Re: interesting exercise
Michael Tobis wrote:
Here is the bloated mess I came up with. I did see that it had to be
recursive, and was proud of myself for getting it pretty much on the
first try, but the thing still reeks of my sorry old fortran-addled
mentality.
Recursion is not necessary, but is much, much clearer.
Here is one non-recursive version from another aging
fortran programmer. I agree it is less clear than most
of the recursive alternatives. No checks for sorted
input etc, these are left as an exercise for the reader.
def permute( s, n ):
def _perm( m, n ):
ilist = [0]*n
while True:
yield ilist
i = n-1
while i = 0 and ilist[i]=m-1: i = i - 1
if i = 0:
ilist = ilist[0:i] + [ilist[i]+1] + [0]*(n-i-1)
else:
return
return [ ''.join([s[i] for i in ilist])
for ilist in _perm(len(s),n) ]
print permute('abc',2) = , permute('abc',2)
print len(permute('13579',3)) = , len(permute('13579',3))
permute('abc',2) = ['aa', 'ab', 'ac', 'ba', 'bb', 'bc',
'ca', 'cb', 'cc']
len(permute('13579',3)) = 125
or even this monstrosity ...
def permute2( s, n ):
return [ ''.join([ s[int(i/len(s)**j)%len(s)]
for j in range(n-1,-1,-1)])
for i in range(len(s)**n) ]
print permute2('abc',2) =, permute2('abc',2)
print len(permute2('13579',3)) =, len(permute2('13579',3))
permute2('abc',2) = ['aa', 'ab', 'ac', 'ba', 'bb', 'bc',
'ca', 'cb', 'cc']
len(permute2('13579',3)) = 125
Charles
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Re: interesting exercise
On May 9, 1:13 am, Charles Sanders [EMAIL PROTECTED]
wrote:
Michael Tobis wrote:
Here is the bloated mess I came up with. I did see that it had to be
recursive, and was proud of myself for getting it pretty much on the
first try, but the thing still reeks of my sorry old fortran-addled
mentality.
Recursion is not necessary, but is much, much clearer.
Here is one non-recursive version from another aging
fortran programmer. I agree it is less clear than most
of the recursive alternatives. No checks for sorted
input etc, these are left as an exercise for the reader.
def permute( s, n ):
def _perm( m, n ):
ilist = [0]*n
while True:
yield ilist
i = n-1
while i = 0 and ilist[i]=m-1: i = i - 1
if i = 0:
ilist = ilist[0:i] + [ilist[i]+1] + [0]*(n-i-1)
else:
return
return [ ''.join([s[i] for i in ilist])
for ilist in _perm(len(s),n) ]
print permute('abc',2) = , permute('abc',2)
print len(permute('13579',3)) = , len(permute('13579',3))
permute('abc',2) = ['aa', 'ab', 'ac', 'ba', 'bb', 'bc',
'ca', 'cb', 'cc']
len(permute('13579',3)) = 125
or even this monstrosity ...
def permute2( s, n ):
return [ ''.join([ s[int(i/len(s)**j)%len(s)]
for j in range(n-1,-1,-1)])
for i in range(len(s)**n) ]
print permute2('abc',2) =, permute2('abc',2)
print len(permute2('13579',3)) =, len(permute2('13579',3))
permute2('abc',2) = ['aa', 'ab', 'ac', 'ba', 'bb', 'bc',
'ca', 'cb', 'cc']
len(permute2('13579',3)) = 125
Charles
Could you explain, this one, actually? Don't forget StopIteration.
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Re: interesting exercise
On May 9, 2:41 pm, [EMAIL PROTECTED] wrote:
On May 9, 1:13 am, Charles Sanders [EMAIL PROTECTED]
wrote:
or even this monstrosity ...
def permute2( s, n ):
return [ ''.join([ s[int(i/len(s)**j)%len(s)]
for j in range(n-1,-1,-1)])
for i in range(len(s)**n) ]
print permute2('abc',2) =, permute2('abc',2)
print len(permute2('13579',3)) =, len(permute2('13579',3))
permute2('abc',2) = ['aa', 'ab', 'ac', 'ba', 'bb', 'bc',
'ca', 'cb', 'cc']
len(permute2('13579',3)) = 125
Charles
Could you explain, this one, actually? Don't forget StopIteration.
Heh, it's cute.
