interesting exercise

[castironpi at gmail.com]

On May 9, 1:13 am, Charles Sanders <C.delete_this.Sand... at BoM.GOv.AU>
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.