[Tutor] Searching backwards from end of string

[Robert Berman]

I think you are looking for a complex solution.

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


Robert

On Fri, 2009-07-10 at 16:24 +0100, Angus Rodgers wrote:
> I'm probably having a bit of an off-day, so please forgive me if
> this is a really silly question (even for a beginner in Python)!
> 
> I'm working through the exercises for Chapter 6 of Wesley Chun's
> book [incidentally, I tried to append another endorsement to the
> list, but my use of Python rather than "+1" may have looked like
> a typo or the work of a lunatic, or perhaps my post is just late
> in appearing - anyway, another happy reader here!], and running
> again into a difficulty which I "cheated" my way through earlier.
> 
> On the earlier occasion:
> 
>  "6-6 Strings. Create the equivalent of string.strip():  Take a
>  string and remove all leading and trailing whitespace. (Use of
>  string.*strip() defeats the purpose of this exercise.)"
> 
> I wrote:
> 
> from string import whitespace
> 
> def lstrip(s):
>     if s:
>         for i, c in enumerate(s):
>             if c not in whitespace:
>                 break
>         return s[i:]
>     else:
>         return s
> 
> def rstrip(s):
>     return lstrip(s[::-1])[::-1]
>     # Note: far from maximally efficient (two unnecessary copies!)
> 
> def strip(s):
>     return lstrip(rstrip(s))
> 
> On the present occasion:
> 
>  "6-12 Strings. ...
>  (b) Create another function called rfindchr() that will find the
>  last occurrence of a character in a string.  Naturally this works
>  similarly to findchr(), but it starts its search from the end of
>  the input string.
>  ..."
> 
> I thought I had better err in the opposite direction, so I wrote:
> 
> def findchr(strng, ch):
>     for i, c in enumerate(strng):
>         if c == ch:
>             return i
>     return -1
> 
> 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
> 
> def subchr(strng, origchar, newchar):
>     chars = list(strng)
>     for i, c in enumerate(chars):
>         if c == origchar:
>             chars[i] = newchar
>     return ''.join(chars)
> 
> I've quoted all my code, in case it's all bad, but it's rfindchr()
> that I'm particularly worried about.