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

## Question:

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

``````>>> random(3)
563

or

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

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
``````

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.

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)])
``````

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)]))
``````

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_}}'
``````
``````import random

fixed_digits = 6

print(random.randrange(111111, 999999, fixed_digits))

out:
271533
``````

I know it’s an old question, but I see many solutions throughout the years. here is my suggestion for a function that creates an n digits random string with default 6.

``````from random import randint

def OTP(n=6):
n = 1 if n< 1 else n
return randint(int("1"*n), int("9"*n))
``````

of course you can change "1"*n to whatever you want the start to be.

Categories: questions Tags: ,
Answers are sorted by their score. The answer accepted by the question owner as the best is marked with
at the top-right corner.