Question

对于一个项目，我需要一种创建数千个随机字符串同时保持较低冲突的方法。我正在寻找他们只有12个字符长和大写。有什么建议吗？

Answer 1

<强> CODE：

from random import choice
from string import ascii_uppercase

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

<强>输出：

5个例子：

QPUPZVVHUNSN
EFJACZEBYQEB
QBQJJEEOYTZY
EOJUSUEAJEEK
QWRWLIWDTDBD

修改

如果您只需要数字，请使用digits常量，而不是ascii_uppercase模块中的string常量。

3个例子：

229945986931 867348810313 618228923380

Answer 2

按Django，您可以在get_random_string模块中使用django.utils.crypto功能。

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

示例：

get_random_string()
u'ngccjtxvvmr9'

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

但如果你不想Django，here是它的独立代码：

代码：

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))

Answer 3

可以制造发电机：

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

然后只需在需要时从生成器中拉出来......在需要时使用next(random_gen)，或者使用random_200 = list(islice(random_gen, 200)) ...

Answer 4

此函数生成具有指定长度的随机字符串大写字母

例如：长度= 6，将生成以下随机序列模式

<强> 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

Answer 5

对于加密强的伪随机字节，您可以使用围绕OpenSSL的pyOpenSSL包装。

它提供bytes函数来收集 bytes 的伪随机序列。

from OpenSSL import rand

b = rand.bytes(7)

_{BTW，12个大写字母比56位熵略多一点。您只需要读取7个字节。}

Answer 6

#!/bin/python3
import random
import string
def f(n: int) -> str:
        bytes(random.choices(string.ascii_uppercase.encode('ascii'),k=n)).decode('ascii')

对于很大的n，运行速度更快。避免str连接。

在Python中生成特定长度的随机字符串的最佳方法是什么？

6 个答案: