def test_vectors_deduplicate():
data = OPS.asarray([[1, 1], [2, 2], [3, 4], [1, 1], [3, 4]], dtype="f")
v = Vectors(data=data, keys=["a1", "b1", "c1", "a2", "c2"])
vocab = Vocab()
vocab.vectors = v
# duplicate vectors do not use the same keys
assert (vocab.vectors.key2row[v.strings["a1"]] !=
vocab.vectors.key2row[v.strings["a2"]])
assert (vocab.vectors.key2row[v.strings["c1"]] !=
vocab.vectors.key2row[v.strings["c2"]])
vocab.deduplicate_vectors()
# there are three unique vectors
assert vocab.vectors.shape[0] == 3
# the uniqued data is the same as the deduplicated data
assert_equal(
numpy.unique(OPS.to_numpy(vocab.vectors.data), axis=0),
OPS.to_numpy(vocab.vectors.data),
)
# duplicate vectors use the same keys now
assert (vocab.vectors.key2row[v.strings["a1"]] == vocab.vectors.key2row[
v.strings["a2"]])
assert (vocab.vectors.key2row[v.strings["c1"]] == vocab.vectors.key2row[
v.strings["c2"]])
# deduplicating again makes no changes
vocab_b = vocab.to_bytes()
vocab.deduplicate_vectors()
assert vocab_b == vocab.to_bytes()
def test_pickle_vocab(strings, lex_attr):
vocab = Vocab(strings=strings)
ops = get_current_ops()
vectors = Vectors(data=ops.xp.zeros((10, 10)), mode="floret", hash_count=1)
vocab.vectors = vectors
vocab[strings[0]].norm_ = lex_attr
vocab_pickled = pickle.dumps(vocab)
vocab_unpickled = pickle.loads(vocab_pickled)
assert vocab.to_bytes() == vocab_unpickled.to_bytes()
assert vocab_unpickled.vectors.mode == "floret"
