returning regex matches as lists

[Jonathan Lukens]

I am in the last phase of building a Django app based on something I
wrote in Java a while back. Right now I am stuck on how to return the
matches of a regular expression as a list *at all*, and in particular
given that the regex has a number of groupings. The only method I've
seen that returns a list is .findall(string), but then I get back the
groups as tuples, which is sort of a problem.

Thank you,
Jonathan

[John Machin]

On Feb 16, 6:07 am, Jonathan Lukens <[e-mail]:

I am in the last phase of building a Django app based on something I
wrote in Java a while back. Right now I am stuck on how to return the
matches of a regular expression as a list *at all*, and in particular
given that the regex has a number of groupings. The only method I've
seen that returns a list is .findall(string), but then I get back the
groups as tuples, which is sort of a problem.

It would help if you explained what you want the contents of the list
to be, why you want a list as opposed to a tuple or a generator or
whatever ... we can't be expected to imagine why getting groups as
tuples is "sort of a problem".

Use a concrete example, e.g.

>>import re
regex = re.compile(r'(\w+)\s+(\d+)')
text = 'python 1 junk xyzzy 42 java 666'
r = regex.findall(text)
r

[('python', '1'), ('xyzzy', '42'), ('java', '666')]

>>>

What would you like to see instead?

[Gabriel Genellina]

En Fri, 15 Feb 2008 17:07:21 -0200, Jonathan Lukens
<jo*************@gmail.comescribió:

I am in the last phase of building a Django app based on something I
wrote in Java a while back. Right now I am stuck on how to return the
matches of a regular expression as a list *at all*, and in particular
given that the regex has a number of groupings. The only method I've
seen that returns a list is .findall(string), but then I get back the
groups as tuples, which is sort of a problem.

Do you want something like this?

pyre.findall(r"([a-z]+)([0-9]+)", "foo bar3 w000 no abc123")
[('bar', '3'), ('w', '000'), ('abc', '123')]
pyre.findall(r"(([a-z]+)([0-9]+))", "foo bar3 w000 no abc123")
[('bar3', 'bar', '3'), ('w000', 'w', '000'), ('abc123', 'abc', '123')]
pygroups = re.findall(r"(([a-z]+)([0-9]+))", "foo bar3 w000 no abc123")
pygroups
[('bar3', 'bar', '3'), ('w000', 'w', '000'), ('abc123', 'abc', '123')]
py[group[0] for group in groups]
['bar3', 'w000', 'abc123']

--
Gabriel Genellina

[Jonathan Lukens]

John,

(1) raw string for improved legibility
ru'(?u)\b([á-ñ]{2,}\s+)([<<"][Á-Ñá-ñ]+)(\s*-?[Á-Ñá-ñ]+)*([>>"])'

This actually escaped my notice after I had posted -- the letters with
diacritics are incorrectly decoded Cyrillic letters -- I suppose I
code use the Unicode escape sequences (the sets [á-ñ] and [Á-Ñá-ñ] are
the Cyrillic equivalents of [a-z] and [A-Za-z]) but then suddenly the
legibility goes out the window again.

(3) what appears between [] is a set of characters, so [<<"] is the
same as [<"] and probably isn't doing what you expect; have you tested
this regex for correctness?

These were angled quotation marks in the original Unicode. Sorry
again. The regex matches everything it is supposed to. The extra
parentheses were because I had somehow missed the .group method and it
had only been returning what was only in the one needed set of
parentheses.

I can't imagine how "not a programmer" implies "interested to know if
there is a more elegant way".

More carefully stated: "I am self-taught have no real training or
experience as a programmer and would be interested in seeing how a
programmer with training
and experience would go about this."

Thank you,
Jonathan