Random string generation with upper case letters and digits


How do I generate a string of size N, made of numbers and uppercase English letters such as:

  • 6U1S75
  • 4Z4UKK
  • U911K4
Asked By: Hellnar



Answer in one line:

''.join(random.choice(string.ascii_uppercase + string.digits) for _ in range(N))

or even shorter starting with Python 3.6 using random.choices():

''.join(random.choices(string.ascii_uppercase + string.digits, k=N))

A cryptographically more secure version: see this post

''.join(random.SystemRandom().choice(string.ascii_uppercase + string.digits) for _ in range(N))

In details, with a clean function for further reuse:

>>> import string
>>> import random
>>> def id_generator(size=6, chars=string.ascii_uppercase + string.digits):
...    return ''.join(random.choice(chars) for _ in range(size))
>>> id_generator()
>>> id_generator(3, "6793YUIO")

How does it work ?

We import string, a module that contains sequences of common ASCII characters, and random, a module that deals with random generation.

string.ascii_uppercase + string.digits just concatenates the list of characters representing uppercase ASCII chars and digits:

>>> string.ascii_uppercase
>>> string.digits
>>> string.ascii_uppercase + string.digits

Then we use a list comprehension to create a list of ‘n’ elements:

>>> range(4) # range create a list of 'n' numbers
[0, 1, 2, 3]
>>> ['elem' for _ in range(4)] # we use range to create 4 times 'elem'
['elem', 'elem', 'elem', 'elem']

In the example above, we use [ to create the list, but we don’t in the id_generator function so Python doesn’t create the list in memory, but generates the elements on the fly, one by one (more about this here).

Instead of asking to create ‘n’ times the string elem, we will ask Python to create ‘n’ times a random character, picked from a sequence of characters:

>>> random.choice("abcde")
>>> random.choice("abcde")
>>> random.choice("abcde")

Therefore random.choice(chars) for _ in range(size) really is creating a sequence of size characters. Characters that are randomly picked from chars:

>>> [random.choice('abcde') for _ in range(3)]
['a', 'b', 'b']
>>> [random.choice('abcde') for _ in range(3)]
['e', 'b', 'e']
>>> [random.choice('abcde') for _ in range(3)]
['d', 'a', 'c']

Then we just join them with an empty string so the sequence becomes a string:

>>> ''.join(['a', 'b', 'b'])
>>> [random.choice('abcde') for _ in range(3)]
['d', 'c', 'b']
>>> ''.join(random.choice('abcde') for _ in range(3))

A simpler, faster but slightly less random way is to use random.sample instead of choosing each letter separately, If n-repetitions are allowed, enlarge your random basis by n times e.g.

import random
import string

char_set = string.ascii_uppercase + string.digits
print ''.join(random.sample(char_set*6, 6))

random.sample prevents character reuse, multiplying the size of the character set makes multiple repetitions possible, but they are still less likely then they are in a pure random choice. If we go for a string of length 6, and we pick ‘X’ as the first character, in the choice example, the odds of getting ‘X’ for the second character are the same as the odds of getting ‘X’ as the first character. In the random.sample implementation, the odds of getting ‘X’ as any subsequent character are only 6/7 the chance of getting it as the first character

Answered By: Anurag Uniyal

I thought no one had answered this yet lol! But hey, here’s my own go at it:

import random

def random_alphanumeric(limit):
    #ascii alphabet of all alphanumerals
    r = (range(48, 58) + range(65, 91) + range(97, 123))
    return reduce(lambda i, s: i + chr(s), r[:random.randint(0, len(r))], "")
Answered By: JWL

I’d do it this way:

import random
from string import digits, ascii_uppercase

legals = digits + ascii_uppercase

def rand_string(length, char_set=legals):

    output = ''
    for _ in range(length): output += random.choice(char_set)
    return output

Or just:

def rand_string(length, char_set=legals):

    return ''.join( random.choice(char_set) for _ in range(length) )
Answered By: Carl Smith

If you need a random string rather than a pseudo random one, you should use os.urandom as the source

from os import urandom
from itertools import islice, imap, repeat
import string

def rand_string(length=5):
    chars = set(string.ascii_uppercase + string.digits)
    char_gen = (c for c in imap(urandom, repeat(1)) if c in chars)
    return ''.join(islice(char_gen, None, length))
Answered By: John La Rooy

This method is slightly faster, and slightly more annoying, than the random.choice() method Ignacio posted.