Not sure what the issue is with StopIteration. He is modeling the
problem as casting an integer to base N. It's terse, but it does way
too much arithmetic and isn't very readable.
The following uses the same idea more readably and (I expect,
untested) more efficiently, if less tersely. It actually is not too
bad for real code, though I much prefer your recursive list
comprehension and will use that.
def getdigits(number,base,length,alphabet):
residual = number
result =
for column in range(length):
residual,digit = divmod(residual,base)
result = alphabet[digit] + result
return result
def permu(alphabet,wordlength):
total = len(alphabet)**wordlength
return [getdigits(index,len(alphabet),wordlength,alphabet)
for index in range(total)]
if __name__ == __main__:
print permu(abc,2)
print len(permu(13579,3))
mt
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Re: interesting exercise
[EMAIL PROTECTED] wrote:
On May 9, 1:13 am, Charles Sanders [EMAIL PROTECTED]
wrote:
[snip]
or even this monstrosity ...
def permute2( s, n ):
return [ ''.join([ s[int(i/len(s)**j)%len(s)]
for j in range(n-1,-1,-1)])
for i in range(len(s)**n) ]
print permute2('abc',2) =, permute2('abc',2)
print len(permute2('13579',3)) =, len(permute2('13579',3))
permute2('abc',2) = ['aa', 'ab', 'ac', 'ba', 'bb', 'bc',
'ca', 'cb', 'cc']
len(permute2('13579',3)) = 125
Charles
Could you explain, this one, actually? Don't forget StopIteration.
As Michael said, simple counting in base n with the
string as the digits. No attempt at clarity or efficiency (I
did say it was a monstrosity).
Also, as a python beginner I didn't know about divmod,
and have no idea what you (castironpi) mean by Don't forget
StopIteration.
Charles
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Re: interesting exercise
On May 9, 7:49 pm, Charles Sanders [EMAIL PROTECTED]
wrote:
[EMAIL PROTECTED] wrote:
On May 9, 1:13 am, Charles Sanders [EMAIL PROTECTED]
wrote:
[snip]
or even this monstrosity ...
def permute2( s, n ):
return [ ''.join([ s[int(i/len(s)**j)%len(s)]
for j in range(n-1,-1,-1)])
for i in range(len(s)**n) ]
print permute2('abc',2) =, permute2('abc',2)
print len(permute2('13579',3)) =, len(permute2('13579',3))
permute2('abc',2) = ['aa', 'ab', 'ac', 'ba', 'bb', 'bc',
'ca', 'cb', 'cc']
len(permute2('13579',3)) = 125
Charles
Could you explain, this one, actually? Don't forget StopIteration.
As Michael said, simple counting in base n with the
string as the digits. No attempt at clarity or efficiency (I
did say it was a monstrosity).
Also, as a python beginner I didn't know about divmod,
and have no idea what you (castironpi) mean by Don't forget
StopIteration.
Charles
Please disregard. I have just learned that:
If the generator exits without yielding another value, a
StopIteration exception is raised.
exception StopIteration
Raised by an iterator's next() method to signal that there are no
further values.
Means normal generator termination.
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Re: interesting exercise
On May 8, 12:24 am, [EMAIL PROTECTED] (Alex Martelli) wrote: [EMAIL PROTECTED] wrote: ... def p(a,b): if list( a ) != sorted( list( a ) ): raise ValueError, String not ordered. if not b: return [''] return [i+j for i in list(a) for j in p(a,b-1)] No need for 2/3 of the list(...) calls. sorted(a) and sorted(list(a)) will ALWAYS be the same sequence; for i in a and for i in list(a) will always iterate on the same sequence [as long as you're not changing a inside the iteration, which, in this case, you aren't]. Alex Ah quite true. list( a ) != sorted( a ). ...for i in a for... -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
Michael Tobis wrote:
I want a list of all ordered permutations of a given length of a set
of tokens. Each token is a single character, and for convenience, they
are passed as a string in ascending ASCII order.
