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

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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

Code:

d

representing the number of digits. How to create a random number with

Code:

d

digits in Python?

Here are three examples:

• Code:

  my_random(2)
  generates
  Code:

  12
  
  
• Code:

  my_random(3)
  generates
  Code:

  389
  
  
• Code:

  my_random(10)
  generates
  Code:

  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

Code:

random

‘s

Code:

randint(10, 99)

, with three digits is

Code:

randint(100,999)

, and with four digits is

Code:

randint(1000,9999)

.

Here’s the same example in Python code:

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

Code:

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

Code:

d

.

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

• Code:

  int('1'+'0'*(d-1))
  creates the start index such as 100 for
  Code:

  d=3
  .
  
  
• Code:

  int('9'*d))
  creates the end index that’s included in
  Code:

  randint()
  such as
  Code:

  999
  for
  Code:

  d=3
  .
  

Here’s the basic Python example:

Code:

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

Code:

randrange()

function from the

Code:

random

module that takes the start and end index as input and generates a random number in between.

Unlike

Code:

randint()

, the end index is excluded in

Code:

randrange()

, so we have an easier way to construct our range for the

Code:

d

-digit random number problem:

Code:

random.randrange(10**(d-1), 10**d)

.

Here’s an example:

Code:

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

Code:

random.randint()

function for each digit. To combine the digits, you convert each digit to a string, pass them into the

Code:

string.join()

function to get one string with

Code:

d

characters, and convert this string back to an integer:

Code:

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

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

Code:

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

Code:

d

digits in Python:

1. Code:

   random.randint(int('1'+'0'*(d-1)), int('9'*d))
   
   
2. Code:

   random.randrange(10**(d-1), 10**d)
   
   
3. Code:

   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!