Tough sorting problem: or, I'm confusing myself

david jensen dmj.ccc at gmail.com
Fri Apr 9 17:03:01 CEST 2010


Hi all,

I'm trying to find a good way of doing the following:

Each n-tuple in combinations( range( 2 ** m ), n ) has a corresponding
value n-tuple (call them "scores" for clarity later). I'm currently
storing them in a dictionary, by doing:

----
res={}
for i in itertools.combinations( range( 2**m ) , n):
    res[ i ] = getValues( i )    # getValues() is computationally
expensive
----

For each (n-1)-tuple, I need to find the two numbers that have the
highest scores versus them. I know this isn't crystal clear, but
hopefully an example will help: with m=n=3:

Looking at only the (1, 3) case, assuming:
getValues( (1, 2, 3) ) == ( -200, 125, 75 )    # this contains the
highest "other" score, where 2 scores 125
getValues( (1, 3, 4) ) == ( 50, -50, 0 )
getValues( (1, 3, 5) ) == ( 25, 300, -325 )
getValues( (1, 3, 6) ) == ( -100, 0, 100 )    # this contains the
second-highest, where 6 scores 100
getValues( (1, 3, 7) ) == ( 80, -90, 10  )
getValues( (1, 3, 8) ) == ( 10, -5, -5 )

I'd like to return ( (2, 125), (6, 100) ).

The most obvious (to me) way to do this would be not to generate the
res dictionary at the beginning, but just to go through each
combinations( range( 2**m), n-1) and try every possibility... this
will test each combination n times, however, and generating those
values is expensive. [e.g. (1,2,3)'s scores will be generated when
finding the best possibilities for (1,2), (1,3) and (2,3)]

What I'm doing now is ugly, and i think is where i'm confusing myself:

----
best2={}
for i in itertools.combinations( range( 2**m), n-1):
    scorelist=[]
    for j in range( 2**m ):
        if j not in i:
            k=list(i)
            k.append(j)
            k=tuple(sorted(k))    #gets the key for looking up the
scores in res
            scorelist.append((j,res[k][k.index(j)]))
    best2[i]=sorted(scorelist,key=lambda x: -x[1])[:2]
----

Am I missing an obviously better way?

Many thanks!

David





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