sample = """ [1]: https://baike.baidu.com/item/%E8%B4%A8%E8%83%BD%E6%96%B9%E7%A8%8B/1884527 "质能方程(质能方程式)_百度百科" [2]: https://www.zhihu.com/question/348249281 "如何理解质能方程 E=mc²? - 知乎" [3]: https://zhuanlan.zhihu.com/p/32597385 "质能方程的推导与理解 - 知乎 - 知乎专栏" 你好,这是必应。质能方程是描述质量与能量之间的当量关系的方程[^1^][1]。用tex格式,质能方程可以写成$$E=mc^2$$,其中$E$是能量,$m$是质量,$c$是光速[^2^][2] [^3^][3]。 """ import re def preprocess_newbing_out(s): pattern = r'\^(\d+)\^' # 匹配^数字^ pattern2 = r'\[(\d+)\]' # 匹配^数字^ sub = lambda m: '\['+m.group(1)+'\]' # 将匹配到的数字作为替换值 result = re.sub(pattern, sub, s) # 替换操作 if '[1]' in result: result += '


' + "
".join([re.sub(pattern2, sub, r) for r in result.split('\n') if r.startswith('[')]) + '
' return result def close_up_code_segment_during_stream(gpt_reply): """ 在gpt输出代码的中途(输出了前面的```,但还没输出完后面的```),补上后面的``` Args: gpt_reply (str): GPT模型返回的回复字符串。 Returns: str: 返回一个新的字符串,将输出代码片段的“后面的```”补上。 """ if '```' not in gpt_reply: return gpt_reply if gpt_reply.endswith('```'): return gpt_reply # 排除了以上两个情况,我们 segments = gpt_reply.split('```') n_mark = len(segments) - 1 if n_mark % 2 == 1: # print('输出代码片段中!') return gpt_reply+'\n```' else: return gpt_reply import markdown from latex2mathml.converter import convert as tex2mathml from functools import wraps, lru_cache def markdown_convertion(txt): """ 将Markdown格式的文本转换为HTML格式。如果包含数学公式,则先将公式转换为HTML格式。 """ pre = '
' suf = '
' if txt.startswith(pre) and txt.endswith(suf): # print('警告,输入了已经经过转化的字符串,二次转化可能出问题') return txt # 已经被转化过,不需要再次转化 markdown_extension_configs = { 'mdx_math': { 'enable_dollar_delimiter': True, 'use_gitlab_delimiters': False, }, } find_equation_pattern = r'\n', '') return content if ('$' in txt) and ('```' not in txt): # 有$标识的公式符号,且没有代码段```的标识 # convert everything to html format split = markdown.markdown(text='---') convert_stage_1 = markdown.markdown(text=txt, extensions=['mdx_math', 'fenced_code', 'tables', 'sane_lists'], extension_configs=markdown_extension_configs) convert_stage_1 = markdown_bug_hunt(convert_stage_1) # re.DOTALL: Make the '.' special character match any character at all, including a newline; without this flag, '.' will match anything except a newline. Corresponds to the inline flag (?s). # 1. convert to easy-to-copy tex (do not render math) convert_stage_2_1, n = re.subn(find_equation_pattern, replace_math_no_render, convert_stage_1, flags=re.DOTALL) # 2. convert to rendered equation convert_stage_2_2, n = re.subn(find_equation_pattern, replace_math_render, convert_stage_1, flags=re.DOTALL) # cat them together return pre + convert_stage_2_1 + f'{split}' + convert_stage_2_2 + suf else: return pre + markdown.markdown(txt, extensions=['fenced_code', 'codehilite', 'tables', 'sane_lists']) + suf sample = preprocess_newbing_out(sample) sample = close_up_code_segment_during_stream(sample) sample = markdown_convertion(sample) with open('tmp.html', 'w', encoding='utf8') as f: f.write(""" My Website """) f.write(sample)