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@@ -5,6 +5,7 @@ from typing import Optional
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import torch
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from ftfy import fix_text
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+from loguru import logger
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from transformers import AutoConfig, AutoModel, AutoModelForCausalLM, AutoTokenizer, PretrainedConfig, PreTrainedModel
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from transformers import VisionEncoderDecoderConfig, VisionEncoderDecoderModel
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@@ -57,9 +58,296 @@ class TokenizerWrapper:
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return toks
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+def fix_latex_left_right(s):
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+ """
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+ 修复LaTeX中的\left和\right命令
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+ 1. 确保它们后面跟有效分隔符
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+ 2. 平衡\left和\right的数量
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+ """
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+ # 白名单分隔符
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+ valid_delims = r'[()\[\]{}/|]|\\{|\\}|\\lceil|\\rceil|\\lfloor|\\rfloor|/|\\backslash|\\uparrow|\\downarrow|\\Uparrow|\\Downarrow|\\||\\.'
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+
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+ # 为\left后缺失有效分隔符的情况添加点
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+ def fix_left_delim(match):
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+ cmd = match.group(1) # \left
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+ rest = match.group(2) if len(match.groups()) > 1 else ""
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+ if not rest or not re.match(f"^({valid_delims})", rest):
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+ return cmd + "."
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+ return match.group(0)
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+
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+ # 为\right后缺失有效分隔符的情况添加点
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+ def fix_right_delim(match):
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+ cmd = match.group(1) # \right
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+ rest = match.group(2) if len(match.groups()) > 1 else ""
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+ if not rest or not re.match(f"^({valid_delims})", rest):
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+ return cmd + "."
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+ return match.group(0)
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+
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+ # 使用更精确的模式匹配\left和\right命令
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+ # 确保它们是独立的命令，不是其他命令的一部分
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+ s = re.sub(r'(\\left)(\S*)', fix_left_delim, s)
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+ s = re.sub(r'(\\right)(\S*)', fix_right_delim, s)
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+
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+ # 更精确地计算\left和\right的数量
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+ left_count = len(re.findall(r'\\left(?![a-zA-Z])', s)) # 不匹配\lefteqn等
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+ right_count = len(re.findall(r'\\right(?![a-zA-Z])', s)) # 不匹配\rightarrow等
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+
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+ if left_count != right_count:
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+ logger.debug(f"latex:{s}")
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+ logger.warning(f"left_count: {left_count}, right_count: {right_count}")
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+
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+ if left_count > right_count:
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+ s += ''.join(['\\right.' for _ in range(left_count - right_count)])
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+ elif right_count > left_count:
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+ # 不再在开头插入\left.，而是在第一个\right前插入
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+ if '\\right' in s:
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+ # 找出所有\right的位置
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+ right_positions = [m.start() for m in re.finditer(r'\\right(?![a-zA-Z])', s)]
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+
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+ # 从前到后为每个缺失的\left处理
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+ new_s = s
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+ offset = 0
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+ for i in range(min(right_count - left_count, len(right_positions))):
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+ pos = right_positions[i] + offset
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+ new_s = new_s[:pos] + '\\left.' + new_s[pos:]
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+ offset += 6 # \left.的长度
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+ s = new_s
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+
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+ return fix_left_right_pairs(s)
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+
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+def fix_left_right_pairs(latex_formula):
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+ """
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+ 检测并修复LaTeX公式中\left和\right不在同一组的情况
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+
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+ Args:
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+ latex_formula (str): 输入的LaTeX公式
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+
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+ Returns:
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+ str: 修复后的LaTeX公式
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+ """
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+ # 用于跟踪花括号嵌套层级
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+ brace_stack = []
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+ # 用于存储\left信息: (位置, 深度, 分隔符)
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+ left_stack = []
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+ # 存储需要调整的\right信息: (开始位置, 结束位置, 目标位置)
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+ adjustments = []
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+
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+ i = 0
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+ while i < len(latex_formula):
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+ # 检查是否是转义字符
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+ if i > 0 and latex_formula[i - 1] == '\\':
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+ backslash_count = 0
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+ j = i - 1
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+ while j >= 0 and latex_formula[j] == '\\':
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+ backslash_count += 1
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+ j -= 1
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+
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+ if backslash_count % 2 == 1:
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+ i += 1
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+ continue
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+
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+ # 检测\left命令
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+ if i + 5 < len(latex_formula) and latex_formula[i:i + 5] == "\\left" and i + 5 < len(latex_formula):
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+ delimiter = latex_formula[i + 5]
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+ left_stack.append((i, len(brace_stack), delimiter))
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+ i += 6 # 跳过\left和分隔符
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+ continue
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+
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+ # 检测\right命令
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+ elif i + 6 < len(latex_formula) and latex_formula[i:i + 6] == "\\right" and i + 6 < len(latex_formula):
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+ delimiter = latex_formula[i + 6]
