Python - random float with limited decimal digits

Question

I just found out how to make random numbers in Python, but if I print them out, they all have 15 decimal digits. How do I fix that? Here is my code:

import random
import os

greaterThan = float(input("Your number will be greater than: "))
lessThan = float(input("Your number will be less than: "))
digits = int(input("Your number will that many decimal digits: "))

os.system('cls')

if digits == 15:
    print(random.uniform(greaterThan, lessThan))

if digits == 14:
    print(random.uniform(greaterThan, lessThan))

if digits == 13:
    print(random.uniform(greaterThan, lessThan))

if digits == 12:
    print(random.uniform(greaterThan, lessThan))

if digits == 11:
    print(random.uniform(greaterThan, lessThan))

if digits == 10:
    print(random.uniform(greaterThan, lessThan))

*(this just continues down to 0)

I know you can do like print("%.2" % someVariable) but the random numbers Python creates with this method are not saved in any variables. At least I think so. I would also like to know if there is a way of letting a variable choose the amount of decimal points, like print("%." + digits % anotherVariable) but I tried tha out and of course it failed.

I really got no idea. Hopefully you can help. Thanks.

Answer 1

If you want to take whatever number you get from random.uniform and truncate it to the specific number of digits you can use the round() function.

It allows you to round to a specific precision. For example:

import random

greaterThan = float(input("Your number will be greater than: "))
lessThan = float(input("Your number will be less than: "))
digits = int(input("Your number will that many decimal digits: "))

rounded_number = round(random.uniform(greaterThan, lessThan), digits)
print(rounded_number)