Re: Why and how there is only one way to do something?
* Chris Mellon [EMAIL PROTECTED] [2005-12-15 06:09]: (Am I dating myself? Do teenagers still put studs on their jackets?) No. They put studs in their lips, tongues, eyebrows, navels, and sexual organs. Oh, and ears. (How quaint.) -- http://mail.python.org/mailman/listinfo/python-list
regexp non-greedy matching bug?
I want to match one or two instances of a pattern in a string. According to the docs for the 're' module ( http://python.org/doc/current/lib/re-syntax.html ) the '?' qualifier is greedy by default, and adding a '?' after a qualifier makes it non-greedy. The *, +, and ? qualifiers are all greedy... Adding ? after the qualifier makes it perform the match in non-greedy or minimal fashion... In the following example, though my re is intended to allow for 1 or 2 instinces of 'foo', there are 2 in the string I'm matching. So, I would expect group(1) and group(3) to both be populated. (When I remove the conditional match on the 2nd foo, the grouping is as I expect.) $ python2.4 Python 2.4.1 (#2, Mar 31 2005, 00:05:10) [GCC 3.3 20030304 (Apple Computer, Inc. build 1666)] on darwin Type help, copyright, credits or license for more information. import re foofoo = re.compile(r'^(foo)(.*?)(foo)?(.*?)$') foofoo.match(s).group(0) 'foobarbazfoobar' foofoo.match(s).group(1) 'foo' foofoo.match(s).group(2) '' foofoo.match(s).group(3) foofoo.match(s).group(4) 'barbazfoobar' foofoo = re.compile(r'^(foo)(.*?)(foo)(.*?)$') foofoo.match(s).group(0) 'foobarbazfoobar' foofoo.match(s).group(1) 'foo' foofoo.match(s).group(2) 'barbaz' foofoo.match(s).group(3) 'foo' foofoo.match(s).group(4) 'bar' So, is this a bug, or just a problem with my understanding? If it's my brain that's broken, what's the proper way to do this with regexps? And, if the above is expected behavior, should I submit a doc bug? It's clear that the ? qualifier (applied to the second foo group) is _not_ greedy in this situation. -John -- http://mail.python.org/mailman/listinfo/python-list
Re: regexp non-greedy matching bug?
[John Hazen] I want to match one or two instances of a pattern in a string. s = 'foobarbazfoobar' foofoo = re.compile(r'^(foo)(.*?)(foo)?(.*?)$') foofoo.match(s).group(1) 'foo' foofoo.match(s).group(3) [Tim Peters] Your problem isn't that (foo)? is not greedy (it is greedy), it's that your first (.*?) is not greedy. Remember that regexps also work left to right. Well, I had the same symptoms when that .* was greedy (it ate up the optional foo), which is why I went to non-greedy there. I guess my error was thinking that greedy trumped non-greedy, rather than left trumping right. (ie, in order for the (foo)? to be maximally greedy, the (.*?) has to be non-maximally non-greedy :) Maybe what you're looking for is ^P(.*P)?.*$ Yes. That works the way I wanted. ( ^(foo)(.*(foo))?.*$ ) Thank you, both for the specific answer, and the general education. -John -- http://mail.python.org/mailman/listinfo/python-list
Re: regexp non-greedy matching bug?
[Mike Meyer]
The thing to understand is that regular expressions are *search*
functions, that return the first parsing that matches. They search a
space of possible matches to each term in the expression. If some term
fails to match, the preceeding term goes on to its next match, and you
try again. The greedy vs. non-greedy describes the order that the
term in question tries matches. If it's greedy, it will try the
longest possible match first. If it's non-greedy, it'll try the
shortest possible match first.
That's a good explanation. Thanks.
[John]
I want to match one or two instances of a pattern in a string.
foofoo = re.compile(r'^(foo)(.*?)(foo)?(.*?)$')
[Mike]
First, this pattern doesn't look for one or two instances of foo in
a string. It looks for a string that starts with foo and maybe has a
second foo in it as well.
Right. In simplifying the expression for public consumption, one of the
terms I dropped was r'^.*?(foo)...'.
To do what you said you want to do, you want to use the split method:
foo = re.compile('foo')
if 2 = len(foo.split(s)) = 3:
print We had one or two 'foo's
Well, this would solve my dumbed down example, but each foo in the
original expression was a stand-in for a more complex term. I was using
match groups to extract the parts of the match that I wanted. Here's an
example (using Tim's correction) that actually demonstrates what I'm
doing:
s = 'zzzfoo123barxxxfoo456baryyy'
s2 = 'zzzfoo123barxxxfooyyy'
foobar2 = re.compile(r'^.*?foo(\d+)bar(.*foo(\d+)bar)?.*$')
print foobar2.match(s).group(1)
123
print foobar2.match(s).group(3)
456
print foobar2.match(s2).group(1)
123
print foobar2.match(s2).group(3)
None
Looking at re.split, it doesn't look like it returns the actual matching
text, so I don't think that fits my need.