It takes advantage of the nature of pseudo-random algorithms, and banks on bitwise and and shift being faster than generating a new random number for each character.

# must be length 32 -- 5 bits -- the question didn't specify using the full set
# of uppercase letters ;)

def generate_with_randbits(size=32):
    def chop(x):
        while x:
            yield x & 31
            x = x >> 5
    return  ''.join(_ALPHABET[x] for x in chop(random.getrandbits(size * 5))).ljust(size, 'A')

…create a generator that takes out 5 bit numbers at a time 0..31 until none left

…join() the results of the generator on a random number with the right bits

With Timeit, for 32-character strings, the timing was:

[('generate_with_random_choice', 28.92901611328125),
 ('generate_with_randbits', 20.0293550491333)]

…but for 64 character strings, randbits loses out 😉

I would probably never use this approach in production code unless I really disliked my co-workers.

edit: updated to suit the question (uppercase and digits only), and use bitwise operators & and >> instead of % and //

Answered By: sea-rob

Simply use Python’s builtin uuid:

If UUIDs are okay for your purposes, use the built-in uuid package.

One Line Solution:

import uuid; uuid.uuid4().hex.upper()[0:6]

In Depth Version:


import uuid
uuid.uuid4() #uuid4 => full random uuid
# Outputs something like: UUID('0172fc9a-1dac-4414-b88d-6b9a6feb91ea')

If you need exactly your format (for example, “6U1S75”), you can do it like this:

import uuid

def my_random_string(string_length=10):
    """Returns a random string of length string_length."""
    random = str(uuid.uuid4()) # Convert UUID format to a Python string.
    random = random.upper() # Make all characters uppercase.
    random = random.replace("-","") # Remove the UUID '-'.
    return random[0:string_length] # Return the random string.

print(my_random_string(6)) # For example, D9E50C
Answered By: Bijan

Based on another Stack Overflow answer, Most lightweight way to create a random string and a random hexadecimal number, a better version than the accepted answer would be:

('%06x' % random.randrange(16**6)).upper()

much faster.

Answered By: Gubbi
>>> import random
>>> str = []
>>> chars = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890'
>>> num = int(raw_input('How long do you want the string to be?  '))
How long do you want the string to be?  10
>>> for k in range(1, num+1):
...    str.append(random.choice(chars))
>>> str = "".join(str)
>>> str

The random.choice function picks a random entry in a list. You also create a list so that you can append the character in the for statement. At the end str is [‘t’, ‘m’, ‘2’, ‘J’, ‘U’, ‘Q’, ‘0’, ‘4’, ‘C’, ‘K’], but the str = "".join(str) takes care of that, leaving you with 'tm2JUQ04CK'.

Hope this helps!

Answered By: A.J. Uppal

This Stack Overflow quesion is the current top Google result for “random string Python”. The current top answer is:

''.join(random.choice(string.ascii_uppercase + string.digits) for _ in range(N))

This is an excellent method, but the PRNG in random is not cryptographically secure. I assume many people researching this question will want to generate random strings for encryption or passwords. You can do this securely by making a small change in the above code:

''.join(random.SystemRandom().choice(string.ascii_uppercase + string.digits) for _ in range(N))

Using random.SystemRandom() instead of just random uses /dev/urandom on *nix machines and CryptGenRandom() in Windows. These are cryptographically secure PRNGs. Using random.choice instead of random.SystemRandom().choice in an application that requires a secure PRNG could be potentially devastating, and given the popularity of this question, I bet that mistake has been made many times already.

If you’re using python3.6 or above, you can use the new secrets module as mentioned in MSeifert’s answer:

''.join(secrets.choice(string.ascii_uppercase + string.digits) for _ in range(N))

The module docs also discuss convenient ways to generate secure tokens and best practices.

Answered By: Randy Marsh

A faster, easier and more flexible way to do this is to use the strgen module (pip install StringGenerator).

