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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