Equivalent of Perl chomp?

[Tim Peters]

[Tim]
> Note that
>
>     s.endswith('\n')
>
> is clearer than
>
>     s[-1:] == '\n'
>
> For real fun, time them <wink>.

[Mark McEahern]
> I timed them with the following code and it occurred to me here's
> a possibly good case where aspect-oriented programming would be nice.
> Anyway, the results first:
>
> 	<function endsWith at 0x100f94d0> : 0.36 seconds.
> 	<function sliceIt at 0x100f9510> : 0.07 seconds.

Well, you're in for some fun of a kind I didn't have in mind <wink>.  Do a
sanity check here:  what's the clock rate on your box?  How many cycles go
by in 0.07 seconds (let alone 0.36!)?  Can you think of *any* non-insane way
to implement these operations that would take so bloody long?

I can't make time to explain timing subtleties today, but here's a hint:
try swapping the order of your calls to endsWith and sliceIt, and see what
you get then.  If you hear your disk grinding <wink>, you might want to
contemplate the sensibility of using a 100MB string.

> The test:
>
>     #! /usr/bin/env python
>
>     import time
>
>     def getString(n):
>         s = "s" * n
>         s = s + '\n'
>         return s
>
>     def endsWith(s):
>         s.endswith('\n')
>
>     def sliceIt(s):
>         s[-1:] == '\n'
>
>     def timeIt(func):
>         n = 100000000
>         s = getString(n)
>         timeIn = time.time()
>         func(s)
>         timeOut = time.time()
>         print "%s : %1.2f seconds." % (func, (timeOut - timeIn))
>
>     timeIt(endsWith)
>     timeIt(sliceIt)