How to generate a random number with a specific amount of digits?

Solution 1:

You can use either of random.randint or random.randrange. So to get a random 3-digit number:

from random import randint, randrange

randint(100, 999)     # randint is inclusive at both ends
randrange(100, 1000)  # randrange is exclusive at the stop

^{* Assuming you really meant three digits, rather than "up to three digits".}

To use an arbitrary number of digits:

from random import randint

def random_with_N_digits(n):
    range_start = 10**(n-1)
    range_end = (10**n)-1
    return randint(range_start, range_end)
    
print random_with_N_digits(2)
print random_with_N_digits(3)
print random_with_N_digits(4)

Output:

33
124
5127

Solution 2:

If you want it as a string (for example, a 10-digit phone number) you can use this:

n = 10
''.join(["{}".format(randint(0, 9)) for num in range(0, n)])

Solution 3:

If you need a 3 digit number and want 001-099 to be valid numbers you should still use randrange/randint as it is quicker than alternatives. Just add the neccessary preceding zeros when converting to a string.

import random
num = random.randrange(1, 10**3)
# using format
num_with_zeros = '{:03}'.format(num)
# using string's zfill
num_with_zeros = str(num).zfill(3)

Alternatively if you don't want to save the random number as an int you can just do it as a oneliner:

'{:03}'.format(random.randrange(1, 10**3))

python 3.6+ only oneliner:

f'{random.randrange(1, 10**3):03}'

Example outputs of the above are:

• '026'
• '255'
• '512'

Implemented as a function that can support any length of digits not just 3:

import random

def n_len_rand(len_, floor=1):
    top = 10**len_
    if floor > top:
        raise ValueError(f"Floor '{floor}' must be less than requested top '{top}'")
    return f'{random.randrange(floor, top):0{len_}}'