Robert Berman wrote:
Hi,
The below script which prints anagrams available for any word available
within a given database. It does work, but it is not very fast. I am
relatively certain there are more Python friendly coding techniques but
I am more concerned with a faster algorithm.
You might consider looking up the permutations only (up to some length
-- there are a lot of permutations for 8+ letter words :)
#from http://www.daniweb.com/code/snippet459.html
def permutate(seq):
"""permutate a sequence and return a list of the permutations"""
if not seq:
return [seq] # is an empty sequence
else:
temp = []
for k in range(len(seq)):
part = seq[:k] + seq[k+1:]
#print k, part # test
for m in permutate(part):
temp.append(seq[k:k+1] + m)
#print m, seq[k:k+1], temp # test
return temp
print permutate('myth')
['myth', 'myht', 'mtyh', 'mthy', 'mhyt', 'mhty', 'ymth', 'ymht', 'ytmh',
'ythm', 'yhmt', 'yhtm', 'tmyh', 'tmhy', 'tymh', 'tyhm', 'thmy', 'thym',
'hmyt', 'hmty', 'hymt', 'hytm', 'htmy', 'htym']
>>> print len(permutate('mythsdfw'))
40320
>>> print len(permutate('mythsdf'))
5040
>>> print len(permutate('mythsd'))
720
>>> print len(permutate('myths'))
120
>>> print len(permutate('myth'))
24
>>> print len(permutate('myt'))
6
so something like:
print [ xx for xx in permutate('myths') if xx in worddict ]
HTH,
Emile
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