map-like function on dict values?

[Huaiyu Zhu]

On Wed, 27 Feb 2002 19:57:21 -0600, Skip Montanaro <skip at pobox.com> wrote:
>
>    Huaiyu> Is there a way to do this
>    Huaiyu> for k, v in dict.items():
>    Huaiyu>     dict[k] = f(v)
>
>    Huaiyu> without involving a Python for loop?
>
>Check out the operator module.  In particular, operator.setitem will
>probably do what you want:
>
>    import operator
>    keys = dict.keys()
>    vals = map(f, dict.values())
>    map(operator.setitem, [dict]*len(keys), keys, vals)

This works.  Thank you.  It speeds up by more than a factor of four againt
for loop iteration.  The disadvantage is that it builds at least five
temporary lists, so it is not usable if memory is constrained.

>
>    Huaiyu> I'd imagine that a function implemented in C could speed things
>    Huaiyu> up quite a bit.  
>
>It all depends on how expensive f() is related to the overhead of a for loop
>and the cost of getting map ramped up.  Don't rely on your instincts.
>Benchmark it.

Well, I can't benchmark it until it is available.  :-( 

What I was alluding to is that a built-in iteration would loop over all the
values held in the dict without touching the keys at all, and certainly
without building temporary lists.  I imagine this would also bypass all the
the hash calculation and conflict resolution.   The fact that your solution
speeds up so much despite building temporary lists shows that a C loop over
values could save even more.

Huaiyu

PS. The test and result:

import operator
from time import time

def iteritems(f, d):
	for k, v in d.items():
		d[k] = f(v)

def mapvalue(f, d):
	keys = d.keys()
	map(operator.setitem, [d]*len(keys), keys, map(f, d.values()))
		
def timeit(test, f, d):
	_time = time()
	test(f, d)
	print time() - _time

if __name__ == '__main__':
	d = {}
	for i in range(1000000):
		d[i] = None
	def f(x): pass

	timeit(iteritems, f, d)
	timeit(mapvalue, f, d)

# result
	
19.8180559874
4.17512202263