Re: [Tutor] Python re without string consumption
On Thu, 25 Jan 2007, Jacob Abraham wrote:
>I would like to thank you for the solution and
> the helper funtion that I have written is as follows.
That's very similar to a solution I coded up for a friend, who was doing
searches in genetic sequences, and had exactly the same problem you did.
My solution:
def myfindall(regex, seq):
resultlist=[]
pos=0
while True:
result = regex.search(seq, pos)
if result is None:
break
resultlist.append(seq[result.start():result.end()])
pos = result.start()+1
return resultlist
Using it:
>>> rexp=re.compile("B.B")
>>> sequence="BABBEBIB"
>>> print myfindall(rexp,sequence)
['BAB', 'BEB', 'BIB']
> But I do hope that future versions of Python include a regular
> expression syntax to handle such cases...
My tongue-in-cheek recommendation was that re.findall be renamed to
re.findmost.
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Re: [Tutor] Python re without string consumption
Jacob Abraham wrote:
> Hi Danny Yoo,
>
>I would like to thank you for the solution and
> the helper funtion that I have written is as follows. But I do hope
> that future versions of Python include a regular expression syntax to
> handle such cases simply because this method seems very process and
> memory intensive. I also notice that fall_back_len is a very crude
> solution.
>
> def searchall(expr, text, fall_back_len=0):
> while True:
> match = re.search(expr, text)
> if not match:
> break
> yield match
> end = match.end()
> text = text[end-fall_back_len:]
>
> for match in searchall("abca", "abcabcabca", 1):
>print match.group()
The string slicing is not needed. The search() method for a compiled re
has an optional pos parameter that tells where to start the search,. You
can start the next search at the next position after the *start* of a
successful search, so fall_back_len is not needed. How about this:
def searchall(expr, text):
searchRe = re.compile(expr)
match = searchRe.search(text)
while match:
yield match
match = searchRe.search(text, match.start() + 1)
Also, if you are just finding plain text, you don't need to use regular
expressions at all, you can use str.find():
def searchall(expr, text):
pos = text.find(expr)
while pos != -1:
yield pos
pos = text.find(expr, pos+1)
(inspired by this recipe:
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/499314)
Kent
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Re: [Tutor] Python re without string consumption
Hi Danny Yoo,
I would like to thank you for the solution and
the helper funtion that I have written is as follows. But I do hope
that future versions of Python include a regular expression syntax to
handle such cases simply because this method seems very process and
memory intensive. I also notice that fall_back_len is a very crude
solution.
def searchall(expr, text, fall_back_len=0):
while True:
match = re.search(expr, text)
if not match:
break
yield match
end = match.end()
text = text[end-fall_back_len:]
for match in searchall("abca", "abcabcabca", 1):
print match.group()
Thanks Again.
Jacob Abraham
Get your own web address.
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Re: [Tutor] Python re without string consumption
> > def searchall(expr, text, fall_back_len=0): > [snip] > > text = text[end-fallbacklen:] oops, those look like two different variables to me. ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Python re without string consumption
Hi Danny Yoo,
I would like to thank you for the solution and the helper funtion that I
have written is as follows. But I do hope that future versions of Python
include a regular expression syntax to handle such cases simply because this
method seems very process and memory intensive. I also notice that
fall_back_len is a very crude solution.
def searchall(expr, text, fall_back_len=0):
while True:
match = re.search(expr, text)
if not match:
break
yield match
end = match.end()
text = text[end-fallbacklen:]
for match in searchall("abca", "abcabcabca", 1):
print match.group()
Thanks Again.
Jacob Abraham
- Original Message
From: Danny Yoo <[EMAIL PROTECTED]>
To: Jacob Abraham <[EMAIL PROTECTED]>
Cc: python
Sent: Thursday, January 25, 2007 12:14:31 PM
Subject: Re: [Tutor] Python re without string consumption
On Wed, 24 Jan 2007, Jacob Abraham wrote:
>>>> import re
>>>> re.findall("abca", "abcabcabca")
> ["abca", "abca"]
>
> While I am expecting.
>
> ["abca", "abca", "abca"]
Hi Jacob,
Just to make sure: do you understand, though, why findall() won't give you
the results you want? The documentation on findall() says:
""" Return a list of all non-overlapping matches of pattern in string. If
one or more groups are present in the pattern, return a list of groups;
this will be a list of tuples if the pattern has more than one group.
Empty matches are included in the result unless they touch the beginning
of another match. New in version 1.5.2. Changed in version 2.4: Added the
optional flags argument. """
It's designed not to return overlapping items.
> How do I modify my regular expression to do the same.
We can just write our own helper function to restart the match. That is,
we can expliciltely call search() ourselves, and pass in a new string
that's a fragment of the old one.
Concretely,
#
>>> import re
>>> text = "abcabcabca"
>>> re.search("abca", text)
<_sre.SRE_Match object at 0x50d40>
>>> re.search("abca", text).start()
0
#
Ok, so we know the first match starts at 0. So let's just restart the
search, skipping that position.
##
>>> re.search("abca", text[1:])
<_sre.SRE_Match object at 0x785d0>
>>> re.search("abca", text[1:]).start()
2
##
There's our second match. Let's continue. We have to be careful, though,
to make sure we're skipping the right number of characters:
###
>>> re.search("abca", text[4:]).start()
2
>>> re.search("abca", text[7:])
>>>
###
And there are no matches after this point.
You can try writing this helper function yourself. If you need help doing
so, please feel free to ask the list for suggestions.
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Re: [Tutor] Python re without string consumption
On Wed, 24 Jan 2007, Jacob Abraham wrote:
import re
re.findall("abca", "abcabcabca")
> ["abca", "abca"]
>
> While I am expecting.
>
> ["abca", "abca", "abca"]
Hi Jacob,
Just to make sure: do you understand, though, why findall() won't give you
the results you want? The documentation on findall() says:
""" Return a list of all non-overlapping matches of pattern in string. If
one or more groups are present in the pattern, return a list of groups;
this will be a list of tuples if the pattern has more than one group.
Empty matches are included in the result unless they touch the beginning
of another match. New in version 1.5.2. Changed in version 2.4: Added the
optional flags argument. """
It's designed not to return overlapping items.
> How do I modify my regular expression to do the same.
We can just write our own helper function to restart the match. That is,
we can expliciltely call search() ourselves, and pass in a new string
that's a fragment of the old one.
Concretely,
#
>>> import re
>>> text = "abcabcabca"
>>> re.search("abca", text)
<_sre.SRE_Match object at 0x50d40>
>>> re.search("abca", text).start()
0
#
Ok, so we know the first match starts at 0. So let's just restart the
search, skipping that position.
##
>>> re.search("abca", text[1:])
<_sre.SRE_Match object at 0x785d0>
>>> re.search("abca", text[1:]).start()
2
##
There's our second match. Let's continue. We have to be careful, though,
to make sure we're skipping the right number of characters:
###
>>> re.search("abca", text[4:]).start()
2
>>> re.search("abca", text[7:])
>>>
###
And there are no matches after this point.
You can try writing this helper function yourself. If you need help doing
so, please feel free to ask the list for suggestions.
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