当前Convention未适配3.10([T]泛型注解导致的问题)

cosyvoice/utils/frontend_utils.py

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+# Copyright (c) 2024 Alibaba Inc (authors: Xiang Lyu, Zhihao Du)
+#
+# 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.
+
+import re
+import regex
+chinese_char_pattern = re.compile(r'[\u4e00-\u9fff]+')
+
+
+# whether contain chinese character
+def contains_chinese(text):
+    return bool(chinese_char_pattern.search(text))
+
+
+# replace special symbol
+def replace_corner_mark(text):
+    text = text.replace('²', '平方')
+    text = text.replace('³', '立方')
+    return text
+
+
+# remove meaningless symbol
+def remove_bracket(text):
+    text = text.replace('（', '').replace('）', '')
+    text = text.replace('【', '').replace('】', '')
+    text = text.replace('`', '').replace('`', '')
+    text = text.replace("——", " ")
+    return text
+
+
+# spell Arabic numerals
+def spell_out_number(text: str, inflect_parser):
+    new_text = []
+    st = None
+    for i, c in enumerate(text):
+        if not c.isdigit():
+            if st is not None:
+                num_str = inflect_parser.number_to_words(text[st: i])
+                new_text.append(num_str)
+                st = None
+            new_text.append(c)
+        else:
+            if st is None:
+                st = i
+    if st is not None and st < len(text):
+        num_str = inflect_parser.number_to_words(text[st:])
+        new_text.append(num_str)
+    return ''.join(new_text)
+
+
+# split paragrah logic：
+# 1. per sentence max len token_max_n, min len token_min_n, merge if last sentence len less than merge_len
+# 2. cal sentence len according to lang
+# 3. split sentence according to puncatation
+def split_paragraph(text: str, tokenize, lang="zh", token_max_n=80, token_min_n=60, merge_len=20, comma_split=False):
+    def calc_utt_length(_text: str):
+        if lang == "zh":
+            return len(_text)
+        else:
+            return len(tokenize(_text))
+
+    def should_merge(_text: str):
+        if lang == "zh":
+            return len(_text) < merge_len
+        else:
+            return len(tokenize(_text)) < merge_len
+
+    if lang == "zh":
+        pounc = ['。', '？', '！', '；', '：', '、', '.', '?', '!', ';']
+    else:
+        pounc = ['.', '?', '!', ';', ':']
+    if comma_split:
+        pounc.extend(['，', ','])
+
+    if text[-1] not in pounc:
+        if lang == "zh":
+            text += "。"
+        else:
+            text += "."
+
+    st = 0
+    utts = []
+    for i, c in enumerate(text):
+        if c in pounc:
+            if len(text[st: i]) > 0:
+                utts.append(text[st: i] + c)
+            if i + 1 < len(text) and text[i + 1] in ['"', '”']:
+                tmp = utts.pop(-1)
+                utts.append(tmp + text[i + 1])
+                st = i + 2
+            else:
+                st = i + 1
+
+    final_utts = []
+    cur_utt = ""
+    for utt in utts:
+        if calc_utt_length(cur_utt + utt) > token_max_n and calc_utt_length(cur_utt) > token_min_n:
+            final_utts.append(cur_utt)
+            cur_utt = ""
+        cur_utt = cur_utt + utt
+    if len(cur_utt) > 0:
+        if should_merge(cur_utt) and len(final_utts) != 0:
+            final_utts[-1] = final_utts[-1] + cur_utt
+        else:
+            final_utts.append(cur_utt)
+
+    return final_utts
+
+
+# remove blank between chinese character
+def replace_blank(text: str):
+    out_str = []
+    for i, c in enumerate(text):
+        if c == " ":
+            if ((text[i + 1].isascii() and text[i + 1] != " ") and
+                    (text[i - 1].isascii() and text[i - 1] != " ")):
+                out_str.append(c)
+        else:
+            out_str.append(c)
+    return "".join(out_str)
+
+
+def is_only_punctuation(text):
+    # Regular expression: Match strings that consist only of punctuation marks or are empty.
+    punctuation_pattern = r'^[\p{P}\p{S}]*$'
+    return bool(regex.fullmatch(punctuation_pattern, text))