Let say I need a 3 digit number, so it would be something like:

>>> random(3)
563

or

>>> random(5)
26748
>> random(2)
56

To get a random 3-digit number:

from random import randint
randint(100, 999)  # randint is inclusive at both ends

(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

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)])

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:

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_}}'

Does 0 count as a possible first digit? If so, then you need random.randint(0,10**n-1). If not, random.randint(10**(n-1),10**n-1). And if zero is never allowed, then you’ll have to explicitly reject numbers with a zero in them, or draw n random.randint(1,9) numbers.

Aside: it is interesting that randint(a,b) uses somewhat non-pythonic “indexing” to get a random number a <= n <= b. One might have expected it to work like range, and produce a random number a <= n < b. (Note the closed upper interval.)

Given the responses in the comments about randrange, note that these can be replaced with the cleaner random.randrange(0,10**n), random.randrange(10**(n-1),10**n) and random.randrange(1,10).

You could write yourself a little function to do what you want:

import random
def randomDigits(digits):
    lower = 10**(digits-1)
    upper = 10**digits - 1
    return random.randint(lower, upper)

Basically, 10**(digits-1) gives you the smallest {digit}-digit number, and 10**digits - 1 gives you the largest {digit}-digit number (which happens to be the smallest {digit+1}-digit number minus 1!). Then we just take a random integer from that range.

I really liked the answer of RichieHindle, however I liked the question as an exercise. Here’s a brute force implementation using strings:)

import random
first = random.randint(1,9)
first = str(first)
n = 5

nrs = [str(random.randrange(10)) for i in range(n-1)]
for i in range(len(nrs))    :
    first += str(nrs[i])

print str(first)

From the official documentation, does it not seem that the sample() method is appropriate for this purpose?

import random

def random_digits(n):
    num = range(0, 10)
    lst = random.sample(num, n)
    print str(lst).strip('[]')

Output:

>>>random_digits(5)
2, 5, 1, 0, 4

You could create a function who consumes an list of int, transforms in string to concatenate and cast do int again, something like this:

import random

def generate_random_number(length):
    return int(''.join([str(random.randint(0,10)) for _ in range(length)]))