On Oct 1, 2:01 pm, Abandoned <[EMAIL PROTECTED]> wrote:
> I want to total score..
> For example
>
> > > dict1={1: 4, 3: 5}... and 2 millions element
> > > dict2={3: 3, 8: 6}... and 3 millions element
>
> result should be dict3={1:4, 3:8, 8:6}
Unless you have some prior knowledge about the dicts (e.g.
len(intersection(d1,d2))), it's hard to beat the following, in pure
Python at least:
def sumdicts(d1,d2):
if len(d1) < len(d2):
d1,d2 = d2,d1
dsum = dict(d1)
for key,value in d2.iteritems():
if key not in dsum:
dsum[key] = value
else:
dsum[key] += value
return dsum
Surprisingly (?), this turns out to be faster than using dict.get(key,
0) instead of the explicit if/else.
George
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