What's the best way to generate random strings of a specific length in Python?

Question:

For a project, I need a method of creating thousands of random strings while keeping collisions low. I’m looking for them to be only 12 characters long and uppercase only. Any suggestions?

Asked By: Brandon

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Answers:

CODE:

from random import choice
from string import ascii_uppercase

print(''.join(choice(ascii_uppercase) for i in range(12)))

OUTPUT:

5 examples:

QPUPZVVHUNSN
EFJACZEBYQEB
QBQJJEEOYTZY
EOJUSUEAJEEK
QWRWLIWDTDBD

EDIT:

If you need only digits, use the digits constant instead of the ascii_uppercase one from the string module.

3 examples:

229945986931
867348810313
618228923380
Answered By: Peter Varo

Could make a generator:

from string import ascii_uppercase
import random
from itertools import islice

def random_chars(size, chars=ascii_uppercase):
    selection = iter(lambda: random.choice(chars), object())
    while True:
        yield ''.join(islice(selection, size))

random_gen = random_chars(12)
print next(random_gen)
# LEQIITOSJZOQ
print next(random_gen)
# PXUYJTOTHWPJ

Then just pull from the generator when they’re needed… Either using next(random_gen) when you need them, or use random_200 = list(islice(random_gen, 200)) for instance…

Answered By: Jon Clements

For cryptographically strong pseudo-random bytes you might use the pyOpenSSL wrapper around OpenSSL.

It provides the bytes function to gather a pseudo-random sequences of bytes.

from OpenSSL import rand

b = rand.bytes(7)

BTW, 12 uppercase letters is a little bit more that 56 bits of entropy. You will only to have to read 7 bytes.

Answered By: Sylvain Leroux

By Django, you can use get_random_string function in django.utils.crypto module.

get_random_string(length=12,
    allowed_chars=u'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789')
    Returns a securely generated random string.

    The default length of 12 with the a-z, A-Z, 0-9 character set returns
    a 71-bit value. log_2((26+26+10)^12) =~ 71 bits

Example:

get_random_string()
u'ngccjtxvvmr9'

get_random_string(4, allowed_chars='bqDE56')
u'DDD6'

But if you don’t want to have Django, here is independent code of it:

Code:

import random
import hashlib
import time

SECRET_KEY = 'PUT A RANDOM KEY WITH 50 CHARACTERS LENGTH HERE !!'

try:
    random = random.SystemRandom()
    using_sysrandom = True
except NotImplementedError:
    import warnings
    warnings.warn('A secure pseudo-random number generator is not available '
                  'on your system. Falling back to Mersenne Twister.')
    using_sysrandom = False


def get_random_string(length=12,
                      allowed_chars='abcdefghijklmnopqrstuvwxyz'
                                    'ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'):
    """
    Returns a securely generated random string.

    The default length of 12 with the a-z, A-Z, 0-9 character set returns
    a 71-bit value. log_2((26+26+10)^12) =~ 71 bits
    """
    if not using_sysrandom:
        # This is ugly, and a hack, but it makes things better than
        # the alternative of predictability. This re-seeds the PRNG
        # using a value that is hard for an attacker to predict, every
        # time a random string is required. This may change the
        # properties of the chosen random sequence slightly, but this
        # is better than absolute predictability.
        random.seed(
            hashlib.sha256(
                ("%s%s%s" % (
                    random.getstate(),
                    time.time(),
                    SECRET_KEY)).encode('utf-8')
            ).digest())
    return ''.join(random.choice(allowed_chars) for i in range(length))
Answered By: Omid Raha

This function generates random string of UPPERCASE letters with the specified length,

eg: length = 6, will generate the following random sequence pattern

YLNYVQ

    import random as r

    def generate_random_string(length):
        random_string = ''
        random_str_seq = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
        for i in range(0,length):
            if i % length == 0 and i != 0:
                random_string += '-'
            random_string += str(random_str_seq[r.randint(0, len(random_str_seq) - 1)])
        return random_string
Answered By: Manoj Selvin
#!/bin/python3
import random
import string
def f(n: int) -> str:
        bytes(random.choices(string.ascii_uppercase.encode('ascii'),k=n)).decode('ascii')

run faster for very big n. avoid str concatenate.

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.

Code:

#!/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:
            check.add(string)
            yield string
            count += 1


for my_count, my_string in enumerate(string_generator(12, 20)):
    print(my_count, my_string)

Output:

0 IESUASWBRHPD
1 JGGO1THKLC9K
2 BW04A5GWBA7K
3 KDQTY72BV1S9
4 FAOL5L28VVMN
5 NLDNNBGHTRTI
6 2RV6TE6BCQ8K
7 B79B8FBPUD07
8 89VXXRHPUN41
9 DFC8QJUY6HRB
10 FXYYDKVQHC5Z
11 57KTZE67RSCU
12 389H1UT7N6CI
13 AKZMN9XITAVB
14 6T9ACH3GDAYG
15 CH8RJUQMTMBE
16 SPQ7E02ZLFD3
17 YD6JFXGIF3YF
18 ZUSA2X6OVNCN
19 JQRH6LR229Y4
Answered By: FifthAxiom
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