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

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'.