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#!/usr/bin/env python3
# Copyright 2021 Xiaomi Corp. (authors: Fangjun Kuang)
#
# See ../../LICENSE for clarification regarding multiple authors
#
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
"""
You can run this file in one of the two ways:
(1) cd icefall; pytest test/test_decode.py
(2) cd icefall; ./test/test_decode.py
"""
import k2
from icefall.decode import Nbest
def test_nbest_from_lattice():
s = """
0 1 1 10 0.1
0 1 5 10 0.11
0 1 2 20 0.2
1 2 3 30 0.3
1 2 4 40 0.4
2 3 -1 -1 0.5
3
"""
lattice = k2.Fsa.from_str(s, acceptor=False)
lattice = k2.Fsa.from_fsas([lattice, lattice])
nbest = Nbest.from_lattice(
lattice=lattice,
num_paths=10,
use_double_scores=True,
nbest_scale=0.5,
)
# each lattice has only 4 distinct paths that have different word sequences:
# 10->30
# 10->40
# 20->30
# 20->40
#
# So there should be only 4 paths for each lattice in the Nbest object
assert nbest.fsa.shape[0] == 4 * 2
assert nbest.shape.row_splits(1).tolist() == [0, 4, 8]
nbest2 = nbest.intersect(lattice)
tot_scores = nbest2.tot_scores()
argmax = tot_scores.argmax()
best_path = k2.index_fsa(nbest2.fsa, argmax)
print(best_path[0])
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