On Sep 28, 3:11�pm, "Gary M. Josack" <[EMAIL PROTECTED]> wrote: > Chris Rebert wrote: > > On Sun, Sep 28, 2008 at 12:59 PM, sotirac <[EMAIL PROTECTED]> 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' -- http://mail.python.org/mailman/listinfo/python-list
