Python 3 os.urandom

Question

Where can I find a complete tutorial or documentation on os.urandom? I need to get get a random int to choose a char from a string of 80 characters.

Why did you accept an answer that doesn't answer the title of your question. Either change your accepted answer or change your question. If you want to know about os.random and its use then you should correct your accepted answer. — Charlie Parker, Jul 03 '16 at 21:09

score 21 · Accepted Answer · edited Nov 04 '12 at 10:53

21

If you just need a random integer, you can use random.randint(a, b) from the random module.

If you need it for crypto purposes, use random.SystemRandom().randint(a, b), which makes use of os.urandom().

Example

import random

r = random.SystemRandom()
s = "some string"
print(r.choice(s)) # print random character from the string
print(s[r.randrange(len(s))]) # same

edited Nov 04 '12 at 10:53

jfs

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answered Nov 04 '12 at 07:59

Tim

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`random.choice` would also probably be a reasonable option. – Danica Nov 04 '12 at 08:00
`random.choice` works too, depending on what he ultimately wants to do with the random character. – Tim Nov 04 '12 at 08:01
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random is a non-cryptographic PRNG, i need a cryptographic quality random – John Nov 04 '12 at 08:50
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You can use `random.SystemRandom` instead of `random`. It makes use of `os.urandom()` and provides all the same functionality as `random`. – Tim Nov 04 '12 at 08:57
Is there no way to get a cryptographically secure integer form the OS though? – Charlie Parker Jul 03 '16 at 21:01
To answer my own question, yes. os.urandom does that. https://docs.python.org/2/library/os.html – Charlie Parker Jul 03 '16 at 21:10
@Tim you can get the seed nor anything from `random.SystemRandom([seed])`, what is the point of it? – Charlie Parker Jul 03 '16 at 21:13

score 8 · Answer 2 · edited Nov 19 '20 at 09:14

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Might not exactly be on topic, but I want to help those coming here from a search engine. To convert os.urandom to an integer I'm using this:

 import os

 rand = int(int(str(os.urandom(4), encoding="UTF-8")).encode('hex'), 16)
 # You can then 'cycle' it against the length.
 rand_char = chars_list[rand % 80] # or maybe '% len(chars_list)'

Note: The range of the index here is up to that of a 4-byte integer. If you want more, change the 4 to a greater value.

The idea was taken from here: https://pythonadventures.wordpress.com/2013/10/04/generate-a-192-bit-random-number/

edited Nov 19 '20 at 09:14

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answered Mar 07 '15 at 12:53

Vladius

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I am assuming that 16 is a argument to the built in function int that tells python that the given string is in base 16 (hexadecial). – Charlie Parker Jul 03 '16 at 21:05
Why did you give urandom 4 and not any other number? I know that 4 indicates a string of n random bytes suitable for cryptographic use, however, I keep seeing examples on the web where 4 is the the chosen number instead of say, 32 or 64. Why didn't you choose 64 say? More numbers more randomness/harder to predict, right? – Charlie Parker Jul 03 '16 at 21:07
Do you might explaining some of the details of your code? I have written above the details I could find out by reading the docs but wanted to know if there are any caveats (and how to fix them) or some important comments to give the readers when trying to re-use your code to optimize randomness and/or security. Thanks. – Charlie Parker Jul 03 '16 at 21:08
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in python3, you need to use `rand=int.from_bytes(os.urandom(4), sys.byteorder)` – eafit Jul 30 '16 at 20:30

Python 3 os.urandom

2 Answers2

Example

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