f

a regexp riddle: re.search(r'(?:(\w+), |and (\w+))+', 'whatever a, bbb, and c') =? ('a', 'bbb', 'c')

HypoNt:

I need to turn a human-readable list into a list():

print re.search(r'(?:(\w+), |and (\w+))+', 'whatever a, bbb, and
c').groups()

That currently returns ('c',). I'm trying to match "any word \w+
followed by a comma, or a final word preceded by and."

The match returns 'a, bbb, and c', but the groups return ('bbb', 'c').
What do I type for .groups() to also get the 'a'?

Please go easy on me (and no RTFM!), because I have only been using
regular expressions for about 20 years...

--
Phlip
http://bit.ly/ZeekLand

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phlip2005 (2215)
11/25/2010 4:46:18 AM
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--- On Thu, 11/25/10, Phlip <phlip2005@gmail.com> wrote:=0A> From: Phlip <p=
hlip2005@gmail.com>=0A> Subject: a regexp riddle: re.search(r'=0A> To: pyth=
on-list@python.org=0A> Date: Thursday, November 25, 2010, 8:46 AM=0A> HypoN=
t:=0A> =0A> I need to turn a human-readable list into a list():=0A> =0A> =
=A0=A0=A0print re.search(r'(?:(\w+), |and=0A> (\w+))+', 'whatever a, bbb, a=
nd=0A> c').groups()=0A> =0A> That currently returns ('c',). I'm trying to m=
atch "any=0A> word \w+=0A> followed by a comma, or a final word preceded by=
and."=0A> =0A> The match returns 'a, bbb, and c', but the groups return=0A=
> ('bbb', 'c').=0A> What do I type for .groups() to also get the 'a'?=0A> =
=0A=0AFirst of all, the 'bbb' coresponds to the first capturing=0Agroup and=
'c' the second. But 'a' is forgotten be cause=0Ait was the first match of =
the first group, but there=0Ais a second match 'bbb'.=0A=0AGenerally, a cap=
turing group only remembers the last match.=0A=0AIt also seems that your re=
may match this: 'and c',=0Awhich does not seem to be your intention.=0ASo =
it may be more intuitively written as:=0A=0Ar'(?:(\w+), )+and (\w+)'=0A=0AI=
'm not sure how to get it done in one step,=0Abut it would be easy to first=
get the whole =0Amatch, then process it with:=0A=0Are.findall(r'(\w+)(?:,|=
$)', the_whole_match)=0A=0Acheers,=0A=0AYingjie=0A=0A=0A=0A   0 Yingjie 11/25/2010 9:44:44 AM Accepting input from a human is frought with dangers and edge cases. ;) Some time ago I wrote a regular expression generator that creates regexen that can parse arbitrarily delimited text, supports quoting (to avoid accidentally separating two elements that should be treated as one), and works in both directions (text<->native). The code that generates the regex is heavily commented: https://github.com/pulp/marrow.util/blob/master/marrow/util/convert.py#L123-234 You should be able to use this as-is and simply handle the optional 'and' on the last element yourself. You can even create an instance of the class with the options you want then get the generated regular expression by running print(parser.pattern). Note that I have friends who use 'and' multiple times when describing lists of things. :P — Alice.   0 utf 11/25/2010 10:00:11 AM Now that I think about it, and can be stripped using a callback function as the 'normalize' argument to my KeywordProcessor class: def normalize(value): value = value.strip() if value.startswith("and"): value = value[3:] return value parser = KeywordProcessor(',', normalize=normalize, result=list) — Alice.   0 utf 11/25/2010 10:04:56 AM Phlip, > I'm trying to match "any word \w+ followed by a comma, or a final word preceded by and." Here's a non-regex solution that handles multi-word values and multiple instances of 'and' (as pointed out by Alice). The posted code could be simplified via list comprehension - I chose the more verbose method to illustrate the logic. def to_list( text ): text = text.replace( ' and ', ',' ) output = list() for item in text.split( ',' ): if item: output.append( item.strip() ) return output test = 'cat, dog, big fish, goat and puppy and horse' print to_list( test ) Outputs: ['cat', 'dog', 'big fish', 'goat', 'puppy', 'horse'] Malcolm   0 python 11/25/2010 2:21:16 PM On 11/24/2010 10:46 PM, Phlip wrote: > HypoNt: > > I need to turn a human-readable list into a list(): > > print re.search(r'(?:(\w+), |and (\w+))+', 'whatever a, bbb, and > c').groups() > > That currently returns ('c',). I'm trying to match "any word \w+ > followed by a comma, or a final word preceded by and." > > The match returns 'a, bbb, and c', but the groups return ('bbb', 'c'). > What do I type for .groups() to also get the 'a'? > > Please go easy on me (and no RTFM!), because I have only been using > regular expressions for about 20 years... A kind of lazy way just uses a pattern for the separators to fuel a call to re.split(). I assume that " and " and " , " are both acceptable in any position: The best I've been able to do so far (due to split's annoying habit of including the matches of any groups in the pattern I have to throw away every second element) is: >>> re.split("\s*(,|and)?\s*", 'whatever a, bbb, and c')[::2] ['whatever', 'a', 'bbb', '', 'c'] That empty string is because of the ", and" which isn't recognise as a single delimiter. A parsing package might give you better results. regards Steve -- Steve Holden +1 571 484 6266 +1 800 494 3119 PyCon 2011 Atlanta March 9-17 http://us.pycon.org/ See Python Video! http://python.mirocommunity.org/ Holden Web LLC http://www.holdenweb.com/   0 Steve 11/25/2010 3:25:52 PM On 25/11/2010 04:46, Phlip wrote: > HypoNt: > > I need to turn a human-readable list into a list(): > > print re.search(r'(?:(\w+), |and (\w+))+', 'whatever a, bbb, and > c').groups() > > That currently returns ('c',). I'm trying to match "any word \w+ > followed by a comma, or a final word preceded by and." > > The match returns 'a, bbb, and c', but the groups return ('bbb', 'c'). > What do I type for .groups() to also get the 'a'? > > Please go easy on me (and no RTFM!), because I have only been using > regular expressions for about 20 years... > Try re.findall: >>> re.findall(r'(\w+), |and (\w+)', 'whatever a, bbb, and c') [('a', ''), ('bbb', ''), ('', 'c')] You can get a list of strings like this: >>> [x or y for x, y in re.findall(r'(\w+), |and (\w+)', 'whatever a, bbb, and c')] ['a', 'bbb', 'c']   0 MRAB 11/25/2010 4:16:15 PM > Accepting input from a human is fraught with dangers and edge cases. > Here's a non-regex solution Thanks all for playing! And as usual I forgot a critical detail: I'm writing a matcher for a Morelia /viridis/ Scenario step, so the matcher must be a single regexp. http://c2.com/cgi/wiki?MoreliaViridis I'm avoiding the current situation, where Morelia pulls out (.*), and the step handler "manually" splits that up with: flags = re.split(r', (?:and )?', flags) That means I already had a brute-force version. A regexp version is always better because, especially in Morelia, it validates input. (.*) is less specific than (\w+). So if the step says: Alice has crypto keys apple, barley, and flax Then the step handler could say (if this worked): def step_user_has_crypto_keys_(self, user, *keys): r'(\w+) has crypto keys (?:(\w+), )+and (\w+)' # assert that user with those keys here That does not work because "a capturing group only remembers the last match". This would appear to be an irritating 'feature' in Regexp. The total match is 'apple, barley, and flax', but the stored groups behave as if each () were a slot, so (\w+)+ would not store "more than one group". Unless there's a Regexp workaround to mean "arbitrary number of slots for each ()", then I /might/ go with this: got = re.findall(r'(?:(\w+), )?(?:(\w+), )?(?:(\w+), )?(?:(\w+), )? (?:(\w+), and )?(\w+)$', 'whatever a, bbb, and c')
print got  #  [('a', '', '', '', 'bbb', 'c')]