For example
permute(abc,2)
should return [aa,ab,ac,ba,bb,bc,ca,cb,cc]
If N is constant and small (e.g hard-coded N of 2), list comprehensions
provide a concise solution:
a = abc
b = [(x,y) for x in a for y in a]
c = [.join(z) for z in b]
print b
print c
Gives:
[('a', 'a'), ('a', 'b'), ('a', 'c'), ('b', 'a'), ('b', ' a'), ('c',
'b'), ('c', 'c')]
['aa', 'ab', 'ac', 'ba', 'bb', 'bc', 'ca', 'cb', 'cc']
Cheers,
Chris
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Re: interesting exercise
sherry wrote: On May 8, 9:31 am, Steven D'Aprano [EMAIL PROTECTED] wrote: On Mon, 07 May 2007 20:45:52 -0700, Michael Tobis wrote: I have a reasonably elegant solution but it's a bit verbose (a couple dozen lines which I'll post later if there is interest). Is there some clever Pythonism I didn't spot? Peering into my crystal ball, I see that your algorithm does the wrong thing, and contains bugs too. You certainly don't need to partion the sequence into three sub-sequences, or do that trick with the metaclass, and it is more efficient to use list.extend() than sum. Hang on... stupid crystal ball... that's somebody else's code. Maybe you should just post your code here, so we can look at it? In the meantime, here's a simple generator to do permutations with repetition: def permute_with_repetitions(seq): if len(seq) = 1: yield list(seq) else: for i, item in enumerate(seq): for tail in permute_with_repetitions(seq[:i] + seq[i+1:]): yield [item] + tail It doesn't do a test for the sequence containing duplicates, and I leave it as anexerciseto do selections of fewer items. -- Steven. Dear Steven I ran into your message quite accidentally while researching about some details on 'Exercise' and thought of sharing some of my findings. I've read at 'http://www.medical-health-care-information.com/Health- living/exercise/index.asp that Swimming, cycling, jogging, skiing, aerobic dancing, walking or any of dozens of other activities can help your heart. Whether it's included in a structured exercise program or just part of your daily routine, all physical activity adds up to a healthier heart. I hope the above is of some help to you as well. Regards, Sherrybove. This takes annoying past annoying to some new level of hell to which even satan himself wouldn't venture. -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
On Mon, 2007-05-07 at 22:20 -0700, sherry wrote: On May 8, 9:31 am, Steven D'Aprano [EMAIL PROTECTED] wrote: [snip Steven's response about a programming exercise...] [snip Sherrybove's spam about physical exercise...] And today's award for the most conspicuous failure of the Turing test goes to the Spambot posting as Sherrybove! (*applause*) Fortunately, it's not programmed to deliver acceptance speeches, so it will accept the award silently. I-know-this-isn't-on-topic-either-but-I-couldn't-resist'ly yours, Carsten. -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
On Tue, 08 May 2007 10:22:05 +, James Stroud wrote: This takes annoying past annoying to some new level of hell to which even satan himself wouldn't venture. And thank you for sharing that piece of spam with us again. It was so much less enjoyable to see it the second time. Seriously James, with more and more people using automated spam filters, it might not be such a wise idea to keep having your name associated with spam content. -- Steven. -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
Steven D'Aprano wrote: On Tue, 08 May 2007 10:22:05 +, James Stroud wrote: This takes annoying past annoying to some new level of hell to which even satan himself wouldn't venture. And thank you for sharing that piece of spam with us again. It was so much less enjoyable to see it the second time. Seriously James, with more and more people using automated spam filters, it might not be such a wise idea to keep having your name associated with spam content. Thank you for the tip. James -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
On May 8, 3:55 pm, James Stroud [EMAIL PROTECTED] wrote:
Steven D'Aprano wrote:
On Tue, 08 May 2007 10:22:05 +, James Stroud wrote:
This takes annoying past annoying to some new level of hell to which
even satan himself wouldn't venture.
And thank you for sharing that piece of spam with us again. It was so much
less enjoyable to see it the second time.
Seriously James, with more and more people using automated spam filters,
it might not be such a wise idea to keep having your name associated with
spam content.
Thank you for the tip.
James
We also have:
p=lambda a,n: [ ''.join( y ) for y in eval('[%%s %s]'%' '.join(['for x
%i in a'%i for i in range(n)]) %'(%s)'%','.join(['x%i'%i for i in
range(n)]) ) ]
p('abc',2)
#fb: ['aa', 'ab', 'ac', 'ba', 'bb', 'bc', 'ca', 'cb', 'cc']
len(p('13579',3))
#fb: 125
edit()
File under obscurities. acb
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Re: interesting exercise
On May 8, 4:55 pm, [EMAIL PROTECTED] wrote:
On May 8, 3:55 pm, James Stroud [EMAIL PROTECTED] wrote:
Steven D'Aprano wrote:
On Tue, 08 May 2007 10:22:05 +, James Stroud wrote:
This takes annoying past annoying to some new level of hell to which
even satan himself wouldn't venture.
