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'.