On Fri, 10 Jul 2009 11:57:21 -0400, Robert Berman
<berma...@cfl.rr.com> wrote:
>I think you are looking for a complex solution.
Hardly.
>How about the following example:
>
>
>In [31]: s1='abcdeefghijkl' #find last 'e'
>
>In [32]: s2=s1[::-1] #reverses s1
>
>In [33]: j=s2.find('e') #finds first 'e' in reversed string
>
>In [36]: ind=len(s1)-j-1 #index into s1 where last occurrence of 'e' is
>
>In [37]: ind
>Out[37]: 5
>
>In [38]: s1[ind]
>Out[38]: 'e'
>
>In [39]: s1
>Out[39]: 'abcdeefghijkl' BINGO. Done.
>
>Is that not a bit simpler
I did explain (perhaps at too great a length, or with too many
irrelevancies):
>On Fri, 2009-07-10 at 16:24 +0100, Angus Rodgers wrote:
>> [...]
>> On the earlier occasion:
>> [...]
>> I wrote:
>> [...]
>> def rstrip(s):
>> return lstrip(s[::-1])[::-1]
>> # Note: far from maximally efficient (two unnecessary copies!)
>> [...]
>> On the present occasion:
>> [...]
>> I thought I had better err in the opposite direction, so I wrote:
>> [...]
>> def rfindchr(strng, ch):
>> # Surely there's a neater (but still efficient) way?
>> n = len(strng)
>> for i in range(n - 1, -1, -1):
>> if strng[i] == ch:
>> return i
>> return -1
I don't even think that's "complicated" (just ugly and clumsy).
I just wondered if there were some simplifying feature of Python
that I had either forgotten or not learned about yet. Python code
(even mine!) is usually neater than this.
I know efficiency is not always a major concern (and it certainly
isn't of any practical importance in a toy example like this),
but it seems downright profligate to make a reversed copy of a
string just in order to avoid searching it backwards, and to make
use of a nifty notation like "[::-1]" - even though my own first
instinct is to do exactly that, to save "development time", and
to make use of a previous solution.
--
Angus Rodgers
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