And thank you for sharing that piece of spam with us again. It was so much
less enjoyable to see it the second time.
Seriously James, with more and more people using automated spam filters,
it might not be such a wise idea to keep having your name associated with
spam content.
Thank you for the tip.
James
We also have:
p=lambda a,n: [ ''.join( y ) for y in eval('[%%s %s]'%' '.join(['for x
%i in a'%i for i in range(n)]) %'(%s)'%','.join(['x%i'%i for i in
range(n)]) ) ]
p('abc',2)
#fb: ['aa', 'ab', 'ac', 'ba', 'bb', 'bc', 'ca', 'cb', 'cc']
len(p('13579',3))
#fb: 125
edit()
File under obscurities. acb
Slightly clearer:
p=lambda a,n:[ ''.join( y ) for y in eval('[(%s) %s]'%(','.join(['x
%i'%i for i in range(n)]),' '.join(['for x%i in a'%i for i in
range(n)])))]
p('abc',2)
#fb: ['aa', 'ab', 'ac', 'ba', 'bb', 'bc', 'ca', 'cb', 'cc']
len(p('13579',3))
#fb: 125
edit()
where for n=4:
('[(%s) %s]'%(','.join(['x%i'%i for i in range(n)]),' '.join(['for x%i
in a'%i for i in range(n)])))
#fb: '[(x0,x1,x2,x3) for x0 in a for x1 in a for x2 in a for x3 in a]'
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Re: interesting exercise
Thanks castironpi and alex, for this: def p(a,b): if not b: return [''] return [i+j for i in a for j in p(a,b-1)] That is a thing of beauty! As usual you guys didn't disappoint. (Validity check of alphabet removed; you didn't check for duplicate characters.) Here is the bloated mess I came up with. I did see that it had to be recursive, and was proud of myself for getting it pretty much on the first try, but the thing still reeks of my sorry old fortran-addled mentality. def validAlphabet(alphabet): if not isinstance(alphabet,str): return False if len(alphabet) 256: return False for index in range(len(alphabet)-1): if not alphabet[index] alphabet[index+1]: return False return True def nextword(word,alphabet): if len(word) == 1: index = alphabet.find(word) if index 0: raise ValueError(bad token found %s % word) if index == len(alphabet) -1: return None else: return alphabet[index+1] else: a = nextword(word[1:],alphabet) if a: return word[0] + a else: b = nextword(word[0],alphabet) if b: return b + (len(word) - 1) * alphabet[0] else: return None def allwords(length,alphabet=01): if not validAlphabet(alphabet): raise ValueError(bad token list %s % alphabet) word = length * alphabet[0] result = [word] while word length * alphabet[-1]: word = nextword(word,alphabet)) result.append(word) return result ## if __name__ == __main__: for item in allwords(5,abc): print item -- http://mail.python.org/mailman/listinfo/python-list
interesting exercise
I want a list of all ordered permutations of a given length of a set of tokens. Each token is a single character, and for convenience, they are passed as a string in ascending ASCII order. For example permute(abc,2) should return [aa,ab,ac,ba,bb,bc,ca,cb,cc] and permute(13579,3) should return a list of 125 elements [111,113, ... ,997,999] permute(axc,N) or permute(2446,N) should raise ValueError as the alphabet is not strictly sorted. I have a reasonably elegant solution but it's a bit verbose (a couple dozen lines which I'll post later if there is interest). Is there some clever Pythonism I didn't spot? thanks mt -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
Michael Tobis wrote: I want a list of all ordered permutations of a given length of a set of tokens. Each token is a single character, and for convenience, they are passed as a string in ascending ASCII order. For example permute(abc,2) should return [aa,ab,ac,ba,bb,bc,ca,cb,cc] and permute(13579,3) should return a list of 125 elements [111,113, ... ,997,999] permute(axc,N) or permute(2446,N) should raise ValueError as the alphabet is not strictly sorted. I have a reasonably elegant solution but it's a bit verbose (a couple dozen lines which I'll post later if there is interest). Is there some clever Pythonism I didn't spot? thanks mt 1. You oughtta go ahead and post it. 2. Have you checked out xpermutations? http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/190465 James -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
En Tue, 08 May 2007 00:45:52 -0300, Michael Tobis [EMAIL PROTECTED]
escribió:
I want a list of all ordered permutations of a given length of a set
of tokens. Each token is a single character, and for convenience, they
are passed as a string in ascending ASCII order.
