How to generate a random UUID which is reproducibl

问题:

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 [2]: uuid.uuid4()
Out[2]: UUID('f6c9ad6c-eea0-4049-a7c5-56253bc3e9c0')

In [3]: uuid.uuid4()
Out[3]: UUID('2fc1b6f9-9052-4564-9be0-777e790af58f')

I would like to be able to generate the same random UUID every time I run a script - that is, I'd like to seed the random generator in uuid4(). Is there a way to do this? (Or achieve this by some other means)?

What I've tried so far

I've to generate a UUID using the uuid.UUID() method with a random 128-bit integer (from a seeded instance of random.Random()) as input:

import uuid
import random

rd = random.Random()
rd.seed(0)
uuid.UUID(rd.getrandbits(128))

However, UUID() seems not to accept this as input:

Traceback (most recent call last):
  File "uuid_gen_seed.py", line 6, in <module>
    uuid.UUID(rd.getrandbits(128))
  File "/usr/lib/python2.7/uuid.py", line 133, in __init__
    hex = hex.replace('urn:', '').replace('uuid:', '')
AttributeError: 'long' object has no attribute 'replace'

Any other suggestions?

回答1:

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'.

回答2:

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

回答3:

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)