Python：从数组生成私钥？

我对python和密码学都是新手。我想从一个变量生成PrivateKey。在寻找了这么多选项后，我才知道我只能从

Random

函数生成密钥对

这是我的变量

a = [[[3, 1, 85, 33, 0, 0, 255, 254, 255, 254, 255, 254, 255, 254, 255, 6, 248, 0, 240, 0, 224, 0, 192, 0, 192, 0, 128, 0, 128, 0, 128, 0, 128, 0, 128, 0, 128, 0, 192, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 108, 181, 11, 254, 92, 186, 205, 94, 27, 192, 36, 126, 26, 164, 13, 159, 114, 59, 162, 95, 50, 189, 78, 95, 99, 24, 140, 124, 72, 165, 204, 156, 114, 40, 99, 28, 57, 25, 141, 122, 107, 39, 226, 122, 24, 169, 35, 186, 93, 39, 162, 242, 24, 185, 229, 120, 38, 169, 205, 57, 31, 184, 141, 153, 78, 152, 12, 86, 92, 152, 13, 22, 77, 155, 226, 182, 66, 163, 163, 16, 89, 157, 227, 180, 67, 160, 204, 142, 101, 165, 226, 244, 45, 167, 99, 20, 117, 157, 76, 44, 38, 181, 225, 44, 107, 32, 75, 179, 110, 32, 163, 16, 38, 165, 101, 48, 32, 43, 14, 76, 38, 50, 204, 205, 50, 163, 227, 172, 38, 46, 98, 108, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [3, 1, 101, 34, 0, 0, 255, 254, 255, 14, 248, 2, 240, 2, 224, 0, 192, 0, 192, 0, 128, 0, 128, 0, 128, 0, 128, 0, 192, 0, 192, 0, 224, 0, 224, 0, 240, 0, 248, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 36, 17, 229, 30, 78, 19, 140, 30, 99, 21, 226, 190, 113, 22, 97, 190, 30, 152, 36, 94, 115, 162, 75, 222, 21, 35, 228, 126, 119, 168, 97, 222, 99, 40, 140, 222, 32, 52, 14, 222, 108, 53, 204, 94, 67, 55, 78, 222, 76, 193, 100, 62, 57, 172, 142, 31, 92, 181, 77, 191, 47, 55, 101, 31, 119, 184, 75, 63, 75, 11, 22, 92, 108, 15, 227, 156, 44, 157, 34, 28, 25, 158, 13, 220, 32, 33, 228, 92, 35, 165, 141, 220, 44, 152, 140, 189, 100, 142, 139, 218, 35, 156, 205, 154, 32, 169, 228, 90, 95, 11, 140, 18, 94, 141, 226, 146, 40, 170, 14, 82, 38, 172, 100, 210, 51, 149, 33, 208, 70, 13, 162, 215, 44, 147, 228, 180, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]]

所以我的问题是

是否可以从序列/数据生成私钥

如果是，有什么建议可以让我继续吗

最后，我应该使用哪种算法（RSA/ECC）

不，仅从序列/数据生成私钥是不可能的。 密钥是使用使用系统熵种子的随机数生成器（RNG）或伪随机数生成器（PRNG）生成的，这使得攻击者更难猜测

以RSA为例，正如您所提到的，您可以从随机函数创建私钥，也可以通过手动设置变量来控制私钥的生成。您可以找到使用'PyCryptodome'包的'Crypto.PublicKey.RSA.construct（）函数手动决定的变量生成私钥的方法。为此，您应该对RSA有很好的了解

可以从对称密码中的用户输入生成密钥'Scrypt'和'Argon2'是帮助您实现这一目标的两个软件包。必须首先将输入转换为二进制。输入可以是字符串或数字。下面是一个使用“Scrypt”包的简单示例。你可以找到更多的细节

算法取决于您的应用程序。尽管如此，如今ECC密码被认为比RSA密码更安全

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它是否仅在Python3.6或更高版本中可用@Asiq KSPyCryptodome和Scrypt的支持始于Python3.4，但它不适用于Python3.5。另外，我是否需要安装

secrets

库，因为我没有找到它？如果您使用的是Linux，请确保使用pip3安装包，并键入以下“sudo apt install-y libssl dev”，然后再次尝试安装scrypt。如果您不使用Linux，请告诉我。”“secrets”是默认的python模块，因此您无需安装它。只要在谷歌上搜索python模块，你就会找到它。我在Raspberry Pi3Maybe上使用Raspbian。这些数据应该代表什么？用这种方式表示一个键不是不可能的，但这会很奇怪，如果你有一个这种格式的键，你可能做错了什么。在这里讨论它不是一个好的选择。请告诉我您的邮箱地址以便进一步讨论。vikashsharma2601@yahoo.com

import scrypt, secrets

password = b'not a number'
salt = secrets.token_bytes(32)

scrypt_key = scrypt.hash(password, salt, N=16384, r=8, p=1, 32)

print('Salt: ', salt)
print('Key: ', scrypt_key)
Output:

Salt:  b'\xdbS\x1e\xa2\x81e\xd3\x948p\xc3lmk\xd6\x8b\xb94\x1c\xd5A/\xa5gZ\xb1\xc2\x15\x99\x9d\xc8\xb8'

Key:  b'J\xc1\xc1"\xfd\x05\xfb\x14J\x96\xea\xe3\x1d\xa6\xbb\x01\xf7sj\x87\xf9\x18%\x00YK\x1f\xe8\xc8\x8d\xff%'