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re sub help

hi

i have a string :
a =
"this\nis\na\ns entence[startdelim]this\nis\nanoth er[enddelim]this\nis\n"

inside the string, there are "\n". I don't want to substitute the '\n'
in between
the [startdelim] and [enddelim] to ''. I only want to get rid of the
'\n' everywhere else.

i have read the tutorial and came across negative/positive lookahead
and i think it can solve the problem.but am confused on how to use it.
anyone can give me some advice? or is there better way other than
lookaheads ...thanks..

Nov 5 '05 #1
7 1358
s9************@ yahoo.com writes:
hi

i have a string :
a =
"this\nis\na\ns entence[startdelim]this\nis\nanoth er[enddelim]this\nis\n"

inside the string, there are "\n". I don't want to substitute the '\n'
in between
the [startdelim] and [enddelim] to ''. I only want to get rid of the
'\n' everywhere else.


Well, I'm not an expert on re's - I've only been using them for three
decades - but I'm not sure this can be done with a single re, as the
pattern you're interested in depends on context, and re's don't handle
that well.

On the

--
Mike Meyer <mw*@mired.or g> http://www.mired.org/home/mwm/
Independent WWW/Perforce/FreeBSD/Unix consultant, email for more information.
Nov 5 '05 #2
s9************@ yahoo.com writes:
hi

i have a string :
a =
"this\nis\na\ns entence[startdelim]this\nis\nanoth er[enddelim]this\nis\n"

inside the string, there are "\n". I don't want to substitute the '\n'
in between
the [startdelim] and [enddelim] to ''. I only want to get rid of the
'\n' everywhere else.


Well, I'm not an expert on re's - I've only been using them for three
decades - but I'm not sure this can be done with a single re, as the
pattern you're interested in depends on context, and re's don't handle
that well.

On the other hand, this is fairly straightforward with simple string
operations:
a = "this\nis\na\ns entence[startdelim]this\nis\nanoth er[enddelim]this\nis\n"
sd = '[startdelim]'
ed = '[enddelim]'
s, r = a.split(sd, 1)
m, e = r.split(ed, 1)
a = s + sd + m.replace('\n', '') + ed + e
a 'this\nis\na\ns entence[startdelim]thisisanother[enddelim]this\nis\n'


<mike

--
Mike Meyer <mw*@mired.or g> http://www.mired.org/home/mwm/
Independent WWW/Perforce/FreeBSD/Unix consultant, email for more information.
Nov 5 '05 #3
thanks for the reply.

i am still interested about using re, i find it useful. am still
learning it's uses.
so i did something like this for a start, trying to get everything in
between [startdelim] and [enddelim]

a =
"this\nis\na\ns entence[startdelim]this\nis\nanoth er[enddelim]this\nis\n"

t = re.compile(r"\[startdelim\](.*)\[enddelim\]")

t.findall(a)
but it gives me []. it's the "\n" that prevents the results.
why can't (.*) work in this case? Or am i missing some steps to "read"
in the "\n"..?
thanks.

Nov 5 '05 #4
<s9************ @yahoo.com> wrote:
i am still interested about using re, i find it useful. am still
learning it's uses.
so i did something like this for a start, trying to get everything in
between [startdelim] and [enddelim]

a =
"this\nis\na\ns entence[startdelim]this\nis\nanoth er[enddelim]this\nis\n"

t = re.compile(r"\[startdelim\](.*)\[enddelim\]")
"*" is greedy (=searches backwards from the right end), so that won't
do the right thing if you have multiple delimiters

to fix this, use "*?" instead.
t.findall(a)
but it gives me []. it's the "\n" that prevents the results.
why can't (.*) work in this case? Or am i missing some steps to "read"
in the "\n"..?


http://docs.python.org/lib/re-syntax.html

(Dot.) In the default mode, this matches any character except
a newline. If the DOTALL flag has been specified, this matches any
character including a newline.

to fix this, pass in re.DOTALL or re.S as the flag argument, or
prepend (?s) to the expression.

</F>

Nov 5 '05 #5
s9************@ yahoo.com writes:
i am still interested about using re, i find it useful. am still
learning it's uses.
so i did something like this for a start, trying to get everything in
between [startdelim] and [enddelim]

a =
"this\nis\na\ns entence[startdelim]this\nis\nanoth er[enddelim]this\nis\n"

t = re.compile(r"\[startdelim\](.*)\[enddelim\]")

t.findall(a)
but it gives me []. it's the "\n" that prevents the results.
why can't (.*) work in this case? Or am i missing some steps to "read"
in the "\n"..?
thanks.