This is what I come, no tricks, no special optimizations, just plain
Python (including your constraints):
def permute(values, n):
def _permute(values, n):
if n==1:
for letter in values:
yield [letter]
else:
for letter in values:
for others in _permute(values, n-1):
yield [letter]+others
if not sorted(values): raise ValueError(unsorted values)
if len(set(values))!=len(values): raise ValueError(duplicate values)
return list(''.join(item) for item in _permute(values, n))
--
Gabriel Genellina
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Re: interesting exercise
On Mon, 07 May 2007 20:45:52 -0700, Michael Tobis wrote: I have a reasonably elegant solution but it's a bit verbose (a couple dozen lines which I'll post later if there is interest). Is there some clever Pythonism I didn't spot? Peering into my crystal ball, I see that your algorithm does the wrong thing, and contains bugs too. You certainly don't need to partion the sequence into three sub-sequences, or do that trick with the metaclass, and it is more efficient to use list.extend() than sum. Hang on... stupid crystal ball... that's somebody else's code. Maybe you should just post your code here, so we can look at it? In the meantime, here's a simple generator to do permutations with repetition: def permute_with_repetitions(seq): if len(seq) = 1: yield list(seq) else: for i, item in enumerate(seq): for tail in permute_with_repetitions(seq[:i] + seq[i+1:]): yield [item] + tail It doesn't do a test for the sequence containing duplicates, and I leave it as an exercise to do selections of fewer items. -- Steven. -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
On Tue, 08 May 2007 01:21:37 -0300, Gabriel Genellina wrote: if not sorted(values): raise ValueError(unsorted values) sorted() doesn't return a flag telling if the values are sorted, it returns a new list containing values sorted. So this line will raise an exception only on an empty sequence. -- Steven. -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
On May 7, 10:45 pm, Michael Tobis [EMAIL PROTECTED] wrote: I want a list of all ordered permutations of a given length of a set of tokens. Each token is a single character, and for convenience, they are passed as a string in ascending ASCII order. For example permute(abc,2) should return [aa,ab,ac,ba,bb,bc,ca,cb,cc] and permute(13579,3) should return a list of 125 elements [111,113, ... ,997,999] permute(axc,N) or permute(2446,N) should raise ValueError as the alphabet is not strictly sorted. I have a reasonably elegant solution but it's a bit verbose (a couple dozen lines which I'll post later if there is interest). Is there some clever Pythonism I didn't spot? thanks mt Post yours. -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
En Tue, 08 May 2007 01:33:51 -0300, Steven D'Aprano [EMAIL PROTECTED] escribió: On Tue, 08 May 2007 01:21:37 -0300, Gabriel Genellina wrote: if not sorted(values): raise ValueError(unsorted values) sorted() doesn't return a flag telling if the values are sorted, it returns a new list containing values sorted. So this line will raise an exception only on an empty sequence. Ouch! -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
On May 7, 11:34 pm, [EMAIL PROTECTED] wrote:
On May 7, 10:45 pm, Michael Tobis [EMAIL PROTECTED] wrote:
I want a list of all ordered permutations of a given length of a set
of tokens. Each token is a single character, and for convenience, they
are passed as a string in ascending ASCII order.
For example
permute(abc,2)
should return [aa,ab,ac,ba,bb,bc,ca,cb,cc]
and permute(13579,3) should return a list of 125 elements
[111,113, ... ,997,999]
permute(axc,N) or permute(2446,N) should raise ValueError as the
alphabet is not strictly sorted.
I have a reasonably elegant solution but it's a bit verbose (a couple
dozen lines which I'll post later if there is interest). Is there some
clever Pythonism I didn't spot?
thanks
mt
Post yours.
