regex problem

[Odd-R.]

Input is a string of four digit sequences, possibly
separated by a -, for instance like this

"1234,2222-8888,4567,"

My regular expression is like this:

rx1=re.compile(r"""\A(\b\d\d\d\d,|\b\d\d\d\d-\d\d\d\d,)*\Z""")

When running rx1.findall("1234,2222-8888,4567,")

I only get the last match as the result. Isn't
findall suppose to return all the matches?

Thanks in advance.

 

[Thomas Guettler]

Am Tue, 26 Jul 2005 09:57:23 +0000 schrieb Odd-R.:

Input is a string of four digit sequences, possibly
separated by a -, for instance like this

"1234,2222-8888,4567,"

My regular expression is like this:

rx1=re.compile(r"""\A(\b\d\d\d\d,|\b\d\d\d\d-\d\d\d\d,)*\Z""")

Hi,

try it without \A and \Z

import re
rx1=re.compile(r"""(\b\d\d\d\d,|\b\d\d\d\d-\d\d\d\d,)""")
print rx1.findall("1234,2222-8888,4567,")
# --> ['1234,', '2222-8888,', '4567,']

Thomas

 

[John Machin]

Input is a string of four digit sequences, possibly
separated by a -, for instance like this

"1234,2222-8888,4567,"

My regular expression is like this:

rx1=re.compile(r"""\A(\b\d\d\d\d,|\b\d\d\d\d-\d\d\d\d,)*\Z""")

When running rx1.findall("1234,2222-8888,4567,")

I only get the last match as the result. Isn't
findall suppose to return all the matches?

For a start, an expression that starts with \A and ends with \Z will
match the whole string (or not match at all). You have only one match.

Secondly, as you have a group in your expression, findall returns what
the group matches. Your expression matches zero or more of what your
group matches, provided there is nothing else at the start/end of the
string. The "zero or more" makes the re engine waltz about a bit; when
the music stopped, the group was matching "4567,".

Thirdly, findall should be thought of as merely a wrapper around a loop
using the search method -- it finds all non-overlapping matches of a
pattern. So the clue to get from this is that you need a really simple
pattern, like the following. You *don't* have to write an expression
that does the looping.

So here's the mean lean no-flab version -- you don't even need the
parentheses (sorry, Thomas).
['1234,', '2222-8888,', '4567,']

HTH,
John

 

[Duncan Booth]

So here's the mean lean no-flab version -- you don't even need the
parentheses (sorry, Thomas).
['1234,', '2222-8888,', '4567,']

No flab? What about all that repetition of \d? A less flabby version:
['1234,', '2222-8888,', '4567,']

 

[John Machin]

John Machin wrote:

So here's the mean lean no-flab version -- you don't even need the
parentheses (sorry, Thomas).

rx1=re.compile(r"""\b\d\d\d\d,|\b\d\d\d\d-\d\d\d\d,""")
rx1.findall("1234,2222-8888,4567,")

['1234,', '2222-8888,', '4567,']

No flab? What about all that repetition of \d? A less flabby version:


['1234,', '2222-8888,', '4567,']

OK, good idea to factor out the prefix and follow it by optional -1234.
However optimising re engines do common prefix factoring, *and* they
rewrite stuff like x{4} as xxxx.

Cheers,
John

 

[Odd-R.]

['1234,', '2222-8888,', '4567,']

Thanks all for good advice. However this last expression
also matches the first four digits when the input is more
than four digits. To resolve this problem, I first do a
match of this,

regex=re.compile(r"""\A(\b\d{4},|\d{4}-\d{4},)*(\b\d{4}|\d{4}-\d{4})\Z""")

If this turns out ok, I do a find all with your expression, and then I get
the desired result.