How can I create a random number that is cryptographically secure in python?


I’m making a project in python and I would like to create a random number that is cryptographically secure, How can I do that? I have read online that the numbers generated by the regular randomizer are not cryptographically secure, and that the function os.urandom(n) returns me a string, and not a number.

Asked By: user2835118



You can get a list of random numbers by just applying ord function over the bytes returned by os.urandom, like this

>>> import os
>>> os.urandom(10)
>>> type(os.urandom(10))
<type 'str'>
>>> map(ord, os.urandom(10))
[65, 120, 218, 135, 66, 134, 141, 140, 178, 25]

Quoting os.urandom documentation,

Return a string of n random bytes suitable for cryptographic use.

This function returns random bytes from an OS-specific randomness source. The returned data should be unpredictable enough for cryptographic applications, though its exact quality depends on the OS implementation. On a UNIX-like system this will query /dev/urandom, and on Windows it will use CryptGenRandom().

Answered By: thefourtheye

Since you want to generate integers in some specific range, it’s a lot easier to use the random.SystemRandom class instead. Creating an instance of that class gives you an object that supports all the methods of the random module, but using os.urandom() under the covers. Examples:

>>> from random import SystemRandom
>>> cryptogen = SystemRandom()
>>> [cryptogen.randrange(3) for i in range(20)] # random ints in range(3)
[2, 2, 2, 2, 1, 2, 1, 2, 1, 0, 0, 1, 1, 0, 0, 2, 0, 0, 0, 0]
>>> [cryptogen.random() for i in range(3)]  # random floats in [0., 1.)
[0.2710009745425236, 0.016722063038868695, 0.8207742461236148]

Etc. Using urandom() directly, you have to invent your own algorithms for converting the random bytes it produces to the results you want. Don’t do that 😉 SystemRandom does it for you.

Note this part of the docs:

class random.SystemRandom([seed])

Class that uses the os.urandom() function for generating random numbers from sources provided by the operating system. Not available on all systems. Does not rely on software state and sequences are not reproducible. Accordingly, the seed() and jumpahead() methods have no effect and are ignored. The getstate() and setstate() methods raise NotImplementedError if called.

Answered By: Tim Peters

If you want an n-bit random number, under Python 2.4+, the easiest method I’ve found is

import random

Note that SystemRandom uses os.urandom(), so the result of this method is only as good as your system’s urandom() implementation.

Answered By: jjlin

Python 3.6 introduces a new secrets module, which “provides access to the most secure source of randomness that your operating system provides.” In order to generate some cryptographically secure numbers, you can call secrets.randbelow().


which will return a number between 0 and n.

Answered By: Cody Piersall

To generate a cryptographically secure pseudorandom integer, you can use the following code:


Where n is an integer and, the larger n is, the larger the integer generated is.

You will have to import os and binascii first.

The result of this code can vary by platform.

Answered By: TheGamePlayer 40
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