Optimize function similiar to dict.update() but adds common values

Gregory Piñero gregpinero at gmail.com
Wed Dec 14 17:47:47 CET 2005


Thanks Peter, those are some really good ideas.  I can't wait to try
them out tonight.

Here's a question about your functions.  if I only look at the keys in
freq2 then won't I miss any keys that are in freq1 and not in freq2? 
That's why I have the two loops in my original function.

-Greg


On 12/14/05, Peter Otten <__peter__ at web.de> wrote:
> Gregory Piñero wrote:
>
> > def add_freqs(freq1,freq2):
> > """Addtwowordfreqdicts"""
> > newfreq={}
> > forkey,valueinfreq1.items():
> > newfreq[key]=value+freq2.get(key,0)
> > forkey,valueinfreq2.items():
> > newfreq[key]=value+freq1.get(key,0)
> > returnnewfreq
>
> > Anyideasondoingthistaskalot faster would be appriciated.
>
> With items() you copy the whole dictionary into a list of tuples;
> iteritems() just walks over the existing dictionary and creates one tuple
> at a time.
>
> With "80% overlap", you are looking up and setting four out of five values
> twice in your for-loops.
>
> Dump the symmetry and try one of these:
>
> def add_freqs2(freq1, freq2):
>     total = dict(freq1)
>     for key, value in freq2.iteritems():
>         if key in freq1:
>             total[key] += value
>         else:
>             total[key] = value
>     return total
>
> def add_freqs3(freq1, freq2):
>     total = dict(freq1)
>     for key, value in freq2.iteritems():
>         try:
>             total[key] += value
>         except KeyError:
>             total[key] = value
>     return total
>
> My guess is that add_freqs3() will perform best.
>
> Peter
> --
> http://mail.python.org/mailman/listinfo/python-list
>


--
Gregory Piñero
Chief Innovation Officer
Blended Technologies
(www.blendedtechnologies.com)



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