Is there any downside to iterating over a range for more times than is required to generate desired output?

Issue

I’m taking a 100 Days of Code in Python, and I’m trying to create a Python password generator by taking in user input for how many letters, numbers, and symbols they’d like in their password.

The program below runs and generates the desired output, but I know there must be a better way than iterating over the range an arbitrary number of times to generate a fixed-length password.

import random
letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
numbers = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
symbols = ['!', '#', '$', '%', '&', '(', ')', '*', '+']

print("Welcome to the PyPassword Generator!")
nr_letters= int(input('How many letters would you like in your password?: ')) 
nr_symbols = int(input('How many symbols would you like?: '))
nr_numbers = int(input('How many numbers would you like?: '))

password = ""

# Generate unshuffled password

# for i in range(1, (nr_letters + 1)):
#   password += random.choice(letters)
# for i in range(1, (nr_symbols + 1)):
#   password += random.choice(numbers)
# for i in range(1, (nr_numbers +1)):
#   password += random.choice(symbols)
# print(password)

letter_counter = 0
symbol_counter = 0
number_counter = 0

# NOTE: This seems dumb but it works so...

for i in range(0, 100):
  random_int = random.randint(0, 2)
  if random_int == 0 and letter_counter < nr_letters:
    password += random.choice(letters)
    letter_counter += 1
  elif random_int == 1 and symbol_counter < nr_symbols:
    password += random.choice(symbols)
    symbol_counter += 1
  elif random_int == 2 and number_counter < nr_numbers:
    password += random.choice(numbers)
    number_counter += 1

print(password)

Is there a cleaner way I can create a shuffled, fixed-length string through a Python for loop?

For the future, is there a major downside to iterating through a loop more times than it takes to generate the desired output?

Solution

I think this is what you are looking for?:

import random
letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
numbers = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
symbols = ['!', '#', '$', '%', '&', '(', ')', '*', '+']

print("Welcome to the PyPassword Generator!")
nr_letters= int(input('How many letters would you like in your password?: ')) 
nr_symbols = int(input('How many symbols would you like?: '))
nr_numbers = int(input('How many numbers would you like?: '))

password = ""

for i in range(1, sum([nr_letters,nr_numbers,nr_symbols])+1):
  if i<= nr_letters:
    password += random.choice(letters)
  elif i > nr_letters and i<=nr_symbols+nr_letters:
    password += random.choice(symbols)
  elif i > nr_symbols+nr_letters:
    password += random.choice(numbers)

password=[x for x in password]
random.shuffle(password)
password="".join(password)
print(password)

Basically, instead of iterating through an arbitrary number, it adds up the number of characters required, then it goes through the range, and adding the numbers/symbols/letters when the requirements are met.

The requirements are basic. While below or equal to the number of letters required, add a letter. Then, you go and add the number of letters and symbols together, and say that it has to meet both above the number of letters, and below letters+symbols. Then, if i is above letters+symbols, add numbers. That’s pretty much the pattern.

The downside is that you will have to shuffle afterwards like I’ve done above. password=[x for x in password] basically turns password into a list. Then you shuffle it with random.shuffle(password), and then join it password="".join(password).

It’s basically your first commented out iterations bundled up in 1.

Answered By – Agent Biscutt

Answer Checked By – Jay B. (AngularFixing Admin)

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