On Sat, Feb 14, 2009 at 3:01 PM, Peter Otten <__pete...@web.de> wrote:
> Gabriel Genellina wrote:
>
> > En Fri, 13 Feb 2009 08:16:00 -0200, S.Selvam Siva <
> s.selvams...@gmail.com>
> > escribió:
> >
> >> I need some help.
> >> I tried to find top n(eg. 5) similar words for a given word, from a
>
Gabriel Genellina wrote:
> En Fri, 13 Feb 2009 08:16:00 -0200, S.Selvam Siva
> escribió:
>
>> I need some help.
>> I tried to find top n(eg. 5) similar words for a given word, from a
>> dictionary of 50,000 words.
>> I used python-levenshtein module,and sample code is as follow.
>>
>> def foo(se
On Feb 13, 5:42 am, "Gabriel Genellina"
wrote:
> You may replace the last steps (sort + slice top 5) by heapq.nlargest - at
> least you won't waste time sorting 49995 irrelevant words...
> Anyway you should measure the time taken by the first part (Levenshtein),
> it may be the most demanding.
Gabriel Genellina wrote:
En Fri, 13 Feb 2009 08:16:00 -0200, S.Selvam Siva
escribió:
I need some help.
I tried to find top n(eg. 5) similar words for a given word, from a
dictionary of 50,000 words.
I used python-levenshtein module,and sample code is as follow.
def foo(searchword):
disdi
En Fri, 13 Feb 2009 08:16:00 -0200, S.Selvam Siva
escribió:
I need some help.
I tried to find top n(eg. 5) similar words for a given word, from a
dictionary of 50,000 words.
I used python-levenshtein module,and sample code is as follow.
def foo(searchword):
disdict={}
for word in sel
Hi all,
I need some help.
I tried to find top n(eg. 5) similar words for a given word, from a
dictionary of 50,000 words.
I used python-levenshtein module,and sample code is as follow.
def foo(searchword):
disdict={}
for word in self.dictionary-words:
distance=Levenshte
