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

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

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[Tut] 3 Best Ways to Generate a Random Number with a Fixed Amount of Digits in Python - xSicKxBot - 09-27-2022

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

5/5 – (1 vote)

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:

Code:
random.randint(int('1'+'0'*(d-1)), int('9'*d))
Code:
random.randrange(10**(d-1), 10**d)
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!