将大量数字乘以 random() (Python)答案

【问题标题】：Multiplying a huge number times random() (Python)将大量数字乘以 random() (Python)
【发布时间】：2012-08-23 02:55:47
【问题描述】：

问题：生成较大的二进制字符串（长度 2000+）。赶快行动吧，因为这个 generateRandom() 函数会在算法中被调用 300,000 次。

尝试的解决方案：生成 3 或 4 个二进制数并将它们全部附加 500 次。这非常慢。

对 random.random() 进行一次调用并将其乘以一个巨大的数字。转换为二进制一次即可完成。这适用于较小的数字，但由于二进制字符串必须有一定的长度，因此要转换为二进制的数字必须非常大（2 ** len(binString)）。

当前代码（适用于较小的数字）：

binaryRepresentation = ''

binaryRepresentation += bin(int(random.random() * (2 ** binLength)))[2:].zfill(binLength)

我需要帮助修复的错误： 此调用会抛出一个带有大数字的“long int too large to convert to float”。有没有办法让整个算法更高效，或者让这个大数可以转换为浮点数？

谢谢！

【问题讨论】：

binLength的值是多少？
@brianbuck：任何超过 1023 的都太大了，OP 正在寻找 2000+。
将浮点数乘以某个大数的方法不是一个好策略。大数的位数比浮点数可表示的位数多得多，因此您不会随机化所有位数。如果要使用bin 函数将大整数转换为“位字符串”，请使用random.randint 而不是random.random。只需bin(random.randint(0, 2**binLength)[2:].zfill(binLength)。
@JohnY: 如果getrandbits() 不使用example，randint() 将产生溢出。
@J.F.Sebastian：看来旧版本的 Python 在randint/randrange 中不一定使用getrandbits。所以是的，可以肯定的是，只需使用getrandbits（我忽略了它，因为我自己从来没有这个用例）。

标签： python optimization largenumber

【解决方案1】：

衡量它是否足够快以满足您的目的，“随机性”可能会随着您的称呼而减少：os.urandom(250)。它产生一个二进制字符串，也就是字节。

为避免“long int too large to convert to float”错误，请勿使用浮点数。

如果您需要带有k 随机位的整数而不是二进制字符串：

import random
r = random.SystemRandom()

n = r.getrandbits(2000) # uses os.urandom() under the hood

获取一串“0”和“1”：

k = 2000
binstr = "{:0{}b}".format(r.getrandbits(k), k)

注意：如果不使用getrandbits，则不能将randint/randrange 用于大数字：

import random

class R(random.Random):
    def random(self): # override random to suppress getrandbits usage
        return random.random()

r = R()
r.randrange(2**2000) # -> OverflowError: long int too large to convert to float

b2a_bin

b2a_bin() extension 允许直接从字节字符串创建二进制字符串（“01”），而无需创建中间 Python 整数。它比纯 Python 类似物快 3-20 倍：

def b2a_bin_bin(data):
    return bin(int.from_bytes(data, 'big', signed=False)
               )[2:].zfill(len(data)*8).encode('ascii', 'strict')

def b2a_bin_format(data):
    n = int.from_bytes(data, 'big', signed=False)
    return "{:0{}b}".format(n, len(data)*8).encode('ascii', 'strict')

用法：

>>> import os
>>> from b2a_bin import b2a_bin
>>> b2a_bin.b2a_bin(b'\x0a')
b'00001010'
>>> b2a_bin(os.urandom(5))
b'1001111011000011111001110010000101111010'

