Steven D'Aprano wrote
> You should always quote enough of the previous poster's message to give
> context. Messages sometimes go missing in Usenet, and people won't have
> the foggiest idea what you are talking about.
one would think that given how many Pythoneers we now have working
for google,
On Thu, 09 Mar 2006 22:01:17 -0800, John wrote:
> Thanks a lot,
>
> This works but is a bit slow, I guess I'll have to live with it.
> Any chance this could be sped up in python?
I don't know. What is it?
You should always quote enough of the previous poster's message to give
context. Messages
Alex Martelli <[EMAIL PROTECTED]> wrote:
> John <[EMAIL PROTECTED]> wrote:
>
> > Thanks a lot,
> >
> > This works but is a bit slow, I guess I'll have to live with it.
> > Any chance this could be sped up in python?
>
> Sure (untested code):
>
> def count_with_overlaps(needle, haystack):
>
John <[EMAIL PROTECTED]> wrote:
> Thanks a lot,
>
> This works but is a bit slow, I guess I'll have to live with it.
> Any chance this could be sped up in python?
Sure (untested code):
def count_with_overlaps(needle, haystack):
count = 0
pos = 0
while True:
where = haystack.
John wrote:
> This works but is a bit slow, I guess I'll have to live with it.
> Any chance this could be sped up in python?
Sure, to a point. Instead of:
def countoverlap(s1, s2):
return len([1 for i in xrange(len(s1)) if s1[i:].startswith(s2)])
Try this version, which takes smaller sl
"John" <[EMAIL PROTECTED]> writes:
> This works but is a bit slow, I guess I'll have to live with it.
> Any chance this could be sped up in python?
Whoops, I meant to say:
len([1 for i in xrange(len(s1)) if s1.startswith(s2,i)])
That avoids creating a lot of small strings.
If s1 is large you
Thanks a lot,
This works but is a bit slow, I guess I'll have to live with it.
Any chance this could be sped up in python?
Thanks once again,
--j
--
http://mail.python.org/mailman/listinfo/python-list
"John" <[EMAIL PROTECTED]> writes:
> if S1 = ""
> and S2 = "AA"
>
> then the count is 3. Is there an easy way to do this in python?
> I was trying to use the "count" function but it does not do
> overlapping counts it seems.
len([1 for i in xrange(len(s1)) if s1[i:].startswith(s2)])
--
http:
I have two strings S1 and S2. I want to know how many times
S2 occurs inside S1.
For instance
if S1 = ""
and S2 = "AA"
then the count is 3. Is there an easy way to do this in python?
I was trying to use the "count" function but it does not do
overlapping counts it seems.
Thanks,
--j
--
ht