The trick is to simply paste in a high number of (?:(\w+), )?
segments, assuming that nobody should plug in too many. Behavior
Driven Development scenarios should be readable and not run-on.
(Morelia has a table feature for when you actually need lots of
arguments.)

Next question: Does re.search() return a match object that I can get
('a', '', '', '', 'bbb', 'c') out of? The calls to groups() and such
always return this crazy ('a', 2, 'bbb', 'c') thing that would disturb
my user-programmers.

--
Phlip

 0
Phlip
11/25/2010 7:57:33 PM
On 25/11/2010 19:57, Phlip wrote:
>> Accepting input from a human is fraught with dangers and edge cases.
>
>> Here's a non-regex solution
>
> Thanks all for playing! And as usual I forgot a critical detail:
>
> I'm writing a matcher for a Morelia /viridis/ Scenario step, so the
> matcher must be a single regexp.
>
>    http://c2.com/cgi/wiki?MoreliaViridis
>
> I'm avoiding the current situation, where Morelia pulls out (.*), and
> the step handler "manually" splits that up with:
>
>    flags = re.split(r', (?:and )?', flags)
>
> That means I already had a brute-force version. A regexp version is
> always better because, especially in Morelia, it validates input. (.*)
> is less specific than (\w+).
>
> So if the step says:
>
>    Alice has crypto keys apple, barley, and flax
>
> Then the step handler could say (if this worked):
>
>    def step_user_has_crypto_keys_(self, user, *keys):
>        r'(\w+) has crypto keys (?:(\w+), )+and (\w+)'
>
>        # assert that user with those keys here
>
[snip]
You could do:

def step_user_has_crypto_keys_(self, user, keys):
r'(\w+) has crypto keys ((?:\w+, )+and \w+)'

to validate and capture, and then split the keys string.

 0
MRAB
11/25/2010 8:45:02 PM
In article <ad755502-bf79-46ea-b7ee-57ac6f7ee1cf@z26g2000prf.googlegroups.com>,
Phlip  <phlip2005@gmail.com> wrote:
>
>Thanks all for playing! And as usual I forgot a critical detail:
>
>I'm writing a matcher for a Morelia /viridis/ Scenario step, so the
>matcher must be a single regexp.

Why?  (You're apparently the author of Morelia, but I don't really
understand it.)
--
Aahz (aahz@pythoncraft.com)           <*>         http://www.pythoncraft.com/

"Think of it as evolution in action."  --Tony Rand

 0
aahz
11/26/2010 7:57:03 PM

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