Random hash in Python

Question

What is the easiest way to generate a random hash (MD5) in Python?

Random as in for anything? Or for an object? If you just want a random MD5, just pick some numbers. — foobarfuzzbizz
– foobarfuzzbizz, Commented Jun 10, 2009 at 16:06
I am renaming files before uploading and want a filename like this: timestamp_randommd5.extension Cheers! — mistero
– mistero, Commented Jun 10, 2009 at 16:15
You could just rename them to timestamp_randomnumber.ext. There really isn't a reason why md5(randomnumber) would be any better than randomnumber itself. — sth
– sth, Commented Jun 11, 2009 at 1:36
best answer for Python 3 is the last one import uuid; uuid.uuid().hex stackoverflow.com/a/20060712/3218806 — maxbellec
– maxbellec, Commented Jul 19, 2016 at 8:39

sth · Accepted Answer · 2023-11-22 13:16:01Z

185

A md5-hash is just a 128-bit value, so if you want a random one:

import random

hash = random.getrandbits(128)

print("hash value: %032x" % hash)

I don't really see the point, though. Maybe you should elaborate why you need this...

edited Nov 22, 2023 at 13:16

answered Jun 10, 2009 at 16:09

sth

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8 Comments

Nicolas Dumazet Over a year ago

+1 for not computing a relatively expensive hash from a random number: this approach is 5x faster.

Jiri Over a year ago

+1 - surely this is better than my answer, can be used also like this: hex(random.getrandbits(128))[2:-1] this gives you same output as md5 hexdigest method.

tzot Over a year ago

the random.seed() call is useless, more or less.

Unknown Over a year ago

I would've used os.urandom because wanting an MD5 hash might mean wanting a secure one.

FogleBird Over a year ago

Here's how to do it with os.urandom: ''.join('%02x' % ord(x) for x in os.urandom(16))

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sebs · Accepted Answer · 2022-08-29 17:11:38Z

170

I think what you are looking for is a universal unique identifier.Then the module UUID in python is what you are looking for.

import uuid
uuid.uuid4().hex

UUID4 gives you a random unique identifier that has the same length as a md5 sum. Hex will represent is as an hex string instead of returning a uuid object.

http://docs.python.org/2/library/uuid.html

https://docs.python.org/3/library/uuid.html

edited Aug 29, 2022 at 17:11

answered Nov 19, 2013 at 0:18

sebs

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Comments

Nick T · Accepted Answer · 2017-07-24 19:41:00Z

98

The secrets module was added in Python 3.6+. It provides cryptographically secure random values with a single call. The functions take an optional nbytes argument, default is 32 (bytes * 8 bits = 256-bit tokens). MD5 has 128-bit hashes, so provide 16 for "MD5-like" tokens.

>>> import secrets

>>> secrets.token_hex(nbytes=16)
'17adbcf543e851aa9216acc9d7206b96'

>>> secrets.token_urlsafe(16)
'X7NYIolv893DXLunTzeTIQ'

>>> secrets.token_bytes(128 // 8)
b'\x0b\xdcA\xc0.\x0e\x87\x9b`\x93\\Ev\x1a|u'

edited Jul 24, 2017 at 19:41

answered Jul 17, 2017 at 22:36

Nick T

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Comments

Zitrax · Accepted Answer · 2016-07-16 13:59:15Z

51

This works for both python 2.x and 3.x

import os
import binascii
print(binascii.hexlify(os.urandom(16)))
'4a4d443679ed46f7514ad6dbe3733c3d'

edited Jul 16, 2016 at 13:59

Zitrax

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answered Mar 22, 2012 at 4:19

Buttons840

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2 Comments

Jake Over a year ago

thanks. this is the best way to make a random hash key.

Clay Over a year ago

works for 2.x and 3.x: binascii.hexlify(os.urandom(16))

Matthew Taylor · Accepted Answer · 2012-01-25 22:10:22Z

26

Yet another approach. You won't have to format an int to get it.

import random
import string

def random_string(length):
    pool = string.letters + string.digits
    return ''.join(random.choice(pool) for i in xrange(length))

Gives you flexibility on the length of the string.

