > implementation. > > Kent > ___ > Tutor maillist - Tutor@python.org > http://mail.python.org/mailman/listinfo/tutor > > -- View this message in context: http://www.nabble.com/Algorithm-tp25107922p25142030.html Sent from the Python - tutor mailing list archive at Nabble.com. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Algorithm
On Tue, Aug 25, 2009 at 5:03 PM, kreglet wrote:
>
> Hello Kent,
>
> Yes I would like to see your implementation. It seems to work but I get
> error at the end.
>
> concat=''.join
>
> mainword="python"
> cmpword="pot"
> # sort the letters in the target word; for example 'python' becomes
> 'hnopty'
> tmplst=[]
> for letters in mainword:
> tmplst.append(letters)
This can be written as
tmplst = list(mainword)
> tmplst.sort()
The above could be greatly simplified to
tmplst = sorted(mainword)
> mainword=concat(tmplst)
You don't really need to convert back to a string, the list would work fine.
> # sort the letters in the word you want to test, for example 'pot' becomes
> 'opt'
> tmplst=[]
> for letters in cmpword:
> tmplst.append(letters)
>
> tmplst.sort()
> cmpword=concat(tmplst)
Same as above
> inword=False
> # walk through test letters looking for them in the target list.
> print cmpword, mainword
> for letter in range(len(cmpword)): #Test
> for let in range(len(mainword)):#Target
You want to walk the two lists together. Nested loops is not the right
structure. Here is my version:
""" Find all words that can be made with the letters of a given word """
def contains(word, letters):
''' Test whether all letters of word are in letters.
Letters must be a sorted list.'''
letterIndex = 0;
for letter in sorted(word):
if letterIndex >= len(letters):
# Ran out of letters
return False
while letter > letters[letterIndex] and letterIndex < len(letters) -1:
# Skip letters that are too small
letterIndex += 1
if letter == letters[letterIndex]:
# A match, keep checking
letterIndex += 1
continue
# No match for current letter
return False
return True
def findWordsIn(targetWord, minLen=3):
letters = sorted(targetWord)
for word in open('/usr/share/dict/words'):
word = word.strip()
if len(word) < minLen:
continue
if contains(word, letters):
print word
findWordsIn('python', 3)
Prints:
hop
hot
hoy
hyp
hypo
not
noy
nth
opt
pho
phon
phony
phot
phyton
poh
pon
pont
pony
pot
poy
python
tho
thon
thy
ton
tony
top
toph
toy
typo
yon
yont
yot
Kent
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[Tutor] how to remove first '/'
Hello, I want to strip the first '/' from the following: '/path/to/file' How can I do this? Dave ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] how to remove first '/'
On Tue, Aug 25, 2009 at 3:55 PM, wrote:
> Hello,
> I want to strip the first '/' from the following:
>
> '/path/to/file'
>
> How can I do this?
>
> Dave
> ___
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>
there are many ways to do this, heres a few:
x = "/path/to/file"
print x.partition("/")[-1]
print x[1:]
print "/".join(x.split("/")[1:])
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Re: [Tutor] how to remove first '/'
On Tue, Aug 25, 2009 at 5:55 PM, wrote: > Hello, > I want to strip the first '/' from the following: > > '/path/to/file' > > How can I do this? > > Dave > ___ > Tutor maillist - Tutor@python.org > http://mail.python.org/mailman/listinfo/tutor > If I understand correctly >>> m = '/path/to/file' >>> m[1:] 'path/to/file' >>> -- لا أعرف مظلوما تواطأ الناس علي هضمه ولا زهدوا في إنصافه كالحقيقة.محمد الغزالي "No victim has ever been more repressed and alienated than the truth" Emad Soliman Nawfal Indiana University, Bloomington ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] how to remove first '/'
Thank you for the replies. Original Message From: Emad Nawfal (عماد نوفل) To: davidwil...@safe-mail.net Cc: tutor@python.org Subject: Re: [Tutor] how to remove first '/' Date: Tue, 25 Aug 2009 18:01:26 -0400 > > > On Tue, Aug 25, 2009 at 5:55 PM, wrote: > > > > Hello, > > I want to strip the first '/' from the following: > > > > '/path/to/file' > > > > How can I do this? > > > > Dave > > ___ > > Tutor maillist - Tutor@python.org > > http://mail.python.org/mailman/listinfo/tutor > > > > If I understand correctly > >>> m = '/path/to/file' > >>> m[1:] > 'path/to/file' > >>> > > > -- > لا أعرف مظلوما تواطأ الناس علي هضمه ولا زهدوا في إنصافه كالحقيقة.محمد > الغزالي > "No victim has ever been more repressed and alienated than the truth" > > Emad Soliman Nawfal > Indiana University, Bloomington > ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Algorithm
