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
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 = 'a1a2a3Aa4a35a6b7b8c9c';
r = r'(\w)(?=(.)\1)'
l
Hi mpeters42 John
With a more complex pattern (like 'a.a': match any character between
two 'a' characters) this will get the length, but not what character is
between the a's.
Lets take this as a starting point for another example
that comes to mind. You have a string of characters
Mirco Wahab wrote:
Py:
import re
tx = 'a1a2a3A4a35a6b7b8c9c'
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
Ben Cartwright wrote:
Yes, and no extra for loops are needed! You can define groups inside
the lookahead assertion:
import re
re.findall(r'(?=(aba))', 'abababababababab')
['aba', 'aba', 'aba', 'aba', 'aba', 'aba', 'aba']
Wow, that was like magic! :)
--
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
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
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
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
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
John Salerno wrote:
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 = 'abababababababab'
pattern =
John Salerno 写道:
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
Right about now somebody usually jumps in and shows you how to do this
without using regex and using string methods instead.
I'll watch.
rd
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Bo Yang wrote:
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.
I try the code , but I give seven results !
Sorry, I meant that findall()
BartlebyScrivener wrote:
Right about now somebody usually jumps in and shows you how to do this
without using regex and using string methods instead.
I'll watch.
rd
Heh heh, I'm sure you're right, but this is more just an exercise for me
in REs, so I'm curious how you might do it,
I have to at least try :)
s = abababababababab
for x in range(len(s)):
... try:
... s.index(aba, x, x + 3)
... except ValueError:
... pass
rd
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BartlebyScrivener wrote:
I have to at least try :)
s = abababababababab
for x in range(len(s)):
... try:
... s.index(aba, x, x + 3)
... except ValueError:
... pass
rd
yeah, looks like index() or find() can be used to do it instead of RE,
but still, i'd
From the Python 2.4 docs:
findall( pattern, string[, flags])
Return a list of all ***non-overlapping*** matches of pattern in
string
By design, the regex functions return non-overlapping patterns.
Without doing some kind of looping, I think you are out of luck.
If you pattern is fixed,
otherwise i might as well just use string
methods
I think you're supposed to use string methods if you can, to avoid the
old adage about having two problems instead of one when using regex.
rd
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[EMAIL PROTECTED] wrote:
string = 'abababababababab'
pat = 'aba'
[pat for s in re.compile('(?='+pat+')').findall(string)]
['aba', 'aba', 'aba', 'aba', 'aba', 'aba', 'aba']
Wow, I have no idea how to read that RE. First off, what does it match?
Should something come before the parentheses,
John Salerno wrote:
[EMAIL PROTECTED] wrote:
string = 'abababababababab'
pat = 'aba'
[pat for s in re.compile('(?='+pat+')').findall(string)]
['aba', 'aba', 'aba', 'aba', 'aba', 'aba', 'aba']
Wow, I have no idea how to read that RE. First off, what does it match?
Should something come
John Salerno wrote:
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?
You can
Exactly,
Now this will work as long as there are no wildcards in the pattern.
Thus, only with fixed strings. But if you have a fixed string, there
is really no need to use regex, as it will complicate you life for no
real reason (as opposed to simple string methods).
With a more complex pattern
John Salerno wrote:
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 = 'abababababababab'
pattern = re.compile(r'ab(?=a)')
Yes, and no extra for loops are needed! You can define groups inside
the lookahead assertion:
import re
re.findall(r'(?=(aba))', 'abababababababab')
['aba', 'aba', 'aba', 'aba', 'aba', 'aba', 'aba']
Wonderful and this works with any regexp, so
import re
def
Murali wrote:
Yes, and no extra for loops are needed! You can define groups inside
the lookahead assertion:
import re
re.findall(r'(?=(aba))', 'abababababababab')
['aba', 'aba', 'aba', 'aba', 'aba', 'aba', 'aba']
Wonderful and this works with any regexp, so
import re
def
Thanks, Ben. Quite an education!
rick
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