Here is a brute-force list comprehension that does not depend on
preserving order between dict.keys() and dict.values():
dict( [ (a[1],a[0]) for a in d.items() ] )
Or for map/lambda lovers:
rev = lambda a: (a[1],a[0])
dict( map( rev, d.items() )
But you still have no control over the order returned by d.items(), so
if your initial dict has duplicate values, there is no telling which
key a duplicate would map to.
The real general-purpose dict inversion results in a dict with each
value containing a list of keys in the original dict that mapped to the
given value. That is:
d = {3: 4, 1: 2, 0:4}
would invert to:
d2 = { 2:[1], 4:[0,3] }
I couldn't cram this into a list comp, so here is a short for loop to
create d2:
d2 = {}
for k,v in d.items():
d2[v] = d2.get(v, list()) + [k]
-- Paul
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