On 12/05/2006 5:13 AM, vbgunz wrote:
> I thought this to be a great exercise so I went the extra length to
> turn it into a function for my little but growing library. I hope you
> enjoy :)
>
Oh, indeed ;-)
>
> def indexer(string, target):
Consider not shadowing the name of the string module.
> '''indexer(string, target) -> [list of target indexes]
>
> enter in a string and a target and indexer will either return a
> list of all targeted indexes if at least one target is found or
> indexer will return None if the target is not found in sequence.
Consider returning [] if the target is not found in the sequence. That
would enable callers to "do nothing gracefully":
for posn in indexer(...
# do something
and it doesn't require a gross rewrite ... merely dedent the last line.
>
> >>> indexer('a long long day is long', 'long')
> [2, 7, 19]
>
> >>> indexer('a long long day is long', 'day')
> [12]
>
> >>> indexer('a long long day is long', 'short')
> None
> '''
>
> res = []
>
Consider evaluating string.count(target) *ONCE*.
> if string.count(target) >= 1:
> res.append(string.find(target))
>
> if string.count(target) >= 2:
> for item in xrange(string.count(target) - 1):
> res.append(string.find(target, res[-1] + 1))
>
> return res
>
>
> if __name__ == '__main__':
> print indexer('a long long day is long', 'long') # -> [2, 7, 19]
> print indexer('a long long day is long', 'day') # -> [12]
> print indexer('a long long day is long', 'short') # -> None
>
When called with the args('abababababababa', 'aba'), it returns [0, 2,
4, 6]. If you *intend* to allow for overlaps, it should return [0, 2, 4,
6, 8, 10, 12]; otherwise it should return [0, 4, 8, 12].
Consider doing something straightforward and understandable, like the
following (tested):
def findallstr(text, target, overlapping=0):
result = []
startpos = 0
if overlapping:
jump = 1
else:
jump = max(1, len(target))
while 1:
newpos = text.find(target, startpos)
if newpos == -1:
return result
result.append(newpos)
startpos = newpos + jump
HTH,
John
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