Generate a 6-character random string with upper case letters and digits:

>>> from strgen import StringGenerator as SG
>>> SG("[ud]{6}").render()

Get a unique list:

>>> SG("[ld]{10}").render_list(5,unique=True)
[u'xqqtmi1pOk', u'zmkWdUr63O', u'PGaGcPHrX2', u'6RZiUbkk2i', u'j9eIeeWgEF']

Guarantee one “special” character in the string:

>>> SG("[ld]{10}&[p]").render()

A random HTML color:

>>> SG("#[h]{6}").render()


We need to be aware that this:

''.join(random.choice(string.ascii_uppercase + string.digits) for _ in range(N))

might not have a digit (or uppercase character) in it.

strgen is faster in developer-time than any of the above solutions. The solution from Ignacio is the fastest run-time performing and is the right answer using the Python Standard Library. But you will hardly ever use it in that form. You will want to use SystemRandom (or fallback if not available), make sure required character sets are represented, use unicode (or not), make sure successive invocations produce a unique string, use a subset of one of the string module character classes, etc. This all requires lots more code than in the answers provided. The various attempts to generalize a solution all have limitations that strgen solves with greater brevity and expressive power using a simple template language.

It’s on PyPI:

pip install StringGenerator

Disclosure: I’m the author of the strgen module.

Answered By: Paul Wolf

A simple one:

import string
import random
character = string.lowercase + string.uppercase + string.digits + string.punctuation
char_len = len(character)
# you can specify your password length here
pass_len = random.randint(10,20)
password = ''
for x in range(pass_len):
    password = password + character[random.randint(0,char_len-1)]
print password
Answered By: Hackaholic

For those of you who enjoy functional python:

from itertools import imap, starmap, islice, repeat
from functools import partial
from string import letters, digits, join
from random import choice

join_chars = partial(join, sep='')
identity = lambda o: o

def irand_seqs(symbols=join_chars((letters, digits)), length=6, join=join_chars, select=choice, breakup=islice):
    """ Generates an indefinite sequence of joined random symbols each of a specific length
    :param symbols: symbols to select,
        [defaults to string.letters + string.digits, digits 0 - 9, lower and upper case English letters.]
    :param length: the length of each sequence,
        [defaults to 6]
    :param join: method used to join selected symbol, 
        [defaults to ''.join generating a string.]
    :param select: method used to select a random element from the giving population. 
        [defaults to random.choice, which selects a single element randomly]
    :return: indefinite iterator generating random sequences of giving [:param length]
    >>> from tools import irand_seqs
    >>> strings = irand_seqs()
    >>> a = next(strings)
    >>> assert isinstance(a, (str, unicode))
    >>> assert len(a) == 6
    >>> assert next(strings) != next(strings)
    return imap(join, starmap(breakup, repeat((imap(select, repeat(symbols)), None, length))))

It generates an indefinite [infinite] iterator, of joined random sequences, by first generating an indefinite sequence of randomly selected symbol from the giving pool, then breaking this sequence into length parts which is then joined, it should work with any sequence that supports getitem, by default it simply generates a random sequence of alpha numeric letters, though you can easily modify to generate other things:

for example to generate random tuples of digits:

>>> irand_tuples = irand_seqs(xrange(10), join=tuple)
>>> next(irand_tuples)
(0, 5, 5, 7, 2, 8)
>>> next(irand_tuples)
(3, 2, 2, 0, 3, 1)

if you don’t want to use next for generation you can simply make it callable:

>>> irand_tuples = irand_seqs(xrange(10), join=tuple)
>>> make_rand_tuples = partial(next, irand_tuples) 
>>> make_rand_tuples()
(1, 6, 2, 8, 1, 9)

if you want to generate the sequence on the fly simply set join to identity.

>>> irand_tuples = irand_seqs(xrange(10), join=identity)
>>> selections = next(irand_tuples)
>>> next(selections)
>>> list(selections)
[6, 3, 8, 2, 2]

As others have mentioned if you need more security then set the appropriate select function:

>>> from random import SystemRandom
>>> rand_strs = irand_seqs(select=SystemRandom().choice)

the default selector is choice which may select the same symbol multiple times for each chunk, if instead you’d want the same member selected at most once for each chunk then, one possible usage:

>>> from random import sample
>>> irand_samples = irand_seqs(xrange(10), length=1, join=next, select=lambda pool: sample(pool, 6))
>>> next(irand_samples)
[0, 9, 2, 3, 1, 6]

we use sample as our selector, to do the complete selection, so the chunks are actually length 1, and to join we simply call next which fetches the next completely generated chunk, granted this example seems a bit cumbersome and it is …

