On 18 Mar 2005 19:41:55 -0800, [EMAIL PROTECTED] wrote:
>Once again my specs were incomplete.
>By largest I mean exactly what you pointed out as in sum(map(len,
>setlist)).
>
But that will not necessarily yield a single setlist taken from the source set
list,
so you still need a selection amongst equals. You can make it explicit, or
you can say you'll take whatever a sort puts at the sorted list extreme, which
you
seem to have done ;-)
>I think this might work--sorting of the initial list should do the
>trick.
>1) sort the sets by size (in decending order)
>2) put the first (largest) into a new list (Lu)
>for s in Lnew[1:]:
> keep=True
> for i in range(len( Lun) ):
> if len(s)==len( s & Lun[i] ):
> keep=False
> break
> if keep==True:
> Lun.append( s )
>
>----------------- here is the complete code
>s1=set(['a','b','c'])
>s2=set(['a','c'])
>s3=set(['a','d','e','f'])
>s4=set(['r','k','l'])
>s5=set(['r','k','l'])
>s6=set(['g', 'h'])
>s7=set(['h', 'i'])
>s8=set(['g', 'h', 'i'])
>
>L=[s1,s2,s3,s4,s5,s6,s7,s8]
>length=[len(s) for s in L]
>L2= sorted(zip(length,range(len(L))))
>Lnew=[L[j] for (i,j) in L2]
>Lnew.reverse()
>Lun=[Lnew[0]] # list with the result
>for s in Lnew[1:]:
> keep=True
> for i in range(len( Lun) ):
> if len(s)==len( s & Lun[i] ):
> keep=False
> break
> if keep==True:
> Lun.append( s )
>
>#----------------
>>>> Lun
>[set(['a', 'e', 'd', 'f']), set(['i', 'h', 'g']), set(['k', 'r', 'l']),
>set(['a', 'c', 'b'])]
>
>Seems like I got it.
>
Does it work on
L = [set('abc'), set('def'), set('abcdef')]
?
I get:
0: []
6: [set(['a', 'c', 'b']), set(['e', 'd', 'f'])]
6: [set(['a', 'c', 'b', 'e', 'd', 'f'])]
Your code above:
>>> L = [set('abc'), set('def'), set('abcdef')]
>>> length=[len(s) for s in L]
>>> L2= sorted(zip(length,range(len(L))))
>>> Lnew=[L[j] for (i,j) in L2]
>>> Lnew.reverse()
>>> Lun=[Lnew[0]] # list with the result
>>> for s in Lnew[1:]:
... keep=True
... for i in range(len( Lun) ):
... if len(s)==len( s & Lun[i] ):
... keep=False
... break
... if keep==True:
... Lun.append( s )
...
>>> Lun
[set(['a', 'c', 'b', 'e', 'd', 'f'])]
But your successful set list is not unique in its success value (6) ;-)
Regards,
Bengt Richter
--
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