Replace the text normalization library
This commit is contained in:
刘鑫
@@ -4,8 +4,6 @@ import os
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import tempfile
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from huggingface_hub import snapshot_download
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from .model.voxcpm import VoxCPMModel
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from .utils.text_normalize import TextNormalizer
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class VoxCPM:
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def __init__(self,
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@@ -25,7 +23,7 @@ class VoxCPM:
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"""
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print(f"voxcpm_model_path: {voxcpm_model_path}, zipenhancer_model_path: {zipenhancer_model_path}, enable_denoiser: {enable_denoiser}")
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self.tts_model = VoxCPMModel.from_local(voxcpm_model_path)
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self.text_normalizer = TextNormalizer()
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self.text_normalizer = None
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if enable_denoiser and zipenhancer_model_path is not None:
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from .zipenhancer import ZipEnhancer
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self.denoiser = ZipEnhancer(zipenhancer_model_path)
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@@ -33,8 +31,9 @@ class VoxCPM:
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self.denoiser = None
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print("Warm up VoxCPMModel...")
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self.tts_model.generate(
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target_text="Hello, this is the first test sentence."
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)
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target_text="Hello, this is the first test sentence.",
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max_len=10,
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)
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@classmethod
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def from_pretrained(cls,
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@@ -145,6 +144,9 @@ class VoxCPM:
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continue
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print("sub_text:", sub_text)
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if normalize:
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if self.text_normalizer is None:
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from .utils.text_normalize import TextNormalizer
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self.text_normalizer = TextNormalizer()
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sub_text = self.text_normalizer.normalize(sub_text)
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wav, target_text_token, generated_audio_feat = self.tts_model.generate_with_prompt_cache(
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target_text=sub_text,
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@@ -3,41 +3,8 @@ import re
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import regex
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import inflect
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from functools import partial
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from tn.chinese.normalizer import Normalizer as ZhNormalizer
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from tn.english.normalizer import Normalizer as EnNormalizer
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from wetext import Normalizer
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def normal_cut_sentence(text):
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# 先处理括号内的逗号，将其替换为特殊标记
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text = re.sub(r'([（(][^）)]*)([,，])([^）)]*[）)])', r'\1&&&\3', text)
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text = re.sub('([。！，？\?])([^’”])',r'\1\n\2',text)#普通断句符号且后面没有引号
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text = re.sub('(\.{6})([^’”])',r'\1\n\2',text)#英文省略号且后面没有引号
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text = re.sub('(\…{2})([^’”])',r'\1\n\2',text)#中文省略号且后面没有引号
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text = re.sub('([. ,。！；？\?\.{6}\…{2}][’”])([^’”])',r'\1\n\2',text)#断句号+引号且后面没有引号
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# 处理英文句子的分隔
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text = re.sub(r'([.,!?])([^’”\'"])', r'\1\n\2', text) # 句号、感叹号、问号后面没有引号
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text = re.sub(r'([.!?][’”\'"])([^’”\'"])', r'\1\n\2', text) # 句号、感叹号、问号加引号后面的部分
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text = re.sub(r'([（(][^）)]*)(&&&)([^）)]*[）)])', r'\1，\3', text)
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text = [t for t in text.split("\n") if t]
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return text
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def cut_sentence_with_fix_length(text : str, length : int):
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sentences = normal_cut_sentence(text)
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cur_length = 0
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res = ""
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for sentence in sentences:
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if not sentence:
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continue