Oh well, as I'm not the first.
def p(a,b):
if list( a ) != sorted( list( a ) ): raise ValueError, String not
ordered.
if not b: return ['']
a = sorted( set( a ) )
return [i+j for i in a for j in p(a,b-1)]
p('abc',3)
#fb: ['aaa', 'aab', 'aac', 'aba', 'abb', 'abc', 'aca', 'acb', 'acc',
'baa', 'bab', 'bac', 'bba', 'bbb', 'bbc', 'bca', 'bcb', 'bcc', 'caa',
'cab', 'cac', 'cba', 'cbb', 'cbc', 'cca', 'ccb', 'ccc']
p('abc',2)
#fb: ['aa', 'ab', 'ac', 'ba', 'bb', 'bc', 'ca', 'cb', 'cc']
len(p(13579,3))
#fb: 125
edit()
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Re: interesting exercise
On May 7, 11:42 pm, [EMAIL PROTECTED] wrote:
On May 7, 11:34 pm, [EMAIL PROTECTED] wrote:
On May 7, 10:45 pm, Michael Tobis [EMAIL PROTECTED] wrote:
I want a list of all ordered permutations of a given length of a set
of tokens. Each token is a single character, and for convenience, they
are passed as a string in ascending ASCII order.
For example
permute(abc,2)
should return [aa,ab,ac,ba,bb,bc,ca,cb,cc]
and permute(13579,3) should return a list of 125 elements
[111,113, ... ,997,999]
permute(axc,N) or permute(2446,N) should raise ValueError as the
alphabet is not strictly sorted.
I have a reasonably elegant solution but it's a bit verbose (a couple
dozen lines which I'll post later if there is interest). Is there some
clever Pythonism I didn't spot?
thanks
mt
Post yours.
Oh well, as I'm not the first.
[working solution snip]
Or even,
def p(a,b):
if list( a ) != sorted( list( a ) ): raise ValueError, String not
ordered.
if not b: return ['']
return [i+j for i in list(a) for j in p(a,b-1)]
p('abc',3)
#fb: ['aaa', 'aab', 'aac', 'aba', 'abb', 'abc', 'aca', 'acb', 'acc',
'baa', 'bab', 'bac', 'bba', 'bbb', 'bbc', 'bca', 'bcb', 'bcc', 'caa',
'cab', 'cac', 'cba', 'cbb', 'cbc', 'cca', 'ccb', 'ccc']
p('abc',2)
#fb: ['aa', 'ab', 'ac', 'ba', 'bb', 'bc', 'ca', 'cb', 'cc']
len(p(13579,3))
#fb: 125
edit()
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Re: interesting exercise
On May 8, 9:31 am, Steven D'Aprano [EMAIL PROTECTED] wrote: On Mon, 07 May 2007 20:45:52 -0700, Michael Tobis wrote: I have a reasonably elegant solution but it's a bit verbose (a couple dozen lines which I'll post later if there is interest). Is there some clever Pythonism I didn't spot? Peering into my crystal ball, I see that your algorithm does the wrong thing, and contains bugs too. You certainly don't need to partion the sequence into three sub-sequences, or do that trick with the metaclass, and it is more efficient to use list.extend() than sum. Hang on... stupid crystal ball... that's somebody else's code. Maybe you should just post your code here, so we can look at it? In the meantime, here's a simple generator to do permutations with repetition: def permute_with_repetitions(seq): if len(seq) = 1: yield list(seq) else: for i, item in enumerate(seq): for tail in permute_with_repetitions(seq[:i] + seq[i+1:]): yield [item] + tail It doesn't do a test for the sequence containing duplicates, and I leave it as anexerciseto do selections of fewer items. -- Steven. Dear Steven I ran into your message quite accidentally while researching about some details on 'Exercise' and thought of sharing some of my findings. I've read at 'http://www.medical-health-care-information.com/Health- living/exercise/index.asp that Swimming, cycling, jogging, skiing, aerobic dancing, walking or any of dozens of other activities can help your heart. Whether it's included in a structured exercise program or just part of your daily routine, all physical activity adds up to a healthier heart. I hope the above is of some help to you as well. Regards, Sherrybove. -- http://mail.python.org/mailman/listinfo/python-list
Re: interesting exercise
[EMAIL PROTECTED] wrote: ... def p(a,b): if list( a ) != sorted( list( a ) ): raise ValueError, String not ordered. if not b: return [''] return [i+j for i in list(a) for j in p(a,b-1)] No need for 2/3 of the list(...) calls. sorted(a) and sorted(list(a)) will ALWAYS be the same sequence; for i in a and for i in list(a) will always iterate on the same sequence [as long as you're not changing a inside the iteration, which, in this case, you aren't]. Alex -- http://mail.python.org/mailman/listinfo/python-list
