Tim Chase wrote:
> >>> indicies = [i for i in xrange(len(mystring)) if
> mystring.startswith(substr, i)]
> >>> indicies
> [4, 8, 10, 17, 22]
>
> is my preferred way of doing this.
it's things like this that makes me wonder why I spent a week speeding
up the string implementation for Python 2
K.S.Sreeram wrote:
> effbot's solution finds overlapping occurrences, whereas your solution
> finds non-overlapping occurrences. So efficiency comparisons are not valid.
oops. my bad. here's a fixed version:
result = []; pos = 0
try:
while 1:
pos = mystring.index(su
Le Mercredi 21 Juin 2006 14:01, Fredrik Lundh a écrit :
> > Another variant, I feel this one more natural as it doesn't contain a
> > C-looking infinite loop
>
> doing things in a convoluted way because you think that non-infinite while-
> loops are not natural? you can get help for that, you know
Maric Michaud:
> I'd love str implement a xsplit(sub, start, end) method, so I could have
> wrote : enumerate(s.xsplit(subs, 0, -1)).
Some of such str.x-methods (or str.i-methods, etc) can be useful
(especially for Py3.0), but keeping APIs simple and compact is very
important, otherwise when you p
> I would like to search for a substring in a string and get the index of
> all occurances.
>
> mystring = 'John has a really nice powerbook.'
> substr = ' ' # space
>
> I would like to get this list:
>[4, 8, 10, 17, 22]
>
> How can I do that without using "for i in mystring" which might b
Maric Michaud wrote:
> Actually it's even more efficient than Lundh's
effbot's solution finds overlapping occurrences, whereas your solution
finds non-overlapping occurrences. So efficiency comparisons are not valid.
e.g:
indices( 'a', 'aa' )
your solution gives: 0,2
effbots's solution: 0,1
Maric Michaud wrote:
> Another variant, I feel this one more natural as it doesn't contain a
> C-looking infinite loop
doing things in a convoluted way because you think that non-infinite while-
loops are not natural? you can get help for that, you know ;-)
> Actually it's even more efficient t
Another variant, I feel this one more natural as it doesn't contain a
C-looking infinite loop (also I made it a generator but this is not the
topic).
In [160]: def indices(s, subs) :
.: last = 0
.: for ind, part in in enumerate(s.split(subs)[:-1]) :
.: yield
Nico Grubert wrote:
> I would like to search for a substring in a string and get the index of
> all occurances.
>
> mystring = 'John has a really nice powerbook.'
> substr = ' ' # space
>
> I would like to get this list:
> [4, 8, 10, 17, 22]
the find and index methods take an optional start ar
mystring = 'John has a really nice powerbook.'
substr = ' ' # space
pos = 0
indices = []
while True:
i = mystring.find(substr,pos)
if i==-1:
break
indices.append(i)
pos = i+1
print indices
> [4, 8, 10, 17, 22]
Pierre
--
http://mail.python.org/mailman/listinfo/python-lis
10 matches
Mail list logo
