generate random digits with length of 5

[Mensanator]

On Sep 28, 3:54�pm, "Aaron \"Castironpi\" Brady"
<castiro... at gmail.com> wrote:
> On Sep 28, 3:44�pm, Mensanator <mensana... at aol.com> wrote:
>
>
>
>
>
> > On Sep 28, 3:11 pm, "Gary M. Josack" <g... at byoteki.com> wrote:
>
> > > Chris Rebert wrote:
> > > > On Sun, Sep 28, 2008 at 12:59 PM, sotirac <soti... at gmail.com> wrote:
>
> > > >> Wondering if there is a better way to generate string of numbers with
> > > >> a length of 5 which also can have a 0 in the front of the number.
>
> > > >> <pre>
> > > >> random_number = random.sample([0,1,2,3,4,5,6,7,8,9], 5) # choose 5
> > > >> elements
> > > >> code = 'this is a string' + str(random_number[0]) +
> > > >> str(random_number[1]) + str(random_number[2]) + str(random_number[3])
> > > >> + str(random_number[4])
>
> > > > code = ''.join(str(digit) for digit in random_number)
>
> > > > Regards,
> > > > Chris
>
> > > >> </pre>
>
> > > >> --
> > > >>http://mail.python.org/mailman/listinfo/python-list
>
> > > will random.randint(10000,99999) work for you?
>
> > It doesn't meet the OP's requirement that the number
> > can start with 0. Also, the method the OP asks about
> > returns a list of unique numbers, so no number can
> > be duplicated. He can get 02468 but not 13345.
>
> > Now, IF it's ok to have an arbitrary number of leading
> > 0s, he can do this:
>
> > >>> str(random.randint(0,99999)).zfill(5)
> > '00089'
> > >>> str(random.randint(0,99999)).zfill(5)
> > '63782'
> > >>> str(random.randint(0,99999)).zfill(5)
> > '63613'
> > >>> str(random.randint(0,99999)).zfill(5)
>
> > '22315'
>
> Is a while loop until there are 5 distinct digits best otherwise?

Of course not.

>
> while 1:
> � a= '%05i'% random.randint( 0, 99999 )
> � if len( set( a ) )== 5: break

How is this better than the OP's original code?