On Sep 28, 3:11�pm, "Gary M. Josack" <(E-Mail Removed)> wrote:

> Chris Rebert wrote:

> > On Sun, Sep 28, 2008 at 12:59 PM, sotirac <(E-Mail Removed)> 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'