【讨论】：

如果我没看错，OP 想要 2000+ bits，而 urandom 返回多个 bytes，所以他只是需要将他所需的位数除以 8。但这绝对是最简单、最随机且可能最快的方法。
urandom 将在/dev/urandom 用完要读取的数据时停止（至少在 linux 上）。
@JohnY：这个问题自相矛盾。第一句“生成大二进制字符串（长度2000+）。”
@Pykler：“u”代表“解锁”/非阻塞。如果没有足够的熵，os.urandom() 应该逐渐使用更多来自 PRNG 的位，因此上面的“随机性可能会减少”。
@J.F.Sebastian：我不完全确定你指的是什么矛盾。围绕术语“二进制字符串”经常存在混淆，因为有些人指的是实际位，而另一些人指的是字符串，其中字符可能是“0”或“1”。可能还有其他我没有想到的情况。但显然，OP 提供了足够的上下文来证明他想要一个字符串（bin 的输出减去前导 '0b' 并根据需要进行填充）。

【解决方案2】：

从 J.F. Sebastian 的答案转为二进制字符串（其中包含 0 和 1 字符的字符串）：

>>> import random
>>> r = random.SystemRandom()
>>> bin(r.getrandbits(2000))[2:].zfill(2000)
'00010110001101100011000011001011011000010001010111000001111001001111010110010110011011110011101101100001111000011111010110100110011011110011111110010111111001110110011011011111011101000010101001001000100010001011011100011101100101010110001101101100100000001001010001101100110001100001101101010010001000010010010101100110100000001001001101000110001011111111111010011111111011001000100011001011011100010100010000010011100100011111011110111101010001001111011010101101010000110110010011000101111110000101001111100111111101101101110001100111011110011001000110110011101001010111100110000111000110011010000000111011001011100000100100111011110111101001110001110010010101100110000111001011100101111110100000100101011000110100111111001100111001000000111010000000101001111011111101111111101101011010010001101101101010011000100101000111000100100011010100111101111010001011110100100011111101101000111011100010000111000111110100001100111010001001010100001010111100001001010110010010100010011010010010011000000011010111110001100001100110101111011100010100100101000101000111110011111011011011111101011001001111010110000101100011111000111010100001000000101111011010011010011101011110000101011011110001111111101100000100000110111100011010110000111110110010100010010111111110010000000100001011110011100001001110110111100001101000011101011011101110001001001110011111111010001000100011010000110011000000000011111011011011101011010010001000101001011111110110000011000000000010101101111100011001010011100010010110111011001001011011010101111011101000001000011110010011111000011001110000100100111010000011100111110101110111010010101110010111100011111001101010010010000001010110011110001110010011011001110011000010010111111100101101100110111100001101100001100000101110000000011111010011000010000000011001101000001001000101001010100001110001011111110100110111000101011101111110010110001111111010100000100011011110011100010000000101001011100000010101101110110001110010111010000111101110100001100101000101010000110111011001001011'
>>> bin(r.getrandbits(2000))[2:].zfill(2000)
'11111011011010000011111101101101001110101011100110100011111011101101111100110001000110110100101010101000110010000101010100011111100111100010000001011011101100011101000001100101000101000010010000001111110101010011001110001001010011000011010100011111110111110010100000111011000000110000100000000110101101101111001101100000010010000100101001111100101010011101011010111110010001111111100101011110001101100111010010101110111000001000100101111011010001111001001010010000011100111101101101111101111010101100001000110011100110010110010101101001011000010101011111010010111000000100101100000100000010010000001000000001110010010100100111001011011111100111111001100000111111110011100001000111111110110001000010010110101100100001000001011110110100000111010101100111111010011111011111011000100101010111111000110111001001100011101101001000100110011001011101100010011010101111000011011111110001010110100100100001010100100100110101100111011110101001001111000010001101001010111111110011110111011111010001110011001000000100000101100001001100101000010011001101001110000101100000110000110101110011000010111110100100100100110010111110011101001000100111110011001101010000101100010011110100100110000111010111001000001101010101000001111001110010111000000101111101000110100101101000100000101100100110111101100010011110111011010111000111111100001110000100111001110001010101111000111100000011111111111110110100011000000111010100111011011100100010110100010110001111010110010100101111010111101110011010110100100010001001001110010110100111010010111001011101100000010011000110011011011010001100010000000010110011101101000111101000011101100001010001010001010111111110101100001110010000001010000000011000000000101111001001100100010110010000000101001010011110110111101111001110001001110111101011111111010101101011101010010111101000000010000101010100000101111010011010001001001101000001001011010110000000000111010001111001110100110000100011110100110110010011000111110100011000110001100001101101001010110001001001001101000011011101001010'
>>> bin(r.getrandbits(2000))[2:].zfill(2000)
'00100010111111011100001001000010100100101000100010101010100001011100101010100000000011111000001001011101010010010000101111111001101101100101011011001011110001001111101011101000101101000001000000110010011001110010101100101000011111001111111011011010001010010000011000000010011001101000100101000001001101001100010100000100100110000010110101011101011111010000010110011001101010010010000111001101101101001001100101111001101110111111111000011000001011011100111010011101001100011101011001110010111101011111010100101101010001011011001001010001111011100100101100100011111010010011110010011101100101001100111011011110011101010101010011001100000100110000001001111001111010011110110110001100110111011011111001110101000100100000011101001100111001100010001101010110000110010101010100011011101101010010100011111100001100100011010111000001011000101110100001111111001001111111101001000000011110110001010101001110110011000010101011010111111011010000011000000111110001110111111110110001110101011000010101001111110111101101010000111000011001111110101000001000011110000101011011001001000100010000110000110011100000100110100101011010011000110001010000101001110101111000110101001010010101000100011010101001110101101111010010101000011111101110110100101000110011010100010110110101010111001010110111011000000010001000110010011000011011110101101000111111101110100100011100000011110100000110111001001101110000011001101101011001000000101001110110011101111111010000000101111011100011010110111110000000001110101101100001001000001010000100101001111110110101000000101101111010010110000111101111011111100001111010001000100110111011010100110111011101011110001001001010100010101011001100100110101101111010011110001010011110000100100011101100101000000001011001010100101010011000011000100001011010010001010100001110101101100000010101011111010011000000000100100000000001011110101011101100101100010110101000111001010000001011000010001111101011111000011100101101110111110000011001000001101010001010000111001100101100111000111000001000000000'

