Antoon Pardon wrote:
> On 2008-12-31, TP <tribulati...@paralleles.invalid> wrote:
>> Hi everybody,
>>
>> I would like to change only the nth occurence of a pattern in a
string. The
>> problem with "replace" method of strings, and "re.sub" is that we
can only
>> define the number of occurrences to change from the first one.
>>
>>>>> v="coucou"
>>>>> v.replace("o","i",2)
>> 'ciuciu'
>>>>> import re
>>>>> re.sub( "o", "i", v,2)
>> 'ciuciu'
>>>>> re.sub( "o", "i", v,1)
>> 'ciucou'
>>
>> What is the best way to change only the nth occurence (occurrence
number n)?
>>
>> Why this default behavior? For the user, it would be easier to put
re.sub or
>> replace in a loop to change the first n occurences.
>
> I would do it as follows:
>
> 1) Change the pattern n times to somethings that doesn't occur in
your string
> 2) Change it back n-1 times
> 3) Change the remaining one to what you want.
>
>>>> v="coucou"
>>>> v.replace('o', 'O', 2).replace('O', 'o', 1).replace('O', 'i')
> 'couciu'
>
Sorry for the last posting, but it did occur to me that str.replace()
could grow another parameter 'start', so it would become:
s.replace(old, new[[, start], end]]) -> string
(In Python 2.x the method doesn't accept keyword arguments, so that
isn't a problem.)
If the possible replacements are numbered from 0, then 'start' is the
first one actually to perform and 'end' the one after the last to perform.
The 2-argument form would be s.replace(old, new) with 'start' defaulting
to 0 and 'end' to None => replacing all occurrences, same as now.
The 3-argument form would be s.replace(old, new, end) with 'start'
defaulting to 0 => equivalent to replacing the first 'end' occurrences,
same as now.
The 4-argument form would be s.replace(old, new, start, end) =>
replacing from the 'start'th to before the 'end'th occurrence,
additional behaviour as requested.
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