[Tim Peters]
>> def lookup1(arg, _marker=object()):
>>     return _marker
>> ...
>> _marker = object()
>> def lookup3(arg):
>>     return _marker
>> ...
>>lookup1  0.427597
>>lookup3  0.404399

[Dieter Maurer] 
> Do you understand why "lookup3" is faster than "lookup1"?
> I had the "impression" that access to the function's local namespace
> should be faster than any other variable access.

It is, although module-global lookup is often much faster now than
Python old-timers "know" it is <wink>:  the critical successful path
thru the dict lookup code for a dict keyed by interned strings is, in
the absence of collisions, _almost_ as lean as an indexed array access
now.  The primary difference in speed remaining is that the dict
lookup endures an extra C-level function call.

I explained a relevant difference last time:  lookup1 has the
additional expense of initializing an additional local variable. 
That's what a default argument becomes, and nothing happens for free. 
Let's time this effect "in isolation":

from itertools import repeat
from time import clock as now

def with(default=5):

def without():

for dummy in range(3):
    for f in with, without:
        start = now()
        for dummy in repeat(None, 1000000):
        finish = now()
        print "%-8s %.6g" % (f.__name__, finish - start)

and typical output:

with     0.374435
without  0.338555
with     0.370469
without  0.339791
with     0.372165
without  0.337355

So adding a default argument indeed has a measurable cost (although
~0.03 seconds across a million calls is nothing to fret about!).  Note
that the difference between lookup1 and lookup3 was less than that on
this box, reflecting that the lookup _inside_ lookup1 goes faster than
the lookup inside lookup3, partly cancelling lookup1's higher call
overhead; the difference in call overhead is greater than the
difference in lookup speed, so lookup1 ends up a net loser.
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