As the founder of SPARE...
Hmm, not a very effective name. A google search didn't fing any obvious
hits (even after adding the python qualifier, and removing spare time
and spare parts hits). (I couldn't find it off your homepage,
either.)
Thanks for your help. If you have any suggestions about a non-re way to
do the above, I'd be interested.
-John
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Re: Background process for ssh port forwarding
Hi Jesse-
def hostforward():
#This is based on the assumption that the passfile is the gnus
#authinfo file, or has a similar format...
f = open(PASS_FILE, r)
f_list = f.read().split(' ')
f.close()
#Now, we get the entry after password (be slicker to make it a
#dictionary, but maybe wouldn't work as well).
pass_index = f_list.index('password') + 1
forwardpass = f_list[pass_index]
#now we connect
command = 'ssh -l %s -L 2022:%s:22 %s' % \
(login, my_server, forwarding_server)
connection = pexpect.spawn(command)
connection.expect('.*assword:')
connection.sendline(forwardpass)
If I end this with 'connection.interact()', I will end up logged in to the
forwarding server. But what I really want is to go on and run rsync to
localhost port 2022, which will forward to my_server port 22. So, how can
I put the ssh connection I set up in hostforward() in the background?
I need to make sure that connection is made before I can run the rsync
command.
I think what's happening is that when you return from 'hostforward', the
connection is being closed because of garbage collection. Python uses
(among other stuff) reference counting to tell it when to delete
objects. After hostforward returns from execution, there are no longer
any references to 'connection', so it gets deleted, which cleans up the
connection.
You probably want to add:
return connection
at the end of hostforward, and call it like:
connection = hostforward()
my_rsync_function()
connection.close() # or whatever the approved pexpect cleanup is
Hope that helps-
John
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Re: Struggling with this concept please help
* George [EMAIL PROTECTED] [2005-09-25 16:20]: Hello everyone I know many have helped but I cannot get this to work out correctly. snip Please help I have never done python before and I can't seem to get the hang of it. You posted code. Is what's in the docstring actual output? If so, what isn't it doing that you want it to? If it's not working correctly, what's it doing? Just saying I cannot get this to work doesn't give us much to go on. If you want specific help, be sure to: 1) post actual code 2) explain what you expect it to do 3) post actual output (or traceback) 4) ask a specific question about what you don't understand about item 3. Even though we know this is schoolwork, I'm OK with offering you a hint or answering a specific question. -John -- http://mail.python.org/mailman/listinfo/python-list
Re: How am I doing?
* Jason [EMAIL PROTECTED] [2005-09-19 16:28]:
I've restructured my code with the assistance of George and Mike which
is now as follows...
import random
class HiScores:
def __init__(self,hiScores):
self.hiScores=[(entry[:5],entry[5:]) for entry in hiScores]
With your redefined hiScores, the above is wrong. I think it should
just be:
self.hiScores=[entry for entry in hiScores]
Your original code used the slicing to pull the score and name out of a
single string. Since you've split the string into its two parts, you
don't need the indexing anymore.
def showScores(self):
for name,score in self.hiScores:
print %s - %s % name,score
The error you cite below is due to trying to zfill an integer, not a
string. I'm not going to modify your code, but I would use integers all
over for the scores, and only turn to a string (and do the zfill) when
outputting it.
def showScores(self):
for name,score in self.hiScores:
score = str(score).zfill(5) #untested
print %s - %s % name,score
def main():
hiScores=[('1','Alpha'),('07500','Beta'),('05000','Gamma'),('02500','Delta'),('0','Epsilon')]
This looks like an indentation error to me. Is hiScores indented in
your version?
a=HiScores(hiScores)
print Original Scores\n---
a.showScores()
while 1:
newScore=random.randint(0,1)
As I said, I would use int's throughout, but to use it as-is, change the
above line to:
newScore=str(random.randint(0,1))
if newScore.zfill(5) a.lastScore():
print Congratulations, you scored %d % newScore
name=raw_input(Please enter your name :)
a.addScore(newScore,name)
a.showScores()
if __name__==__main__:
main()
snip
1) The most important is that when run, the program crashes with
AttributeError: 'int' object has no attribute 'zfill'
I've read as many websites as I can about zfill and I can't see why on
earth it's failing.
I think I explained this above. You're trying to call a string method
on an integer.