Answered By: Samy Vilar
>>> import string 
>>> import random

the following logic still generates 6 character random sample

>>> print ''.join(random.sample((string.ascii_uppercase+string.digits),6))

No need to multiply by 6

>>> print ''.join(random.sample((string.ascii_uppercase+string.digits)*6,6))

Answered By: user128956
import uuid
lowercase_str = uuid.uuid4().hex  

lowercase_str is a random value like 'cea8b32e00934aaea8c005a35d85a5c0'

uppercase_str = lowercase_str.upper()

uppercase_str is 'CEA8B32E00934AAEA8C005A35D85A5C0'

Answered By: Savad KP

(1) This will give you all caps and numbers:

import string, random
for x in range(8):
    if random.choice([1,2]) == 1:
        passkey += passkey.join(random.choice(string.ascii_uppercase))
        passkey += passkey.join(random.choice(string.digits))
print passkey 

(2) If you later want to include lowercase letters in your key, then this will also work:

import string, random
for x in range(8):
    if random.choice([1,2]) == 1:
        passkey += passkey.join(random.choice(string.ascii_letters))
        passkey += passkey.join(random.choice(string.digits))
print passkey  
Answered By: Jacob
import string
from random import *
characters = string.ascii_letters + string.punctuation  + string.digits
password =  "".join(choice(characters) for x in range(randint(8, 16)))
print password
Answered By: Natasha

Use Numpy’s random.choice() function

import numpy as np
import string        

if __name__ == '__main__':
    length = 16
    a = np.random.choice(list(string.ascii_uppercase + string.digits), length)                

Documentation is here http://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.random.choice.html

Answered By: Mudit Jain

From Python 3.6 on you should use the secrets module if you need it to be cryptographically secure instead of the random module (otherwise this answer is identical to the one of @Ignacio Vazquez-Abrams):

from secrets import choice
import string

''.join([choice(string.ascii_uppercase + string.digits) for _ in range(N)])

One additional note: a list-comprehension is faster in the case of str.join than using a generator expression!

Answered By: MSeifert

this is a take on Anurag Uniyal ‘s response and something that i was working on myself.

import random
import string

oneFile = open('‪Numbers.txt', 'w')
userInput = 0
key_count = 0
value_count = 0
chars = string.ascii_uppercase + string.digits + string.punctuation

for userInput in range(int(input('How many 12 digit keys do you want?'))):
    while key_count <= userInput:
        key_count += 1
        number = random.randint(1, 999)
        key = number

        text = str(key) + ": " + str(''.join(random.sample(chars*6, 12)))
        oneFile.write(text + "n")
Answered By: Stopit Donk

I found this to be simpler and cleaner.

str_Key           = ""
str_FullKey       = "" 
str_CharacterPool = "01234ABCDEFfghij~>()"
for int_I in range(64): 
    str_Key = random.choice(str_CharacterPool) 
    str_FullKey = str_FullKey + str_Key 

Just change the 64 to vary the length, vary the CharacterPool to do alpha only alpha numeric or numeric only or strange characters or whatever you want.

Answered By: M T Head

I would like to suggest you next option:

import crypt
n = 10
crypt.crypt("any sring").replace('/', '').replace('.', '').upper()[-n:-1]

Paranoic mode:

import uuid
import crypt
n = 10
crypt.crypt(str(uuid.uuid4())).replace('/', '').replace('.', '').upper()[-n:-1]
Answered By: mrvol

Two methods :

import random, math

def randStr_1(chars:str, length:int) -> str:
    chars *= math.ceil(length / len(chars))
    chars = letters[0:length]
    chars = list(chars)

    return ''.join(chars)

def randStr_2(chars:str, length:int) -> str:
    return ''.join(random.choice(chars) for i in range(chars))

Benchmark :

from timeit import timeit

setup = """
import os, subprocess, time, string, random, math

def randStr_1(letters:str, length:int) -> str:
    letters *= math.ceil(length / len(letters))
    letters = letters[0:length]
    letters = list(letters)
    return ''.join(letters)

def randStr_2(letters:str, length:int) -> str:
    return ''.join(random.choice(letters) for i in range(length))

print('Method 1 vs Method 2', ', run 10 times each.')

for length in [100,1000,10000,50000,100000,500000,1000000]:
    print(length, 'characters:')

    eff1 = timeit("randStr_1(string.ascii_letters, {})".format(length), setup=setup, number=10)
    eff2 = timeit("randStr_2(string.ascii_letters, {})".format(length), setup=setup, number=10)
    print('t{}s : {}s'.format(round(eff1, 6), round(eff2, 6)))
    print('tratio = {} : {}n'.format(eff1/eff1, round(eff2/eff1, 2)))

Output :

Method 1 vs Method 2 , run 10 times each.

