将字符串转换为随机但确定性可重复的均匀概率 [英] Convert string to random but deterministically repeatable uniform probability

问题描述

如何转换字符串，例如用户ID加盐，看起来很随机，但实际上在半开放范围[0.0，1.0)中具有确定性可重复的均匀概率?此表示表示输出≥0.0且< 1.0.无论输入分布如何，输出分布都必须是均匀的.例如，如果输入字符串为"a3b2Foobar"，则输出概率可以重复为0.40341504.

跨语言和跨平台算法的可重复性是可取的.我倾向于使用哈希函数，除非有更好的方法.这是我所拥有的:

>>> in_str = 'a3b2Foobar'
>>> (int(hashlib.sha256(in_str.encode()).hexdigest(), 16) % 1e8) / 1e8
0.40341504

我正在使用最新的稳定Python3.请注意，这个问题与解决方案

使用哈希

密码哈希可能是[0，MAX_HASH]范围内的均匀分布的整数.因此，通过将其除以MAX_HASH + 1，可以将其缩放为[0，1)范围内的浮点数.

import hashlib

Hash = hashlib.sha512
MAX_HASH_PLUS_ONE = 2**(Hash().digest_size * 8)

def str_to_probability(in_str):
    """Return a reproducible uniformly random float in the interval [0, 1) for the given string."""
    seed = in_str.encode()
    hash_digest = Hash(seed).digest()
    hash_int = int.from_bytes(hash_digest, 'big')  # Uses explicit byteorder for system-agnostic reproducibility
    return hash_int / MAX_HASH_PLUS_ONE  # Float division

>>> str_to_probability('a3b2Foobar')
0.3659629991207491

注意:

• 内置 hash 不能使用该方法，因为它可以保留输入的 分布，例如使用hash(123).另外，重新启动Python时，它可以返回不同的值，例如与hash('123').
• 因为浮点数就足够了，所以不必使用模.

使用随机

random 模块可以与in_str一起使用，同时解决了有关线程安全和连续性的问题.

使用这种方法，不仅要考虑跨语言的可重复性，而且还要考虑多个未来版本的Python的可重复性.因此不建议这样做.

import random

def str_to_probability(in_str):
    """Return a reproducible uniformly random float in the interval [0, 1) for the given seed."""
    return random.Random(in_str).random()

>>> str_to_probability('a3b2Foobar')
0.4662507245848473

How do I convert a string, e.g. a user ID plus salt, to a random looking but actually a deterministically repeatable uniform probability in the semi-open range [0.0, 1.0)? This means that the output is ≥ 0.0 and < 1.0. The output distribution must be uniform irrespective of the input distribution. For example, if the input string is 'a3b2Foobar', the output probability could repeatably be 0.40341504.

Cross-language and cross-platform algorithmic reproducibility is desirable. I'm inclined to use a hash function unless there is a better way. Here is what I have:

>>> in_str = 'a3b2Foobar'
>>> (int(hashlib.sha256(in_str.encode()).hexdigest(), 16) % 1e8) / 1e8
0.40341504

I'm using the latest stable Python 3. Please note that this question is similar but not exactly identical to the related question to convert an integer to a random but deterministically repeatable choice.

解决方案

Using hash

A cryptographic hash is assumably a uniformly distributed integer in the range [0, MAX_HASH]. Accordingly, it can be scaled to a floating-point number in the range [0, 1) by dividing it by MAX_HASH + 1.

import hashlib

Hash = hashlib.sha512
MAX_HASH_PLUS_ONE = 2**(Hash().digest_size * 8)

def str_to_probability(in_str):
    """Return a reproducible uniformly random float in the interval [0, 1) for the given string."""
    seed = in_str.encode()
    hash_digest = Hash(seed).digest()
    hash_int = int.from_bytes(hash_digest, 'big')  # Uses explicit byteorder for system-agnostic reproducibility
    return hash_int / MAX_HASH_PLUS_ONE  # Float division

>>> str_to_probability('a3b2Foobar')
0.3659629991207491

Notes:

• The built-in hash method must not be used because it can preserve the input's distribution, e.g. with hash(123). Alternatively, it can return values that differ when Python is restarted, e.g. with hash('123').
• Using modulo is not necessary as float division is sufficient.

Using random

The random module can be used with in_str as its seed, while addressing concerns surrounding both thread safety and continuity.

With this approach, not only is cross-language reproducibility a concern, but reproducibility across multiple future versions of Python could also be a concern. It is therefore not recommended.

import random

def str_to_probability(in_str):
    """Return a reproducible uniformly random float in the interval [0, 1) for the given seed."""
    return random.Random(in_str).random()

>>> str_to_probability('a3b2Foobar')
0.4662507245848473

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