2) The output of the predefined hiscores is now...
1 - Alpha ()
07500 - Beta ()
05000 - Gamma ()
02500 - Delta ()
0 - Epsilon ()
Are you sure it's not:
('1', 'Alpha') - ()
etc. ?
I think this is a result of your still doing indexing to separate the
score and the name, even though you've already separated the name and
score into tuples in the predefined list:
hiScores=[('1','Alpha'),('07500','Beta'),('05000','Gamma'),('02500','Delta'),('0','Epsilon')]
s=hiScores[0]
s
('1', 'Alpha')
s[:5]
('1', 'Alpha')
s[5:]
()
p = (s[:5],s[5:])
p
(('1', 'Alpha'), ())
print %s - %s % p
('1', 'Alpha') - ()
Why are there the pairing parenthesis there? George very kindly showed
me another way which was to have...
def showScores(self):
for entry in self.hiScores:
print entry[0:5], - ,entry[5:]
But using that method output the entire list in it's full format (sorry
if that's not the correct terminology). But give me a small plus mark
for changing code and not simply copying George :)
3) The hardest thing to 'understand' is the line...
self.hiScores=[(entry[:5],entry[5:]) for entry in hiScores]
This is now slicing into the tuple for each entry, instead of into the
string, so your results are unexpected. slicing past the end of a tuple
returns the empty tuple (which I think is the '()' you're getting in
your output.
TIA
HTH-
John
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Re: Sockets: code works locally but fails over LAN
* n00m [EMAIL PROTECTED] [2005-08-31 05:45]: import socket, thread host, port = '192.168.0.3', 1434 s1 = socket.socket(socket.AF_INET, socket.SOCK_STREAM) s2 = socket.socket(socket.AF_INET, socket.SOCK_STREAM) s2.connect((host, 1433)) s1.bind((host, port)) I think the problem is that you're using the same host for both the send and recieve sockets. Wouldn't you want one host to be the localhost, and one to be sql server? If it doesn't work even when only the VBscript is on a different box, are you running an OS with a firewall that you have to disable? -John -- http://mail.python.org/mailman/listinfo/python-list
Re: To the python-list moderator
This is probably a fairly bad way of contacting the python-list admins... Perhaps. Feel free to forward if you know a better way. I tried sending a question to [EMAIL PROTECTED], and got another 'don't call us, we'll maybe call you'. Maybe [EMAIL PROTECTED] But public posting allows comments by others as to similar problems, like Gregory's. And mine. FWIW, I'm posting direct to the mailing list with mutt/postfix. Since I *do* have concrol over my headers, it would be nice to see what header is 'suspicious', so I could possibly modify it. It'll be interesting to see if this post makes it to the list... -John -- http://mail.python.org/mailman/listinfo/python-list
Re: Why is this?
[[]]*2 [[], []] [[], []] == [[]]*2 True Same effect. But try the 'is' operator, to see if they are actually the same instances of 'empty list': [[], []] is [[]]*2 True Just curious, did you actually cut and paste this from a real interactive session? (I think not.) My interpreter (yes, I know it's old) gives: $ python Python 2.3 (#1, Sep 13 2003, 00:49:11) [GCC 3.3 20030304 (Apple Computer, Inc. build 1495)] on darwin Type help, copyright, credits or license for more information. [[], []] is [[]]*2 False Which is, I think, the point you were trying to make. -John -- http://mail.python.org/mailman/listinfo/python-list
Re: subscribe
* Brian Lee [EMAIL PROTECTED] [2005-07-26 06:34]: Subscribe OK. You're now subscribed to thousands of spam lists. Congratulations! ;) You can subscribe to the python-list here: http://mail.python.org/mailman/listinfo/python-list Good luck- John -- http://mail.python.org/mailman/listinfo/python-list
Re: socket programming
* Helge Aksdal [EMAIL PROTECTED] [2005-07-19 11:23]: if i then change to a console window, and telnet to this server it sends me to another one. That's probably why my program dies, how can i get my code to handle this? Trying xxx.xxx.xxx.xxx telnet: connect to address xxx.xxx.xxx.xxx: Connection refused Trying xxx.xxx.xxx.xxx Connected to xx.x. Escape character is '^]'. I don't think the server is sending you to another one. I think perhaps DNS returns two IP addresses for the name, and after the first one fails, your telnet client tries the second? I'm just guessing, since you didn't include what command you ran to do the telnet. As Grant mentioned, the telnet protocol doesn't do redirects. -John -- http://mail.python.org/mailman/listinfo/python-list
Re: Python Programming Contest