On Tue, Aug 25, 2009 at 2:14 PM, kreglet wrote:
>
> Wayne,
> I appreciate your patience with me. I still can't get this to work:
>
>
> from operator import itemgetter
> class testwords:
> def __init__(self):
>self.lettercount={}
> self.inword=False
>self.mainword=""
>self.cmpword=""
>
> def countletters(word):
>lc.lettercount = {}
>for letter in word:
>lc.lettercount[letter] =lc.lettercount.get(letter,0) + 1
>print sorted(lc.lettercount.iteritems(), key=itemgetter(1))
>
> def comparewords(cmpword, mainword):
>for letter in cmpword:
>if mainword.get(letter):
>print letter, cmpword[letter], mainword[letter]
> if cmpword[letter] >mainword[letter]:
> lc.inword=False
>
>else:
> if cmpword[letter] <=mainword[letter]:
> lc.inword=True
>
>
> lc=testwords()
>
> lc.mainword="batty"
> lc.cmpword="bat"
>
> countletters(lc.mainword)
> mainword = lc.lettercount
>
> countletters(lc.cmpword)
> cmpword = lc.lettercount
>
> comparewords(cmpword, mainword)
>
> if lc.inword==True:
>print lc.cmpword + " IS in: " + lc.mainword
> if lc.inword==False:
>print lc.cmpword + " IS NOT in: " + lc.mainword
This is a bit redundant - since lc.inword returns True or False you can
simply test:
if lc.inword:
#do stuff
else:
#do other stuff
That's really the proper way to do it. Also, AFAIK two "if" statements take
longer than an if/else statement. Of course we're talking about ms or less,
but it still reads better to have an if/else.
It appears that I've not been coding enough lately - and my explanation has
been a bit of a failure.
I just did a quick test and this code seems to work correctly:
def countletters(word):
lettercount = {}
for letter in word:
lettercount[letter] = lettercount.get(letter, 0) + 1
return lettercount
def comparewords(cmpword, mainword):
for letter in cmpword:
if mainword.get(letter):
if cmpword[letter] > mainword[letter]:
return False
else:
return False
return True
word1 = countletters('python')
word2 = countletters('pyz')
print comparewords(word2, word1)
at least "py" was in "python", so was "pyn", but 'pyz' was not.
HTH,
Wayne
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Re: [Tutor] Algorithm
On Tue, Aug 25, 2009 at 9:55 PM, Wayne wrote:
> I just did a quick test and this code seems to work correctly:
> def countletters(word):
> lettercount = {}
> for letter in word:
> lettercount[letter] = lettercount.get(letter, 0) + 1
> return lettercount
> def comparewords(cmpword, mainword):
> for letter in cmpword:
> if mainword.get(letter):
> if cmpword[letter] > mainword[letter]:
> return False
This could be
if cmpword[letter] > mainword.get(letter, 0):
The whole loop could be simplified a bit using dict.items():
for letter, count in cmpword.items():
if count > mainword.get(letter, 0):
return False
Kent
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Re: [Tutor] Algorithm
Wayne, Kent, Alan, and Mac Thank you for all the help and suggestions . Wayne and Kent, Both your solutions work great. The extra comments you gave on the code that I tried will help reduce the line count and make the code more readable in the rest of the project. Apparently, I still have a lot to learn. I really appreciate the time you took to help and teach me. Thanx, kreglet -- View this message in context: http://www.nabble.com/Algorithm-tp25107922p25145813.html Sent from the Python - tutor mailing list archive at Nabble.com. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Ptyhon GUI doubt
Reddy Etikela, Rajasekhar wrote: Hi Rich, I am using my organization machine. I am not aware of the firewall which we are using and don't have the access to it. You can just ignore the message if you cannot the administrator to adjust the firewall's setting. Just be aware that when IDLE is running without subprocess, the execution environment between IDLE itself and your own program is shared (although IDLE does its best to clean things up). This means your program can mess up IDLE and a previous runs of the program may affect the later runs. It may also be problematic to work with a windowing toolkit or threading as they may compete with IDLE's own GUI thread. To work around this, test your program directly from the real python interpreter (i.e. from the command line/terminal, instead of via IDLE). ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
