2nd iteration of a character

[Philippe Rousselot]

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 ?

[Jeff Epler]

You probably want to write this in terms of find with a start argument
specified. This minimally-tested function seems to do the job.

def findn(haystack, needle, n, start=0):
while 1:
idx = haystack.find(needle, start)
if idx == -1 or n==1: break
n -= 1
start = idx+1
return idx

findn("axbxcx", "x", 2) 3 findn("axbxcx", "z", 2)

-1

Jeff

[Bengt Richter]

On Sat, 20 Sep 2003 23:27:23 +0200, Philippe Rousselot <ma*****@rousselot.rg> wrote:

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 ?

Online tutorials first ;-)

s='01x34x678x012345x789x'
def find_nth(s, ss, n): ... i = -len(ss)
... for j in xrange(n):
... i += len(ss)
... i = s.find(ss,i)
... if i<0: return i
... return i<0 and -1 or i
... find_nth(s,'x', 1) 2 find_nth(s,'x', 2) 5 find_nth(s,'x', 3) 9 [find_nth(s,'x', i) for i in xrange(5)] [-1, 2, 5, 9, 16] [find_nth(s,'x7', i) for i in xrange(5)] [-1, 16, -1, -1, -1] [find_nth(s,'01', i) for i in xrange(5)] [-1, 0, 10, -1, -1] [find_nth(s,'34', i) for i in xrange(5)] [-1, 3, 13, -1, -1] [find_nth(s,'xx', i) for i in xrange(5)]

[-1, -1, -1, -1, -1]

BTW, the above is not the fastest possible code, just a quick first prototype,
and not tested beyond what you see above ;-)
Regards,
Bengt Richter

[Emile van Sebille]

> 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 ?

You've already got some good answers, so how about (-:

def nth(s,ss,n): return n<len(s.split(ss)) and len(ss.join(s.split(ss)[:n]))
.... s = 'this is a test this is a test'
nth(s,'t',7) False nth(s,'t',5) 25 nth(s,'i',2) 5 nth(s,'x',2) False

Emile van Sebille
em***@fenx.com

[Anthony McDonald]

"Philippe Rousselot" <ma*****@rousselot.rg> wrote in message
news:bk***********@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
for a in Ifind("123456789012345678901234567890", "1234567890"): .... print a
....
0
10
20
for a in Ifind("123456789012345678901234567890", "nope"): .... print a
....

Anthony McDonald

[Bengt Richter]

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

"Philippe Rousselot" <ma*****@rousselot.rg> wrote in message
news:bk***********@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