list.sort(func) speed

[mackstann]

On Sat, Aug 30, 2003 at 08:27:19PM -0400, Terry Reedy wrote:
> Google list for decorate sort undecorate pattern

Awesome, thanks!

I tried it three ways (including original), and timed each one a few
times.

-------------------

# 1. original way

origlist.sort(self._basenameCmp)

# 2. decorate sort undecorate w/ list comp.

newlist = [ (os.path.basename(i), i) for i in origlist ]
newlist.sort()
origlist = [ i[1] for i in newlist ]

# 3. decorate sort decorate w/ zip and map

newlist = zip(map(os.path.basename, origlist), xrange(len(origlist)))
newlist.sort()
origlist = [ origlist[index] for basename,index in newlist ]

-------------------

And here are the times:

1. Around 300ms
2. 75-80ms
3. 70-75ms

And after further tweaking, I arrived at what seems to be the simplest
and fastest way to go about it (variation on #3):

newlist = [ pair[::-1] for pair in
          enumerate(map(os.path.basename, oldlist)) ]
newlist.sort()
origlist = [ origlist[index] for basename,index in newlist ]


Thanks,
-- 
m a c k s t a n n  mack @ incise.org  http://incise.org
What this world needs is a good five-dollar plasma weapon.