Re: Reg Ex help
>>> se = SE.SE (' "~/[A-Za-z0-9_]+/CHECKEDOUT~==" | /= CHECKEDOUT=')
>>> se
('/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT')
'dbg_for_python'
If I understand your problem, this might be a solution. It is a stream
editor I devised on the impression that it could handle in a simple manner a
number of relatively simple problems on this list for which no
commensurately simple methodologies seem to exist. I intend to propose it to
the group when I finish the doc. Meantime who do I propose it to?
Frederic
- Original Message -
From: "don" <[EMAIL PROTECTED]>
Newsgroups: comp.lang.python
To:
Sent: Thursday, May 11, 2006 7:39 PM
Subject: Reg Ex help
> I have a string from a clearcase cleartool ls command.
>
> /main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
> from /main/parallel_branch_1/release_branch_1.0/4
>
> I want to write a regex that gives me the branch the file was
> checkedout on ,in this case - 'dbg_for_python'
>
> Also if there is a better way than using regex, please let me know.
>
> Thanks in advance,
> Don
>
> --
> http://mail.python.org/mailman/listinfo/python-list
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Re: Reg Ex help
bruno at modulix wrote:
> parts = s.replace(' ', '/').strip('/').split('/')
> branch = parts[parts.index('CHECKEDOUT') - 1]
>
> Edward Elliott wrote:
>>
>> marker = s.index('/CHECKEDOUT')
>> branch = s [s.rindex('/', 0, marker) + 1 : marker]
>
> Much cleaner than mine. I shouldn't try to code when it's time to bed !-)
Not terribly readable though, hard to tell what the magic slice indexes
mean. Yours is easier to follow. I think I'd just use a regex though.
--
Edward Elliott
UC Berkeley School of Law (Boalt Hall)
complangpython at eddeye dot net
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Re: Reg Ex help
Edward Elliott wrote:
(snip)
>>don a écrit :
>>
>>>Also if there is a better way than using regex, please let me know.
>>
(snip)
>
> I wouldn't call these better (or worse) than regexes, but a slight variation
> on the above:
>
> marker = s.index('/CHECKEDOUT')
> branch = s [s.rindex('/', 0, marker) + 1 : marker]
Much cleaner than mine. I shouldn't try to code when it's time to bed !-)
--
bruno desthuilliers
python -c "print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for
p in '[EMAIL PROTECTED]'.split('@')])"
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Re: Reg Ex help
P.S.
This is how it works:
>>> s ="/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
>>> from /main/parallel_branch_1/release_branch_1.0/4"
>>> s
'/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
from /main/parallel_branch_1/release_branch_1.0/4'
>>> s.split()
['/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT',
'from', '/main/parallel_branch_1/release_branch_1.0/4']
>>> s.split()[0]
'/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT'
>>> s.split()[0].split('/')
['', 'main', 'parallel_branch_1', 'release_branch_1.0',
'dbg_for_python', 'CHECKEDOUT']
>>> s.split()[0].split('/')[-1]
'CHECKEDOUT'
>>> s.split()[0].split('/')[-2]
'dbg_for_python'
>>> s ="/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
>>> from /main/parallel_branch_1/release_branch_1.0/4"
>>> s.split()[0].split('/')[-2]
'dbg_for_python'
>>>
- Paddy.
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Re: Reg Ex help
If you have strings of all the CHECKEDOUT (is this from the lsco
command?), then the following might work for you:
>>> s ="/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
>>> from /main/parallel_branch_1/release_branch_1.0/4"
>>> s.split()[0].split('/')[-2]
'dbg_for_python'
>>>
- Paddy.
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Re: Reg Ex help
Bruno Desthuilliers wrote:
> don a écrit :
>> Also if there is a better way than using regex, please let me know.
>
> s ="/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
> from /main/parallel_branch_1/release_branch_1.0/4"
> parts = s.replace(' ', '/').strip('/').split('/')
> branch = parts[parts.index('CHECKEDOUT') - 1]
I wouldn't call these better (or worse) than regexes, but a slight variation
on the above:
marker = s.index('/CHECKEDOUT')
branch = s [s.rindex('/', 0, marker) + 1 : marker]
This version will throw exceptions when the marker isn't found, which may or
may not be preferable under the circumstances.