Newlines are magic to regular expressions. You use the flags in re to
change that. In this case, you want . to match them, so you use the
DOTALL flag:
a = "this\nis\na\ns entence[startdelim]this\nis\nanoth er[enddelim]this\nis\n"
t = re.compile(r"\[startdelim\](.*)\[enddelim\]", re.DOTALL)
t.findall(a) ['this\nis\nanot her']


<mike
--
Mike Meyer <mw*@mired.or g> http://www.mired.org/home/mwm/
Independent WWW/Perforce/FreeBSD/Unix consultant, email for more information.
Nov 5 '05 #6
s9************@ yahoo.com wrote:
hi

i have a string :
a =
"this\nis\na\ns entence[startdelim]this\nis\nanoth er[enddelim]this\nis\n"

inside the string, there are "\n". I don't want to substitute the '\n'
in between
the [startdelim] and [enddelim] to ''. I only want to get rid of the
'\n' everywhere else.


Here is a solution using re.sub and a class that maintains state. It works when the input text contains multiple startdelim/enddelim pairs.

import re

a = "this\nis\na\ns entence[startdelim]this\nis\nanoth er[enddelim]this\nis\n" * 2

class subber(object):
def __init__(self):
self.delimiterS een = False

def __call__(self, m):
text = m.group()
if text == 'startdelim':
self.delimiterS een = True
return text

if text == 'enddelim':
self.delimiterS een = False
return text

if self.delimiterS een:
return text

return ''

delimRe = re.compile('\n| startdelim|endd elim')

newText = delimRe.sub(sub ber(), a)
print repr(newText)
Kent
Nov 5 '05 #7
On 4 Nov 2005 22:49:03 -0800, s9************@ yahoo.com wrote:
hi

i have a string :
a =
"this\nis\na\n sentence[startdelim]this\nis\nanoth er[enddelim]this\nis\n"

inside the string, there are "\n". I don't want to substitute the '\n'
in between
the [startdelim] and [enddelim] to ''. I only want to get rid of the
'\n' everywhere else.

i have read the tutorial and came across negative/positive lookahead
and i think it can solve the problem.but am confused on how to use it.
anyone can give me some advice? or is there better way other than
lookaheads ...thanks..


Sometimes splitting and processing the pieces selectively can be a solution, e.g.,
if delimiters are properly paired, splitting (with parens to keep matches) should
give you a repeating pattern modulo 4 of
<"everywhere else" as you said><first delim><between> <second delim> ...
a = "this\nis\na\ns entence[startdelim]this\nis\nanoth er[enddelim]this\nis\n"
import re
splitter = re.compile(r'(? s)(\[startdelim\]|\[enddelim\])')
sp = splitter.split( a)
sp ['this\nis\na\ns entence', '[startdelim]', 'this\nis\nanot her', '[enddelim]', 'this\nis\n'] ''.join([(lambda s:s, lambda s:s.replace('\n ',''))[not i%4](s) for i,s in enumerate(sp)]) 'thisisasentenc e[startdelim]this\nis\nanoth er[enddelim]thisis' print ''.join([(lambda s:s, lambda s:s.replace('\n ',''))[not i%4](s) for i,s in enumerate(sp)]) thisisasentence[startdelim]this
is
another[enddelim]thisis

I haven't checked for corner cases, but HTH
Maybe I'll try two pairs of delimiters:
a += "2222\n33\n4\n5 5555555[startdelim]6666\n77\n88888 88[enddelim]9999\n00\n"
sp = splitter.split( a)
print ''.join([(lambda s:s, lambda s:s.replace('\n ',''))[not i%4](s) for i,s in enumerate(sp)]) thisisasentence[startdelim]this
is
another[enddelim]thisis222233455 555555[startdelim]6666
77
8888888[enddelim]999900

which came from sp ['this\nis\na\ns entence', '[startdelim]', 'this\nis\nanot her', '[enddelim]', 'this\nis\n2222 \n33
\n4\n55555555', '[startdelim]', '6666\n77\n8888 888', '[enddelim]', '9999\n00\n']

Which had the replacing when not i%4 was true
for i,s in enumerate(sp): print '%6s: %r'%(not i%4,s)

...
True: 'this\nis\na\ns entence'
False: '[startdelim]'
False: 'this\nis\nanot her'
False: '[enddelim]'
True: 'this\nis\n2222 \n33\n4\n555555 55'
False: '[startdelim]'
False: '6666\n77\n8888 888'
False: '[enddelim]'
True: '9999\n00\n'

Regards,
Bengt Richter
Nov 6 '05 #8

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