change only the nth occurrence of a pattern in a string
roy at panix.com
Wed Dec 31 16:23:05 CET 2008
In article <0scs26-7a5.ln1 at rama.fbx.proxad.net>,
TP <Tribulations at 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).
v = "coucoucoucou"
pattern = "o"
n = 2
parts = re.split(pattern, v)
first = parts[:n]
last = parts[n:]
j1 = pattern.join(first)
j2 = pattern.join(last)
print "i".join([j1, j2])
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