有了这个基准：

import random
import time

def run(n):
    r = random.SystemRandom()
    for i in xrange(n):
        if i%30000 == 0: print i
        bin(r.getrandbits(2000))[2:].zfill(2000)

s = time.time()
run(300000)
e = time.time()
print "Took %.2fs" % (e-s,)

结果是Took 12.32s

仅在没有任何字符串转换的情况下获取随机位（仅r.getrandbits(2000)）需要7.77s，因此如果您能找到一种方法将随机位用作long，那么您将节省一些时间。

使用 os.urandom(250) 重新运行基准测试（无需额外处理）只需要 3.59s，因此这似乎是最快的选择。

【讨论】：

您可以使用"{:0{}b}".format(r.getrandbits(k), k) 生成二进制字符串（带有“0”和“1”的字符串）。
我实际上尝试过，但它最终慢了一两秒 (13.5s)。实际上在python 2.6中格式化字符串不起作用所以我最终不得不做"{0:02000b}".format(r.getrandbits(2000))
在my measurements bin 中也比"".format 稍微快一点。

【解决方案3】：

random.randrange真的太慢了吗？让我们看看它到底有多慢。

import random

word_size = 2048
word_max = 2 ** word_size

def random_bits(n):
    """
    Return a string consisting of `n` zeroes and ones (chosen randomly).
    """
    def words():
        s, m, r = word_size, word_max, n % word_size
        for _ in range(n // s):
            yield bin(random.randrange(m))[2:].zfill(s)
        yield bin(random.randrange(2 ** r))[2:].zfill(r)
    return ''.join(words())

>>> from timeit import Timer
>>> Timer(lambda:random_bits(2000)).timeit(number=300000)
9.680696964263916

10 秒对于选择 6 亿个随机位似乎并不荒谬。因此，也许您可以更多地说明您的速度要求。这真的太慢了吗？

【讨论】：