Reg Ex help

[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.

Thanks in advance,
Don

[James Thiele]

Re: Reg Ex help

don wrote:[color=blue]
> 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[/color]

Not regex, but does this do what you want?[color=blue][color=green][color=darkred]
>>> 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][/color][/color][/color]
'dbg_for_python '

[Tim Chase]

Re: Reg Ex help

> /main/parallel_branch _1/release_branch_ 1.0/dbg_for_python/CHECKEDOUT[color=blue]
> 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.[/color]

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

[Aaron Barclay]

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:
[color=blue]
>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
>
>
>[/color]

[Mirco Wahab]

Re: Reg Ex help

Hi don[color=blue]
> 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.[/color]

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.

[Bruno Desthuilliers]

Re: Reg Ex help

don a écrit :[color=blue]
> 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.[/color]

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('CH ECKEDOUT') - 1]

[Edward Elliott]

Re: Reg Ex help

Bruno Desthuilliers wrote:[color=blue]
> don a écrit :[color=green]
>> Also if there is a better way than using regex, please let me know.[/color]
>
> 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('CH ECKEDOUT') - 1][/color]

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.

[Paddy]

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:
[color=blue][color=green][color=darkred]
>>> 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][/color][/color][/color]
'dbg_for_python '[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

- Paddy.

[Paddy]

Re: Reg Ex help

P.S.

This is how it works:
[color=blue][color=green][color=darkred]
>>> s ="/main/parallel_branch _1/release_branch_ 1.0/dbg_for_python/CHECKEDOUT from /main/parallel_branch _1/release_branch_ 1.0/4"
>>> s[/color][/color][/color]
'/main/parallel_branch _1/release_branch_ 1.0/dbg_for_python/CHECKEDOUT
from /main/parallel_branch _1/release_branch_ 1.0/4'[color=blue][color=green][color=darkred]
>>> s.split()[/color][/color][/color]
['/main/parallel_branch _1/release_branch_ 1.0/dbg_for_python/CHECKEDOUT',
'from', '/main/parallel_branch _1/release_branch_ 1.0/4'][color=blue][color=green][color=darkred]
>>> s.split()[0][/color][/color][/color]
'/main/parallel_branch _1/release_branch_ 1.0/dbg_for_python/CHECKEDOUT'[color=blue][color=green][color=darkred]
>>> s.split()[0].split('/')[/color][/color][/color]
['', 'main', 'parallel_branc h_1', 'release_branch _1.0',
'dbg_for_python ', 'CHECKEDOUT'][color=blue][color=green][color=darkred]
>>> s.split()[0].split('/')[-1][/color][/color][/color]
'CHECKEDOUT'[color=blue][color=green][color=darkred]
>>> s.split()[0].split('/')[-2][/color][/color][/color]
'dbg_for_python '[color=blue][color=green][color=darkred]
>>> 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][/color][/color][/color]
'dbg_for_python '[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

- Paddy.

[bruno at modulix]

Re: Reg Ex help

Edward Elliott wrote:
(snip)[color=blue][color=green]
>>don a écrit :
>>[color=darkred]
>>>Also if there is a better way than using regex, please let me know.[/color]
>>[/color][/color]
(snip)[color=blue]
>
> 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][/color]

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 'onurb@xiludom. gro'.split('@')])"

[Edward Elliott]

Re: Reg Ex help

bruno at modulix wrote:
[color=blue]
> parts = s.replace(' ', '/').strip('/').split('/')
> branch = parts[parts.index('CH ECKEDOUT') - 1]
>
> Edward Elliott wrote:[color=green]
>>
>> marker = s.index('/CHECKEDOUT')
>> branch = s [s.rindex('/', 0, marker) + 1 : marker][/color]
>
> Much cleaner than mine. I shouldn't try to code when it's time to bed !-)[/color]

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

[Anthra Norell]

Re: Reg Ex help

>>> se = SE.SE ('<EAT> "~/[A-Za-z0-9_]+/CHECKEDOUT~==" | /= CHECKEDOUT=')[color=blue][color=green][color=darkred]
>>> se[/color][/color][/color]
('/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" <abhidon@gmail. com>
Newsgroups: comp.lang.pytho n
To: <python-list@python.org >
Sent: Thursday, May 11, 2006 7:39 PM
Subject: Reg Ex help

[color=blue]
> 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[/color]