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if cur_length > length or cur_length + len(sentence) > length:
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yield res
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res = ""
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cur_length = 0
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res += sentence
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cur_length += len(sentence)
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if res:
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yield res
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chinese_char_pattern = re.compile(r'[\u4e00-\u9fff]+')
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# whether contain chinese character
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@@ -195,8 +162,8 @@ def clean_text(text):
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class TextNormalizer:
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def __init__(self, tokenizer=None):
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self.tokenizer = tokenizer
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self.zh_tn_model = ZhNormalizer(remove_erhua=False, full_to_half=False, remove_interjections=False, overwrite_cache=True)
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self.en_tn_model = EnNormalizer()
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self.zh_tn_model = Normalizer(lang="zh", operator="tn", remove_erhua=True)
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self.en_tn_model = Normalizer(lang="en", operator="tn")
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self.inflect_parser = inflect.engine()
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def normalize(self, text, split=False):
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@@ -207,38 +174,12 @@ class TextNormalizer:
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text = text.replace("=", "等于") # 修复 ”550 + 320 等于 870 千卡。“ 被错误正则为 ”五百五十加三百二十等于八七十千卡.“
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if re.search(r'([\d$%^*_+≥≤≠×÷?=])', text): # 避免 英文连字符被错误正则为减
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text = re.sub(r'(?<=[a-zA-Z0-9])-(?=\d)', ' - ', text) # 修复 x-2 被正则为 x负2
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text = self.zh_tn_model.normalize(text)
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text = re.sub(r'(?<=[a-zA-Z0-9])-(?=\d)', ' - ', text) # 修复 x-2 被正则为 x负2
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text = self.zh_tn_model.normalize(text)
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text = replace_blank(text)
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text = replace_corner_mark(text)
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text = remove_bracket(text)
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text = re.sub(r'[，,]+$', '。', text)
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else:
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text = self.en_tn_model.normalize(text)
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text = spell_out_number(text, self.inflect_parser)
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if split is False:
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return text
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if __name__ == "__main__":
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text_normalizer = TextNormalizer()
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text = r"""今天我们学习一元二次方程。一元二次方程的标准形式是：
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ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0
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其中，aaa、bbb 和 ccc 是常数，xxx 是变量。这个方程的解可以通过求根公式来找到。
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一元二次方程的解法有几种：
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- 因式分解法：通过将方程因式分解来求解。我们首先尝试将方程表达成两个括号的形式，解决方程的解。比如，方程x2−5x+6=0x^2 - 5x + 6 = 0x2−5x+6=0可以因式分解为(x−2)(x−3)=0(x - 2)(x - 3) = 0(x−2)(x−3)=0，因此根为2和3。
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- 配方法：通过配方将方程转化为完全平方的形式，从而解出。我们通过加上或减去适当的常数来完成这一过程，使得方程可以直接写成一个完全平方的形式。
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- 求根公式：我们可以使用求根公式直接求出方程的解。这个公式适用于所有的一元二次方程，即使我们无法通过因式分解或配方法来解决时，也能使用该公式。
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公式：x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac这个公式可以帮助我们求解任何一元二次方程的根。
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对于一元二次方程，我们需要了解判别式。判别式的作用是帮助我们判断方程的解的个数和性质。判别式 Δ\DeltaΔ 由下式给出：Δ=b2−4ac\Delta = b^2 - 4acΔ=b2−4ac 根据判别式的值，我们可以知道：
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- 如果 Δ>0\Delta > 0Δ>0，方程有两个不相等的实数解。这是因为判别式大于0时，根号内的值是正数，所以我们可以得到两个不同的解。
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- 如果 Δ=0\Delta = 0Δ=0，方程有一个实数解。这是因为根号内的值为零，导致两个解相等，也就是说方程有一个解。
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- 如果 Δ<0\Delta < 0Δ<0，方程没有实数解。这意味着根号内的值是负数，无法进行实数运算，因此方程没有实数解，可能有复数解。"""
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texts = ["这是一个公式 (a+b)³=a³+3a²b+3ab²+b³ S=(a×b)÷2", "这样的发展为AI仅仅作为“工具”这一观点提出了新的挑战，", "550 + 320 = 870千卡。", "解一元二次方程：3x^2+x-2=0", "你好啊"]
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texts = [text]
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for text in texts:
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text = text_normalizer.normalize(text)
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print(text)
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for t in cut_sentence_with_fix_length(text, 15):
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print(t)
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return text
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