Random number having 11 to 20 digits

Question

How to generate random numbers in python having(11 to 20 digits) that should not start with 0 in beginning?

eg:23418915901

def key_more_than_10():
    v=random.randint(11,20)
    char_set = string.digits+string.digits+string.digits
    s= ''.join(random.sample(char_set,v))
    return s

I used this code but since string.digits(give numbers between 0 to 9) it doesnt help me?

There should be an equal possibility to have 11 digits number and a 20 digits number.

Answer 1

Simple:

>>> help(random.randint)
Help on method randint in module random:

randint(self, a, b) method of random.Random instance
    Return random integer in range [a, b], including both end points.

>>> random.randint(10000000000,99999999999999999999)
35677750742418249489

Of course, if you actually want a string (not sure why you would), you could wrap a str() around the whole thing:

>>> str(random.randint(1000000000,99999999999999999999))
'91138793919270489115'

Answer 2

How about a literal translation of what you want into code:

import random
DIGITS = '0123456789'

ndigits = random.randint(11, 20)
digits = [random.choice(DIGITS[1:])] + [random.choice(DIGITS) for i in xrange(ndigits-1)]
print int(''.join(digits))

The number of digits in the almost random result should be uniformly distributed between 11..20. Not quite random because restricting the first digit to a something other than a zero precludes it.

Answer 3

I would choose randrange for this

Help on method randrange in module random:

    randrange(self, start, stop=None, step=1, int=<type 'int'>, default=None, maxwidth=9007199254740992L) method of random.Random instance
        Choose a random item from range(start, stop[, step]).

    This fixes the problem with randint() which includes the
    endpoint; in Python this is usually not what you want.
    Do not supply the 'int', 'default', and 'maxwidth' arguments.

so

random.randrange(1e11, 1e20)

will do what you want. For a string use this

str(random.randrange(1e11, 1e20))