100 characters:
    0.001411s : 0.00179s
    ratio = 1.0 : 1.27

1000 characters:
    0.013857s : 0.017603s
    ratio = 1.0 : 1.27

10000 characters:
    0.13426s : 0.151169s
    ratio = 1.0 : 1.13

50000 characters:
    0.709403s : 0.855136s
    ratio = 1.0 : 1.21

100000 characters:
    1.360735s : 1.674584s
    ratio = 1.0 : 1.23

500000 characters:
    6.754923s : 7.160508s
    ratio = 1.0 : 1.06

1000000 characters:
    11.232965s : 14.223914s
    ratio = 1.0 : 1.27

The performance of first method is better.

Answered By: Bi Ao
import random
list  =[r'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','s','0','1','2','3','4','5','6','7','8','9','0']

    for i in range(1,128):

Here length of string can be changed in for loop i.e for i in range(1,length)
It is simple algorithm which is easy to understand. it uses list so you can discard characters that you do not need.

Answered By: lawlie8

I have gone though almost all of the answers but none of them looks easier. I would suggest you to try the passgen library which is generally used to create random passwords.

You can generate random strings of your choice of length, punctuation, digits, letters and case.

Here’s the code for your case:

from passgen import passgen
string_length = int(input())
random_string = passgen(length=string_length, punctuation=False, digits=True, letters=True, case='upper')
Answered By: Underoos

Sometimes 0 (zero) & O (letter O) can be confusing. So I use

import uuid
Answered By: Aseem

Generate random 16-byte ID containig letters, digits, ‘_’ and ‘-‘


Answered By: socketpair
import string, random
lower = string.ascii_lowercase
upper = string.ascii_uppercase
digits = string.digits
special = '!"£$%^&*.,@#/?'

def rand_pass(l=4, u=4, d=4, s=4):
    p = []
    [p.append(random.choice(lower)) for x in range(l)]
    [p.append(random.choice(upper)) for x in range(u)]
    [p.append(random.choice(digits)) for x in range(d)]
    [p.append(random.choice(special)) for x in range(s)]
    return "".join(p)

# @5U,@A4yIZvnp%51
Answered By: Pedro Lobito

I was looking at the different answers and took time to read the documentation of secrets

The secrets module is used for generating cryptographically strong random numbers suitable for managing data such as passwords, account authentication, security tokens, and related secrets.

In particularly, secrets should be used in preference to the default pseudo-random number generator in the random module, which is designed for modelling and simulation, not security or cryptography.

Looking more into what it has to offer I found a very handy function if you want to mimic an ID like Google Drive IDs:

Return a random URL-safe text string, containing nbytes random bytes. The text is Base64 encoded, so on average each byte results in approximately 1.3 characters. If nbytes is None or not supplied, a reasonable default is used.

Use it the following way:

import secrets
import math

def id_generator():
    id = secrets.token_urlsafe(math.floor(32 / 1.3))
    return id


Output a 32 characters length id:


I know this is slightly different from the OP’s question but I expect that it would still be helpful to many who were looking for the same use-case that I was looking for.

Answered By: Antonin GAVREL

If you want an easy-to-use but highly customisable key generator, use key-generator pypi package.

Here is the GitHub repo where you can find the complete documentation.

You can customise it to give a string jist like you want with many more options. Here’s an example:

from key_generator.key_generator import generate

custom_key = generate(2, ['-', ':'], 3, 10, type_of_value = 'char', capital = 'mix', seed = 17).get_key()
print(custom_key)  # ZLFdHXIUe-ekwJCu

Hope this helps 🙂

Disclaimer: This uses the key-generator library which I made.

Answered By: Sahith Kurapati

you can now use a new library (python >= 3.6) chancepy here

from chancepy import Chance

random_string = Chance.string(length=10, pool="someLettersAndNumbers123")
Answered By: Mr. Nun.