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Re: Reg Ex help
don a écrit :
> I have a string from a clearcase cleartool ls command.
>
> /main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
> from /main/parallel_branch_1/release_branch_1.0/4
>
> I want to write a regex that gives me the branch the file was
> checkedout on ,in this case - 'dbg_for_python'
>
> Also if there is a better way than using regex, please let me know.
s ="/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
from /main/parallel_branch_1/release_branch_1.0/4"
parts = s.replace(' ', '/').strip('/').split('/')
branch = parts[parts.index('CHECKEDOUT') - 1]
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Re: Reg Ex help
Hi don > I have a string from a clearcase cleartool ls command. > /main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT > from /main/parallel_branch_1/release_branch_1.0/4 > I want to write a regex that gives me the branch the file was > checkedout on ,in this case - 'dbg_for_python' > Also if there is a better way than using regex, please let me know. This is a good situation where Regex come into play, because all other solutions won't catch on different string structures easily. If you know that you will need the string before CHECKEDOUT, you can, for example use some nice positive lookahead (mentioned today here) pseudo: take all strings between / ... / and return 'em if the next thing is CHECKEDOUT (or something else): / ([^/]+) / (?=CHECKEDOUT) The ([^/]+) means ^/ (not /) in a character class, [^/]+ one or more than one times and ([^/]+) capture it by (..) The code: import re t = '/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT from /main/p...' r = r'/([^/]+)/(?=CHECKEDOUT)' # print re.search(r, t).group(1) would do the job, independent of the structure of string - except the /CHECKEDOUT thing (which has to be there) If there are 'better ways' - that depends on 'better ways for whom?'. If you can handle the Railgun, why bother with the Pistols ;-) Regards M. -- http://mail.python.org/mailman/listinfo/python-list
Re: Reg Ex help
Hi don,
there may well be a better way then regex, although I find them usefull
and use them a lot.
The way they work would be dependant on knowing some things. For
example, if the dir you are after is always 4
deep in the structure you could try something like...
path =
'/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
from /main/parallel_branch_1/release_branch_1.0/4'
p = re.compile('/\S*/\S*/\S*/(\S*)/')
m = re.search(p, path)
.
if m:
print m.group(1)
This is a good reference...
http://www.amk.ca/python/howto/regex/
Hope that helps,
aaron.
don wrote:
>I have a string from a clearcase cleartool ls command.
>
>/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
>from /main/parallel_branch_1/release_branch_1.0/4
>
>I want to write a regex that gives me the branch the file was
>checkedout on ,in this case - 'dbg_for_python'
>
>Also if there is a better way than using regex, please let me know.
>
>Thanks in advance,
>Don
>
>
>
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Re: Reg Ex help
> /main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
> from /main/parallel_branch_1/release_branch_1.0/4
>
> I want to write a regex that gives me the branch the file was
> checkedout on ,in this case - 'dbg_for_python'
>
> Also if there is a better way than using regex, please let me know.
Well, if you have it all in a single string:
s =
"/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
from /main/parallel_branch_1/release_branch_1.0/4"
you can do
branch = s.split("/")[4]
which returns the branch, assuming the path from root is the
same for each item in question.
If not, you can tinker with something like
r = re.compile(r'/([^/]*)/CHECKEDOUT')
m = r.match(s)
and which should make m.groups(1) the resulting item. You
don't give much detail regarding what is constant (the
number of subdirectories in the path? the CHECKEDOUT
portion?, etc) so it's kinda hard to figure out what is most
globally applicable.
-tkc
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Re: Reg Ex help
don wrote:
> I have a string from a clearcase cleartool ls command.
>
> /main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT
> from /main/parallel_branch_1/release_branch_1.0/4
>
> I want to write a regex that gives me the branch the file was
> checkedout on ,in this case - 'dbg_for_python'
>
> Also if there is a better way than using regex, please let me know.
>
> Thanks in advance,
> Don
Not regex, but does this do what you want?
>>> s = "/main/parallel_branch_1/release_branch_1.0/dbg_for_python/CHECKEDOUT"
>>> s = s + " from /main/parallel_branch_1/release_branch_1.0/4"
>>> s.split('/')[4]
'dbg_for_python'
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