How to generate a random UUID which is reproducible (with a seed) in Python

Question

The uuid4() function of Python\'s module uuid generates a random UUID, and seems to generate a different one every time:

In [1]: import uuid

In          


        

Answer 1

Since the straight-forward solution hasn't been posted yet to generate consistent version 4 UUIDs:

import random
import uuid

rnd = random.Random()
rnd.seed(123)

uuid = uuid.UUID(int=rnd.getrandbits(128), version=4)

where you can see then:

>>> uuid.version
4

This doesn't just "mock" the version information. It creates a proper UUIDv4:

The version argument is optional; if given, the resulting UUID will have its variant and version number set according to RFC 4122, overriding bits in the given hex, bytes, bytes_le, fields, or int.

Python 3.8 docs

Answer 2

Based on alex's solution, the following would provide a proper UUID4:

random.seed(123210912)
a = "%32x" % random.getrandbits(128)
rd = a[:12] + '4' + a[13:16] + 'a' + a[17:]
uuid4 = uuid.UUID(rd)

Answer 3

Gonna add this here if anyone needs to monkey patch in a seeded UUID. My code uses uuid.uuid4() but for testing I wanted consistent UUIDs. The following code is how I did that:

import uuid
import random

# -------------------------------------------
# Remove this block to generate different
# UUIDs everytime you run this code.
# This block should be right below the uuid
# import.
rd = random.Random()
rd.seed(0)
uuid.uuid4 = lambda: uuid.UUID(int=rd.getrandbits(128))
# -------------------------------------------

# Then normal code:

print(uuid.uuid4().hex)
print(uuid.uuid4().hex)
print(uuid.uuid4().hex)
print(uuid.uuid4().hex)

Answer 4

This is based on a solution used here:

import hashlib
import uuid

m = hashlib.md5()
m.update(seed.encode('utf-8'))
new_uuid = uuid.UUID(m.hexdigest())

Answer 5

Faker makes this easy

>>> from faker import Faker
>>> f1 = Faker()
>>> f1.seed(4321)
>>> print(f1.uuid4())
cc733c92-6853-15f6-0e49-bec741188ebb
>>> print(f1.uuid4())
a41f020c-2d4d-333f-f1d3-979f1043fae0
>>> f1.seed(4321)
>>> print(f1.uuid4())
cc733c92-6853-15f6-0e49-bec741188ebb

Answer 6

Almost there:

uuid.UUID(int=rd.getrandbits(128))

This was determined with the help of help:

>>> help(uuid.UUID.__init__)
Help on method __init__ in module uuid:

__init__(self, hex=None, bytes=None, bytes_le=None, fields=None, int=None, version=None) unbound uuid.UUID method
    Create a UUID from either a string of 32 hexadecimal digits,
    a string of 16 bytes as the 'bytes' argument, a string of 16 bytes
    in little-endian order as the 'bytes_le' argument, a tuple of six
    integers (32-bit time_low, 16-bit time_mid, 16-bit time_hi_version,
    8-bit clock_seq_hi_variant, 8-bit clock_seq_low, 48-bit node) as
    the 'fields' argument, or a single 128-bit integer as the 'int'
    argument.  When a string of hex digits is given, curly braces,
    hyphens, and a URN prefix are all optional.  For example, these
    expressions all yield the same UUID:

    UUID('{12345678-1234-5678-1234-567812345678}')
    UUID('12345678123456781234567812345678')
    UUID('urn:uuid:12345678-1234-5678-1234-567812345678')
    UUID(bytes='\x12\x34\x56\x78'*4)
    UUID(bytes_le='\x78\x56\x34\x12\x34\x12\x78\x56' +
                  '\x12\x34\x56\x78\x12\x34\x56\x78')
    UUID(fields=(0x12345678, 0x1234, 0x5678, 0x12, 0x34, 0x567812345678))
    UUID(int=0x12345678123456781234567812345678)

    Exactly one of 'hex', 'bytes', 'bytes_le', 'fields', or 'int' must
    be given.  The 'version' argument is optional; if given, the resulting
    UUID will have its variant and version set according to RFC 4122,
    overriding the given 'hex', 'bytes', 'bytes_le', 'fields', or 'int'.