generate random digits with length of 5
Mensanator
mensanator at aol.com
Sun Sep 28 22:44:00 CEST 2008
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'
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