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