None of the answers so far guarantee presence of certain categories of characters like upper, lower, digits etc; so other answers may result in passwords that do not have digits, etc. Surprised that such a function is not part of standard lib. Here is what I use:

def random_password(*, nchars = 7, min_nupper = 3, ndigits = 3, nspecial = 3, special=string.punctuation):
    letters = random.choices(string.ascii_lowercase, k=nchars)
    letters_upper = random.choices(string.ascii_uppercase, k=min_nupper)
    digits = random.choices(string.digits, k=ndigits)
    specials = random.choices(special, k=nspecial)

    password_chars = letters + letters_upper + digits + specials

    return ''.join(password_chars)
Answered By: Oliver

I used this method to generate random string of length n from a -> z

import random
s = ''.join(random.choice([chr(i) for i in range(ord('a'),ord('z'))]) for _ in range(10))

Answered By: Nishan

Security Oriented Approach

Our recommendation for anything security related is to avoid "rolling you own" and to use the secrets module which is specifically vetted for security.

This is from the best practices section of the docs:

import string
import secrets
alphabet = string.ascii_letters + string.digits
password = ''.join(secrets.choice(alphabet) for i in range(8))

Since you specifically asked for uppercase letters, you can either substitute ascii_uppercase for ascii_letters, or just uppercase the password with:

password = password.upper()

Standard Approach Not Aiming for Security

The canonical approach to this problem (as specified) uses the choices() function in the random module:

>>> from random import choices
>>> from string import ascii_uppercase, digits

>>> population = ascii_uppercase + digits
>>> str.join('', choices(population, k=6))
Answered By: Raymond Hettinger

A random generator function without duplicates using a set to store values which have been generated before. Note this will cost some memory with very large strings or amounts and it probably will slow down a bit. The generator will stop at a given amount or when the maximum possible combinations are reached.


#!/usr/bin/env python

from typing import Generator
from random import SystemRandom as RND
from string import ascii_uppercase, digits

def string_generator(size: int = 1, amount: int = 1) -> Generator[str, None, None]:
    Return x random strings of a fixed length.

    :param size: string length, defaults to 1
    :type size: int, optional
    :param amount: amount of random strings to generate, defaults to 1
    :type amount: int, optional
    :yield: Yield composed random string if unique
    :rtype: Generator[str, None, None]
    CHARS = list(ascii_uppercase + digits)
    LIMIT = len(CHARS) ** size
    count, check, string = 0, set(), ''
    while LIMIT > count < amount:
        string = ''.join(RND().choices(CHARS, k=size))
        if string not in check:
            yield string
            count += 1

for my_count, my_string in enumerate(string_generator(6, 20)):
    print(my_count, my_string)


0 RS9N3P
3 96FF97
4 38JJZ3
5 Q3214A
7 QMT05E
8 X1ZFP0
9 MZF442
10 10L9AZ
12 S7PA43
14 YX7Y0G
19 HQF01Q

If your string always needs to contain letters and numbers, use this:

#!/usr/bin/env python

from typing import Generator
from random import SystemRandom as RND
from string import ascii_uppercase, digits

def string_generator(size: int = 2, amount: int = 1) -> Generator[str, None, None]:
    Return x random strings of a fixed length.

    :param size: string length, defaults to 1
    :type size: int, optional
    :param amount: amount of random strings to generate, defaults to 1
    :type amount: int, optional
    :yield: Yield composed random string if unique
    :rtype: Generator[str, None, None]
    if size < 2:
            'Since the string must contain letters and numbers,',
            'its size must be at least two characters',
    CHARS = list(ascii_uppercase + digits)
    LIMIT = len(CHARS) ** size
    count, check, string = 0, set(), ''
    while LIMIT > count < amount:
        while 1:
            string = ''.join(RND().choices(CHARS, k=size))
            if not (string.isalpha() or string.isnumeric()):
        if string not in check:
            yield string
            count += 1

for my_count, my_string in enumerate(string_generator(6, 20)):
    print(my_count, my_string)


1 3W7AI1
2 8QMO10
5 50X2CC
6 1LI8WA
7 39CKET
8 R4IM0E
10 65CBIU
13 06VFCW
14 VE9MJ8
16 Y5ROJ0
18 TQ709S
Answered By: FifthAxiom
Categories: questions Tags: , ,
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