reusing parts of a string in RE matches?

[John Salerno]

I probably should find an RE group to post to, but my news server at
work doesn't seem to have one, so I apologize. But this is in Python
anyway :)

So my question is, how can find all occurrences of a pattern in a
string, including overlapping matches? I figure it has something to do
with look-ahead and look-behind, but I've only gotten this far:

import re
string = 'ababababababab ab'
pattern = re.compile(r'ab (?=a)')
m = pattern.findall (string)

This matches all the 'ab' followed by an 'a', but it doesn't include the
'a'. What I'd like to do is find all the 'aba' matches. A regular
findall() gives four results, but really there are seven.

Is there a way to do this with just an RE pattern, or would I have to
manually add the 'a' to the end of the matches?

Thanks.

[John Salerno]

Mirco Wahab wrote:

Py:
import re
tx = 'a1a2a3A4a35a6b 7b8c9c'
rg = r'(\w)(?=(.\1)) '
print re.findall(rg, tx)

The only problem seems to be (and I ran into this with my original
example too) that what gets returned by this code isn't exactly what you
are looking for, i.e. the numbers '1', '2', etc. You get a list of
tuples, and the second item in this tuple contains the number, but also
the following \w character.

So there still seems to be some work that must be done when dealing with
overlapping patterns/look-ahead/behind.

Oh wait, a thought just hit me. Instead of doing it as you did:

rg = r'(\w)(?=(.\1)) '

Could you do:

rg = r'(\w)(?=(.)\1) '

That would at least isolate the number, although you'd still have to get
it out of the list/tuple.

[Mirco Wahab]

Hi John

rg = r'(\w)(?=(.)\1) '

That would at least isolate the number, although you'd still have to get
it out of the list/tuple.

I have no idea how to do this
in Python in a terse way - but
I'll try ;-)

In Perl, its easy. Here, the
"match construct" (\w)(?=(.)\1)
returns all captures in a
list (a 1 a 2 a 4 b 7 c 9)
because we capture 2 fields per
comparision:
The first is (\w), needed for
backreference, the second is
the dot (.), which finds the
number in the center (or any-
thing else).

So, in perl you 'filter' by the
'grep' function on each list
element: grep{ /\d/ } - this
means, only numbers (\d) will
pass through:

$_ = 'a1a2a3Aa4a35a6 b7b8c9c';
print grep{/\d/} /(\w)(?=(.)\1)/g;
#prints => 1 2 4 7 9

I'll try to fiddle somthing out
that works in Python too ...

Regards

M.

[John Salerno]

Mirco Wahab wrote:

I have no idea how to do this
in Python in a terse way - but
I'll try ;-)

In Perl, its easy. Here, the
"match construct" (\w)(?=(.)\1)
returns all captures in a
list (a 1 a 2 a 4 b 7 c 9)

Ah, I see the difference. In Python you get a list of tuples, so there
seems to be a little extra work to do to get the number out.

[Mirco Wahab]

Hi John

Ah, I see the difference. In Python you get a list of tuples, so there
seems to be a little extra work to do to get the number out.

Dohh, after two cups of coffee
ans several bars of chocolate
I eventually mad(e) it ;-)

In Python, you have to deconstruct
the 2D-lists (here: long list of
short lists [a,2] ...) by
'slicing the slice':

char,num = list[:][:]

in a loop and using the apropriate element then:

import re

t = 'a1a2a3Aa4a35a6 b7b8c9c';
r = r'(\w)(?=(.)\1) '
l = re.findall(r, t)

for a,b in (l[:][:]) : print b

In the moment, I find this syntax
awkward and arbitary, but my mind
should change if I'm adopted more
to this in the end ;-)

Regards,

M.

[Mirco Wahab]

Hi John

Ah, I see the difference. In Python you get a list of
tuples, so there seems to be a little extra work to do
to get the number out.

Dohh, after two cups of coffee
ans several bars of chocolate
I eventually mad(e) it ;-)

In Python, you have to deconstruct
the 2D-lists (here: long list of
short lists [a,2] ...) by
'slicing the slice':

char,num = list[:][:]

in a loop and using the apropriate element then:

import re

t = 'a1a2a3Aa4a35a6 b7b8c9c';
r = r'(\w)(?=(.)\1) '
l = re.findall(r, t)

for a,b in l : print b

(l sould implicitly be decoded
sequentially as l[:][->a : ->b]
in the loop context.)

In the moment, I find this syntax
somehow hard to remember, but my
mind should change if I'm adopted
more to this in the end ;-)

Regards,

M.

[Fredrik Lundh]

Mirco Wahab wrote:

In Python, you have to deconstruct
the 2D-lists (here: long list of
short lists [a,2] ...) by
'slicing the slice':

char,num = list[:][:]

in a loop and using the apropriate element then:

import re

t = 'a1a2a3Aa4a35a6 b7b8c9c';
r = r'(\w)(?=(.)\1) '
l = re.findall(r, t)

for a,b in (l[:][:]) : print b

In the moment, I find this syntax
awkward and arbitary, but my mind
should change if I'm adopted more
to this in the end ;-)

in contemporary Python, this is best done by a list comprehension:

l = [m[1] for m in re.findall(r, t)]

or, depending on what you want to do with the result, a generator
expression:

g = (m[1] for m in re.findall(r, t))

or

process(m[1] for m in re.findall(r, t))

if you want to avoid creating the tuples, you can use finditer instead:

l = [m.group(2) for m in re.finditer(r, t)]
g = (m.group(2) for m in re.finditer(r, t))

finditer is also a good tool to use if you need to do more things with
each match:

for m in re.finditer(r, t):
s = m.group(2)
... process s in some way ...

the code body will be executed every time the RE engine finds a match,
which can be useful if you're working on large target strings, and only
want to process the first few matches.

for m in re.finditer(r, t):
s = m.group(2)
if s == something:
break
... process s in some way ...

</F>

[Mirco Wahab]

Hi Fredrik

you brought up some terse and
somehow expressive lines with
their own beauty ...

[this] is best done by a list comprehension:
l = [m[1] for m in re.findall(r, t)]

or, [...] a generator expression:
g = (m[1] for m in re.findall(r, t))

or
process(m[1] for m in re.findall(r, t))

... avoid creating the tuples, ... finditer instead:
l = [m.group(2) for m in re.finditer(r, t)]
g = (m.group(2) for m in re.finditer(r, t))

finditer is also a good tool to use
for m in re.finditer(r, t):
s = m.group(2)
... process s in some way ...

.... which made me wish to internalize such wisdom too ;-)

This looks almost beautiful, it made me stand up and
go to some large book stores in order to grab a good
book on python.

Sadly, there were none (except one small 'dictionary',
ISBN: 3826615123). I live in a fairly large city
in Germany w/three large bookstores in the center,
where one can get loads of PHP and Java books, lots
of C/C++ and the like - even some Ruby books (some
"Rails" too) on display (WTF).

Not that I wouldn't order books (I do that all the
time for 'original versions') but it makes one
sad-faced to see the small impact of the Python
language here today on bookstore-tournarounds ...

Thanks & regards

Mirco