* Brian Quinlan [EMAIL PROTECTED] [2005-07-15 02:08]: You can find the first problem here: http://www.sweetapp.com/pycontest/contest1 I have one question about the problem. Is the cost we are to minimize the cost of arriving in the target city at all, or the cost of arriving at the target city at the end of the final day of the schedule? (If you were traveling to a conference, for example, you'd have a specific arrival time, and a cost to stay in the destination city until that time. But, if you were going to sight-see, then you could arrive at any time, and begin your itinerary upon arrival.) Say I can find a combination of flights that gets me to the target at the end of day 3 for 390 units, and a combination that gets me there at the end of day 4 for 400. If you count the hostel cost from day 3 to day 4, the first combination costs 410. So, which is preferred? -John P.S. I just realized this may be answered be the test suite, but I'm still at the thinking stage. -- http://mail.python.org/mailman/listinfo/python-list
Re: poker card game revisited (code included)
[Erik Max Francis] Searching for straights and flushes is much better done by masks. Interesting. I've been thinking about playing with this stuff too, but hadn't considered masks. So each card has a representation: n bits for rank, then m bits for suit. 10 0001 = 2 clubs 01 1000 = K spades ... [flupke] As for straights, if i understand correctly, you make all possible straights of the cards in the hand and then see if one matches? Yeah, I originally thought that's what you meant, too. But if you take the scheme above, and sort the cards before you merge them into the hand-bits aggregate, then you can just have *one* mask for straights, and shift it down by a bit each time you check. So the best straight mask (A high) would be (ignoring suit bits): 10 01 001000 000100 10 Then this could be shifted right for a K-high straight: 01 001000 000100 10 01 I guess that means that an Ace has to have the following representation, since it can be both at the high and low end of a straight: 11 0100 = A hearts But this may mess with bit-counting shortcuts for other calculations... This is interesting to think about. Thanks for the pointer. I'm also going to look at pokersource, though I may put that off until I at least partially re-invent the wheel for learning purposes. -John -- http://mail.python.org/mailman/listinfo/python-list
Re: Python Challenge ahead [NEW] for riddle lovers
* [EMAIL PROTECTED] [EMAIL PROTECTED] [2005-05-02 05:10]: Any hints on level 6 ? There is zip in the page source, ZIp in the image data, but have no clue what to do with them - tried many things :). Have you found the zip file yet? (I did, and I still can't figure it out. I'm wondering if there are comments embedded in the jpg that none of my viewers will show me. Hmm, maybe there's a python module that does that... I'll play with it more after I sleep on it.) -John -- http://mail.python.org/mailman/listinfo/python-list
Re: Python Challenge ahead [NEW] for riddle lovers
* Reinhold Birkenfeld [EMAIL PROTECTED] [2005-05-02 06:19]: John Hazen wrote: Have you found the zip file yet? (I did, and I still can't figure it out) It's best when you use the zipfile module. Cool. I had just unzipped the zipfile, and was looking at the files directly. The comments weren't added to the files in the OS, so I was only thinking about jpeg comments (since the photo was also a zip). I love the way this challenge is making me poke into pieces of the library I've never used before. :) -John -- http://mail.python.org/mailman/listinfo/python-list
Re: Python Challenge ahead [NEW] for riddle lovers
* Dan Bishop [EMAIL PROTECTED] [2005-05-02 21:09]: Dan Christensen wrote: Reinhold Birkenfeld [EMAIL PROTECTED] writes: Dan Christensen wrote: Roel Schroeven [EMAIL PROTECTED] writes: There's no level 12 yet though. Now there's a 12 and a 13 (at least!). Anyone solved 12? Hints? **hint below** Look at the filename for the first image you see, You mean evil1.jpg? yup. and make a guess at another filename. I've made tons of guesses, but none of them worked :-( Hmmm. The title of the html page is dealing evil. If you were playing war instead of solitaire, how would you change how you deal the cards? -John P.S. - I think I've found everything, but still haven't figured out what to do with it. No hints yet, though. I want to think about it some more... -- http://mail.python.org/mailman/listinfo/python-list
Re: Dr. Dobb's Python-URL! - weekly Python news and links (Apr 11)
* Dave Brueck [EMAIL PROTECTED] [2005-04-14 07:49]: [Roel Schroeven] Not that it really matters, but does anybody know why the weekly Python news always arrives twice? Does it only happen to me, or does it happen to others too? It's not that it irritates me or anything, I'm just being curious. [Dave Brueck] I just thought it was for emphasis. My theory has always been that it's posted once to python-list, and once to usenet, and the bridge dutifully sends each message to the other domain. (And, yes. (Obviously) I get two copies, as well. I've meen meaning to ask the same question.) -John -- http://mail.python.org/mailman/listinfo/python-list