>>> random_string(64)
'XTgDkdxHK7seEbNDDUim9gUBFiheRLRgg7HyP18j6BZU5Sa7AXiCHP1NEIxuL2s0'

answered Jan 25, 2012 at 22:10

Matthew Taylor

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3 Comments

ranchalp Over a year ago

I would probably change string.letters to 'abcdf' to reflect hexadecimal digits. But great solution!

404pio Over a year ago

''.join(random.sample(string.ascii_letters + string.digits, 8)) more pythonic?

Alex Over a year ago

string.ascii_letters for python 3 instead of string.letters

tzot · Accepted Answer · 2011-05-29 17:52:51Z

6

Another approach to this specific question:

import random, string

def random_md5like_hash():
    available_chars= string.hexdigits[:16]
    return ''.join(
        random.choice(available_chars)
        for dummy in xrange(32))

I'm not saying it's faster or preferable to any other answer; just that it's another approach :)

edited May 29, 2011 at 17:52

answered Jun 11, 2009 at 22:52

tzot

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Comments

Sam · Accepted Answer · 2013-07-26 18:03:28Z

5

import uuid
from md5 import md5

print md5(str(uuid.uuid4())).hexdigest()

answered Jul 26, 2013 at 18:03

Sam

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Comments

gizzmole · Accepted Answer · 2017-06-05 14:39:03Z

5

import os, hashlib
hashlib.md5(os.urandom(32)).hexdigest()

edited Jun 5, 2017 at 14:39

answered May 30, 2017 at 23:44

gizzmole

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1 Comment

not2qubit Over a year ago

outputs something like: eca224d7443ec56efd984c08904cbe4b

Pasha Podolsky · Accepted Answer · 2022-05-04 06:55:53Z

4

The most proper way is to use random module

import random
format(random.getrandbits(128), 'x')

Using secrets is an overkill. It generates cryptographically strong randomness sacrifying performance.

All responses that suggest using UUID are intrinsically wrong because UUID (even UUID4) are not totally random. At least they include fixed version number that never changes.

import uuid
>>> uuid.uuid4()
UUID('8a107d39-bb30-4843-8607-ce9e480c8339')
>>> uuid.uuid4()
UUID('4ed324e8-08f9-4ea5-bc0c-8a9ad53e2df6')

All MD5s containing something other than 4 at 13th position from the left will be unreachable this way.

edited May 4, 2022 at 6:55

answered May 4, 2022 at 5:04

Pasha Podolsky

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3 Comments

Nick T Over a year ago

os.urandom(128//8) takes me 5x as long, or 0.25 microseconds longer to compute. If you care enough that you need to save 1 second every 4 million hashes you generate, you should get a faster RNG like PCG. If you think it will ever matter that your 'hashes' are unguessable/unpredictable, you should use a CSPRNG as it's trivially easy to reconstruct the RNG state after using your call ~160 times.

not2qubit Over a year ago

I don't understand your MD5 comment. Please explain.

Pasha Podolsky Over a year ago

With UUID4 you will generate a string that always contains 4 at 13th position. It is an issue.

save_jeff · Accepted Answer · 2021-10-07 22:32:53Z

0


from hashlib import md5
plaintext = input('Enter the plaintext data to be hashed: ') # Must be a string, doesn't need to have utf-8 encoding
ciphertext = md5(plaintext.encode('utf-8')).hexdigest()
print(ciphertext)

It should also be noted that MD5 is a very weak hash function, also collisions have been found (two different plaintext values result in the same hash) Just use a random value for plaintext.

edited Oct 7, 2021 at 22:32

save_jeff

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answered Feb 24, 2019 at 19:23

Eric Jin

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2 Comments

AS Mackay Over a year ago

Requiring user input doesn't help with the "easiest" aspect of the original question...

ingyhere Over a year ago

Did you check your code? It's missing a paren on line 3.

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