2nd iteration of a character

[Bengt Richter]

On Tue, 23 Sep 2003 12:38:36 +0200, "Anthony McDonald" <tonym1972/at/club-internet/in/fr> wrote:

>"Philippe Rousselot" <mailing at rousselot.rg> wrote in message
>news:bkigte$1e3b$1 at biggoron.nerim.net...
>> Hi,
>>
>> How do I find the second iteration of a character in a string using python
>>
>> this gives me the first one
>> f1 = string.find(line,sub)
>>
>> this the last one
>> f2 = string.rfind(line,sub)
>>
>> but for the nth one ?
>>
>> Thanks in advance
>>
>> Philippe
>>
>> PS : any good book to recommend ?
>
>Most of the answers you've been given take the form (haystack, needle,
>iteration), so I thought it'd be nice to write a little iterator.
>
>class Ifind:
>    def __init__(self, haystack, needle):
>        self.haystack = haystack
>        self.needle = needle
>        self.pos = 0
>    def __iter__(self):
>        return self
>    def next(self):
>        found = self.haystack[self.pos:].find(self.needle)
>        if found == -1:
>            raise StopIteration
>        found = self.pos + found
>        self.pos = found + 1
>        return found
>
>>>> for a in Ifind("123456789012345678901234567890", "3"):
>...  print a
>...
>2
>12
>22

Carrying on with that theme ...

You could use the re module, e.g., (observing that a constant string with no magic characters
is a simple regular expression, though you could of course make up more complex things with re).

 >>> import re
 >>> for mo in re.finditer("3", "123456789012345678901234567890"): print mo.start()
 ...
 2
 12
 22

>>>>
>>>> for a in Ifind("123456789012345678901234567890", "1234567890"):
>...  print a
>...
>0
>10
>20

 >>> for mo in re.finditer("1234567890", "123456789012345678901234567890"): print mo.start()
 ...
 0
 10
 20

>>>>
>>>> for a in Ifind("123456789012345678901234567890", "nope"):
>...  print a
>...

 >>> for mo in re.finditer("nope", "123456789012345678901234567890"): print mo.start()
 ...
 >>>

Or if you do want to roll your own find-based iterable, the generator version seems easier to me:

 >>> def ifind(haystack, needle):
 ...     pos=0; skip = len(needle)
 ...     while 1:
 ...         pos = haystack.find(needle, pos)
 ...         if pos<0: break
 ...         yield pos
 ...         pos += skip
 ...
 >>> for i in ifind("123456789012345678901234567890", "3"): print i
 ...
 2
 12
 22
 >>> for i in ifind("123456789012345678901234567890", "1234567890"): print i
 ...
 0
 10
 20
 >>> for i in ifind("123456789012345678901234567890", "nope"): print i
 ...
 >>>   

Regards,
Bengt Richter