I wrote:
> Another generator solution, based on computing a permutation from its
> rank according to some natural order. Written for strings.
>
> def perms(s) :
> def nth(n,L,k=1) :
> if k>len(L) :
> if n :
> raise StopIteration
> return ''
>
Another generator solution, based on computing a permutation from its rank
according to some natural order. Written for strings.
def perms(s) :
def nth(n,L,k=1) :
if k>len(L) :
if n :
raise StopIteration
return ''
return nth(n/k,L,
Girish Sahani wrote:
>>>thanks lawrence...however this func doesnt return the permuted strings,
>>>so
>>>i added some code as follows to generate a list of all the permutations:
>>>
>>>def permute(seq):
>>>l = []
>>>if len(seq) == 0:
>>>yield []
>>>else:
>>>for i in rang
>> thanks lawrence...however this func doesnt return the permuted strings,
>> so
>> i added some code as follows to generate a list of all the permutations:
>>
>> def permute(seq):
>> l = []
>> if len(seq) == 0:
>> yield []
>> else:
>> for i in range(0,len(seq)):
>>
Girish Sahani wrote:
>>In article <[EMAIL PROTECTED]>,
>> "Girish Sahani" <[EMAIL PROTECTED]> wrote:
>>
>>
>>> I want to generate all permutations of a string.
>>
>>def permute(Seq) :
>>"""generator which yields successive permutations of the elements
>>of Seq."""
>>if len(Seq) == 0 :
>
Girish Sahani wrote:
> Hi guys,
> I want to generate all permutations of a string. I've managed to
> generate all cyclic permutations. Please help :)
>
http://gflanagan.net/site/python/05/Johnson.html
Gerard
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Girish Sahani wrote:
> I want to generate all permutations of a string. I've managed to
> generate all cyclic permutations. Please help :)
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/496724
anton
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In article <[EMAIL PROTECTED]>,
Girish Sahani <[EMAIL PROTECTED]> wrote:
>Hi guys,
> I want to generate all permutations of a string. I've managed to
>generate all cyclic permutations. Please help :)
>
>def permute(string):
>l= []
>l.append(string)
>string1 = ''
>for i in range(0,l
Le Jeudi 22 Juin 2006 10:07, Girish Sahani a écrit :
> Hi guys,
> I want to generate all permutations of a string. I've managed to
> generate all cyclic permutations. Please help :)
>
Here is mine, maybe more versatile :
def permute(iterable) :
if len(iterable) == 1 : yield iterable
e
> In article <[EMAIL PROTECTED]>,
> "Girish Sahani" <[EMAIL PROTECTED]> wrote:
>
>> I want to generate all permutations of a string.
>
> def permute(Seq) :
> """generator which yields successive permutations of the elements
> of Seq."""
> if len(Seq) == 0 :
> yield ()
> el
Mind, that Lawrence's solution may contain doubles:
>>> [ i for i in permute("aa") ]
[('a', 'a'), ('a', 'a')]
If you don't want this, use sets.
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In article <[EMAIL PROTECTED]>,
"Girish Sahani" <[EMAIL PROTECTED]> wrote:
> I want to generate all permutations of a string.
def permute(Seq) :
"""generator which yields successive permutations of the elements
of Seq."""
if len(Seq) == 0 :
yield ()
else :
for i
Hi guys,
I want to generate all permutations of a string. I've managed to
generate all cyclic permutations. Please help :)
def permute(string):
l= []
l.append(string)
string1 = ''
for i in range(0,len(string)-1,1):
string1 = string[1:len(string)] + string[:1]
l.ap
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