迭代所有可能组合的 Python III 生成器 32 字节数据，有没有更快的方法？

如何解决迭代所有可能组合的 Python III 生成器 32 字节数据，有没有更快的方法？

我怎样才能使这个 Python III 代码更有效率？我正在尝试遍历 32 字节的所有可能组合，该脚本以十六进制完成，我对任何数据类型的可能性持开放态度。我似乎找不到比使用生成器然后将整数转换为_bytes(32,'big') 更快的方法。

输出将发送到 hashlib.new("sha256") 函数。我似乎无法找到一种只用字节而不是转换来做到这一点的方法，因为我需要对所有可能的组合进行暴力破解。您是否发现我的概念有任何问题或我遗漏了什么？

感谢您提供的任何帮助。

def standard_noraml_z(z1):
    standard_normal=[[0.00,0.01,0.02,0.03,0.04,0.05,0.06,0.07,0.08,0.09],[0.0,0.0000,0.0040,0.0080,0.0120,0.0160,0.0199,0.0239,0.0279,0.0319,0.0359],[0.1,0.0398,0.0438,0.0478,0.0517,0.0557,0.0596,0.0636,0.0675,0.0714,0.0753],[0.2,0.0793,0.0832,0.0871,0.0910,0.0948,0.0987,0.1026,0.1064,0.1103,0.1141],[0.3,0.1179,0.1217,0.1255,0.1293,0.1331,0.1368,0.1406,0.1443,0.1480,0.1517],[0.4,0.1554,0.1591,0.1628,0.1664,0.1700,0.1736,0.1772,0.1809,0.1844,0.1879],[0.5,0.1915,0.1950,0.1985,0.2019,0.2054,0.2088,0.2123,0.2157,0.2190,0.2224],[0.6,0.2257,0.2291,0.2324,0.2357,0.2389,0.2422,0.2454,0.2486,0.2517,0.2549],[0.7,0.2580,0.2611,0.2642,0.2673,0.2704,0.2734,0.2764,0.2794,0.2823,0.2852],[0.8,0.2881,0.2910,0.2939,0.2967,0.2995,0.3023,0.3051,0.3078,0.3106,0.3133],[0.9,0.3159,0.3186,0.3212,0.3238,0.3264,0.3289,0.3315,0.3340,0.3365,0.3389],[1.0,0.3413,0.3438,0.3461,0.3485,0.3508,0.3531,0.3554,0.3577,0.3599,0.3621],[1.1,0.3643,0.3665,0.3686,0.3708,0.3729,0.3749,0.3770,0.3790,0.3810,0.3830],[1.2,0.3849,0.3869,0.3888,0.3907,0.3925,0.3944,0.3962,0.3980,0.3997,0.4015],[1.3,0.4032,0.4049,0.4066,0.4082,0.4099,0.4115,0.4131,0.4147,0.4162,0.4177],[1.4,0.4192,0.4207,0.4222,0.4236,0.4251,0.4265,0.4279,0.4292,0.4306,0.4319],[1.5,0.4332,0.4345,0.4357,0.4370,0.4382,0.4394,0.4406,0.4418,0.4429,0.4441],[1.6,0.4452,0.4463,0.4474,0.4484,0.4495,0.4505,0.4515,0.4525,0.4535,0.4545],[1.7,0.4554,0.4564,0.4573,0.4582,0.4591,0.4599,0.4608,0.4616,0.4625,0.4633],[1.8,0.4641,0.4649,0.4656,0.4664,0.4671,0.4678,0.4686,0.4693,0.4699,0.4706],[1.9,0.4713,0.4719,0.4726,0.4732,0.4738,0.4744,0.4750,0.4758,0.4761,0.4767],[2.0,0.4772,0.4778,0.4783,0.4788,0.4793,0.4799,0.4803,0.4808,0.4812,0.4817],[2.1,0.4821,0.4826,0.4830,0.4834,0.4838,0.4842,0.4846,0.4850,0.4854,0.4857],[2.2,0.4861,0.4864,0.4868,0.4871,0.4875,0.4878,0.4881,0.4884,0.4887,0.4890],[2.3,0.4893,0.4896,0.4898,0.4901,0.4904,0.4906,0.4909,0.4911,0.4913,0.4916],[2.4,0.4918,0.4920,0.4922,0.4925,0.4927,0.4929,0.4931,0.4932,0.4934,0.4936],[2.5,0.4938,0.4940,0.4941,0.4943,0.4945,0.4946,0.4948,0.4949,0.4951,0.4952],[2.6,0.4953,0.4955,0.4956,0.4957,0.4959,0.4960,0.4961,0.4962,0.4963,0.4964],[2.7,0.4965,0.4966,0.4967,0.4968,0.4969,0.4970,0.4971,0.4972,0.4973,0.4974],[2.8,0.4974,0.4975,0.4976,0.4977,0.4978,0.4979,0.4980,0.4981],[2.9,0.4981,0.4982,0.4983,0.4984,0.4985,0.4986,0.4986],[3.0,0.49865,0.4987,0.4988,0.4989,0.4990]]
    z=str(z1)
    zy=z[1:3]
    zx=z[4]
    y=0
    x=0
    for i in range(len(standard_normal)):
        if standard_normal[i][0]==zy:
            y=i
    for i in range(len(standard_normal[0])):
        if standard_normal[0][i]==zx:
            x=i+1
    return standard_normal[y][x]

print(standard_noraml_z(1.34))

输出：

import hashlib

import binascii

strike_counter = 0

def hex_permutations(digits=16):

    x = 0

    max_ = 16 ** digits

    while x < max_:

        yield x.to_bytes(32,'big') # might be little endian

        x += 1

for permutations in hex_permutations(4): # change the number here for more.

    strike_counter += 1

    #print("strike_counter: ",strike_counter,permutations)

    print("strike counter: ",strike_counter)

else:

    print("strike counter: ",permutations)

    pr = binascii.hexlify(permutations)

    print("permutations hexlify: ",pr)

    print("length: ",len(pr))