*-unpacking

[David Mertz]

Alex Martelli <aleax at aleax.it> wrote previously:
|I'm not sure what this 'more' is about, actually.  Greg's case is
|currently best solved [IMHO] with
|    args = cmdstring.split()
|    command = args.pop(0)

Part of his description was "I might or might not have more in the
list."  My 'more' stuff was just illustrating handling that extra stuff.
Admittedly, that's not directly part of what 'car,*cdr=lst' solves.

|Actually, I don't -- both args.reverse() and args.pop(0) are
|O(len(args))

Hmmm...  I am pretty sure I have found Alex in an outright mistake.  A
shocking thing, if so :-).

|[alex at lancelot python2.3]$ python timeit.py -s'ags=range(1000)'
|'x=ags[:].pop(0)'
|10000 loops, best of 3: 24.5 usec per loop
|[alex at lancelot python2.3]$ python timeit.py -s'ags=range(1000)'
|'ags.reverse(); x=ags[:].pop()'
|10000 loops, best of 3: 23.2 usec per loop

The second operation does a '.reverse()' for every '.pop()', which is
both the wrong behavior, and not the relevant thing to time.

  >>> from time import clock
  >>> def reverse_then_pop(l):
  ...     start = clock()
  ...     l.reverse()
  ...     while 1:
  ...         try: l.pop()
  ...         except: break
  ...     print "%.2f seconds" % (clock()-start)
  ...
  >>> def pop_from_left(l):
  ...     start = clock()
  ...     while 1:
  ...         try: l.pop(0)
  ...         except: break
  ...     print "%.2f seconds" % (clock()-start)
  ...
  >>> reverse_then_pop(range(10000))
  0.03 seconds
  >>> pop_from_left(range(10000))
  0.39 seconds
  >>> reverse_then_pop(range(20000))
  0.06 seconds
  >>> pop_from_left(range(20000))
  1.47 seconds
  >>> reverse_then_pop(range(50000))
  0.14 seconds
  >>> pop_from_left(range(50000))
  11.73 seconds
  >>> reverse_then_pop(range(100000))
  0.29 seconds
  >>> pop_from_left(range(100000))
  140.77 seconds
  >>> reverse_then_pop(range(500000))
  1.41 seconds
  >>> pop_from_left(range(500000))
  ...abandoned after a few hours :-)...

Yours, David...

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