In article <0scs26-7a5....@rama.fbx.proxad.net>,
TP <tribulati...@paralleles.invalid> wrote:
> Hi everybody,
>
> I would like to change only the nth occurence of a pattern in a string.
It's a little ugly, but the following looks like it works. The gist is to
split the string on your pattern, then re-join the pieces using the
original delimiter everywhere except for the n'th splice. Split() is a
wonderful tool. I'm a hard-core regex geek, but I find that most things I
might have written a big hairy regex for are easier solved by doing split()
and then attacking the pieces.
There may be some fencepost errors here. I got the basics working, and
left the details as an exercise for the reader :-)
This version assumes the pattern is a literal string. If it's really a
regex, you'll need to put the pattern in parens when you call split(); this
will return the exact text matched each time as elements of the list. And
then your post-processing gets a little more complicated, but nothing
that's too bad.
This does a couple of passes over the data, but at least all the operations
are O(n), so the whole thing is O(n).
#!/usr/bin/python
import re
v = "coucoucoucou"
pattern = "o"
n = 2
parts = re.split(pattern, v)
print parts
first = parts[:n]
last = parts[n:]
print first
print last
j1 = pattern.join(first)
j2 = pattern.join(last)
print j1
print j2
print "i".join([j1, j2])
print v
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