hi
i need to do something like this
eg given a number (as a string) = "123"
there are a few combination i want to find with this string, ie
"132","321","23 1","312" and "213". so there are 6 combinations
altogether. i remember there's a formula for this, but forgot. Does
python have any modules/functions to do this?
thanks 3 1457 ei***********@y ahoo.com wrote: hi i need to do something like this eg given a number (as a string) = "123" there are a few combination i want to find with this string, ie "132","321","23 1","312" and "213". so there are 6 combinations altogether. i remember there's a formula for this, but forgot. Does python have any modules/functions to do this?
thanks
factorial = [ 1,
1,
2,
6,
24,
120,
720,
5040,
40320,
362880,
3628800,
39916800,
479001600,
6227020800,
87178291200,
1307674368000,
20922789888000,
355687428096000 ,
640237370572800 0 ]
def permutation( iterable ):
length = len(iterable)
count = factorial[length]
for i in range(count):
sequence = list(iterable[:])
index = i
N = count
result = []
for j in range( length, 0, -1):
N = N / j
choice, index = index // N, index % N
result += [ sequence.pop(ch oice) ]
yield result
a = [ ''.join(perm) for perm in permutation('12 3')]
print a
['123', '132', '213', '231', '312', '321']
Gerard
Gerard Flanagan wrote: ei***********@y ahoo.com wrote: hi i need to do something like this eg given a number (as a string) = "123" there are a few combination i want to find with this string, ie "132","321","23 1","312" and "213". so there are 6 combinations altogether. i remember there's a formula for this, but forgot. Does python have any modules/functions to do this?
thanks
factorial = [ 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000 , 640237370572800 0 ]
def permutation( iterable ): length = len(iterable) count = factorial[length] for i in range(count): sequence = list(iterable[:]) index = i N = count result = [] for j in range( length, 0, -1): N = N / j choice, index = index // N, index % N result += [ sequence.pop(ch oice) ] yield result
a = [ ''.join(perm) for perm in permutation('12 3')]
print a
['123', '132', '213', '231', '312', '321']
Which is actually not a good algorithm at all, having just tried
'123456' - it's better for randomly-accessing a particular permutation.
There's a better 'algo' here: http://gflanagan.net/site/python/05/Johnson.html
for getting all perms, though there's no getting away from the
factorial - IIRC, NPermutation(7) took less than a minute,
NPermutation(8) took 2 minutes and NPermutation(9) took 10 minutes on
my machine.
Also, search this group for Jack Diederich and probstat.
HTH
Gerard This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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