# generate random digits with length of 5

Aaron "Castironpi" Brady castironpi at gmail.com
Sun Sep 28 22:54:15 CEST 2008

```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
> 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?

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

```