[Tut] 3 Best Ways to Generate a Random Number with a Fixed Amount of Digits in Python

3 Best Ways to Generate a Random Number with a Fixed Amount of Digits in Python

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Coding Challenge

Challenge: Given an integer d representing the number of digits. How to create a random number with d digits in Python?

Here are three examples:

• my_random(2) generates 12
• my_random(3) generates 389
• my_random(10) generates 8943496710

I’ll discuss three interesting methods to accomplish this easily in Python—my personal favorite is Method 2!

Shortest Solution with randint()

Let’s start with an easy hand-coded observation:

The easiest way to create a random number with two digits is to use random‘s randint(10, 99), with three digits is randint(100,999), and with four digits is randint(1000,9999).

Here’s the same example in Python code:

from random import randint # Create random number with two digits (d=2):

print(randint(10, 99)) # Create random number with three digits (d=3):

print(randint(100, 999)) # Create random number with three digits (d=3):

print(randint(1000, 9999))

This solution can be generalized by using the one-liner random.randint(int('1'+'0'*(d-1)), int('9'*d)) that generates the start and end values on the fly, based on the number of digits d.

I used simple string arithmetic to define the start and end index of the random range:

• int('1'+'0'*(d-1)) creates the start index such as 100 for d=3.
• int('9'*d)) creates the end index that’s included in randint() such as 999 for d=3.

Here’s the basic Python example:

import random def my_random(d): ''' Generates a random number with d digits ''' return random.randint(int('1'+'0'*(d-1)), int('9'*d)) for i in range(1, 10): print(my_random(i)) '''

Output:

8

82

296

5909

90957

227691

1348638

61368798

160959002 '''

Cleanest Solution with randrange()

The cleanest solution is based on the randrange() function from the random module that takes the start and end index as input and generates a random number in between.

Unlike randint(), the end index is excluded in randrange(), so we have an easier way to construct our range for the d-digit random number problem: random.randrange(10**(d-1), 10**d).

Here’s an example:

import random def my_random(d): ''' Generates a random number with d digits ''' return random.randrange(10**(d-1), 10**d) for i in range(1, 10): print(my_random(i)) '''

Output:

7

64

872

2440

39255

979369

6897920

83589118

707920991 '''

An Iterative Solution Aggregating Outputs of Single-Digit Random Function Calls

You can also use a one-liner to repeatedly execute the random.randint() function for each digit. To combine the digits, you convert each digit to a string, pass them into the string.join() function to get one string with d characters, and convert this string back to an integer:

int(''.join(str(random.randint(0,9)) for _ in range(d)))

Here’s this exact approach in a Python code snippet:

import random def my_random(d): ''' Generates a random number with d digits ''' return int(''.join(str(random.randint(0,9)) for _ in range(d))) for i in range(1, 10): print(my_random(i)) '''

Output:

6

92

135

156

95865

409722

349673

31144072

439469934 '''

Summary

Thanks for reading through the whole article—I hope you got some value out of it.

Here’s again a summary of how to best generate a random number with d digits in Python:

1. random.randint(int('1'+'0'*(d-1)), int('9'*d))
2. random.randrange(10**(d-1), 10**d)
3. int(''.join(str(random.randint(0,9)) for _ in range(d)))

Personally, I like Method 2 the most because it’s short, concise, and very efficient!

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https://www.sickgaming.net/blog/2022/09/...in-python/