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+
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+ if left_stack:
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+ left_pos, left_depth, left_delim = left_stack.pop()
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+
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+ # 如果\left和\right不在同一花括号深度
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+ if left_depth != len(brace_stack):
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+ # 找到\left所在花括号组的结束位置
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+ target_pos = find_group_end(latex_formula, left_pos, left_depth)
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+ if target_pos != -1:
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+ # 记录需要移动的\right
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+ adjustments.append((i, i + 7, target_pos))
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+
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+ i += 7 # 跳过\right和分隔符
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+ continue
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+
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+ # 处理花括号
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+ if latex_formula[i] == '{':
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+ brace_stack.append(i)
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+ elif latex_formula[i] == '}':
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+ if brace_stack:
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+ brace_stack.pop()
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+
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+ i += 1
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+
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+ # 应用调整，从后向前处理以避免索引变化
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+ if not adjustments:
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+ return latex_formula
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+
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+ result = list(latex_formula)
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+ adjustments.sort(reverse=True, key=lambda x: x[0])
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+
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+ for start, end, target in adjustments:
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+ # 提取\right部分
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+ right_part = result[start:end]
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+ # 从原位置删除
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+ del result[start:end]
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+ # 在目标位置插入
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+ result.insert(target, ''.join(right_part))
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+
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+ return ''.join(result)
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+
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+
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+def find_group_end(text, pos, depth):
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+ """查找特定深度的花括号组的结束位置"""
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+ current_depth = depth
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+ i = pos
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+
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+ while i < len(text):
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+ if text[i] == '{' and (i == 0 or not is_escaped(text, i)):
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+ current_depth += 1
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+ elif text[i] == '}' and (i == 0 or not is_escaped(text, i)):
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+ current_depth -= 1
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+ if current_depth < depth:
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+ return i
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+ i += 1
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+
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+ return -1 # 未找到对应结束位置
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+
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+
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+def is_escaped(text, pos):
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+ """检查字符是否被转义"""
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+ backslash_count = 0
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+ j = pos - 1
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+ while j >= 0 and text[j] == '\\':
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+ backslash_count += 1
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+ j -= 1
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+
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+ return backslash_count % 2 == 1
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+
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+
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+def fix_unbalanced_braces(latex_formula):
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+ """
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+ 检测LaTeX公式中的花括号是否闭合，并删除无法配对的花括号
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+
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+ Args:
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+ latex_formula (str): 输入的LaTeX公式
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+
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+ Returns:
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+ str: 删除无法配对的花括号后的LaTeX公式
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+ """
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+ stack = [] # 存储左括号的索引
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+ unmatched = set() # 存储不匹配括号的索引
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+ i = 0
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+
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+ while i < len(latex_formula):
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+ # 检查是否是转义的花括号
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+ if latex_formula[i] in ['{', '}']:
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+ # 计算前面连续的反斜杠数量
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+ backslash_count = 0
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+ j = i - 1
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+ while j >= 0 and latex_formula[j] == '\\':
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+ backslash_count += 1
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+ j -= 1
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+
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+ # 如果前面有奇数个反斜杠，则该花括号是转义的，不参与匹配
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+ if backslash_count % 2 == 1:
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+ i += 1
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+ continue
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+
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+ # 否则，该花括号参与匹配
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+ if latex_formula[i] == '{':
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+ stack.append(i)
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+ else: # latex_formula[i] == '}'
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+ if stack: # 有对应的左括号
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+ stack.pop()
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+ else: # 没有对应的左括号
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+ unmatched.add(i)
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+
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+ i += 1
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+
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+ # 所有未匹配的左括号
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+ unmatched.update(stack)
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+
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+ # 构建新字符串，删除不匹配的括号
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+ return ''.join(char for i, char in enumerate(latex_formula) if i not in unmatched)
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+
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+
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+def process_latex(input_string):
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+ """
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+ 处理LaTeX公式中的反斜杠：
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+ 1. 如果\后跟特殊字符(#$%&~_^\\{})或空格，保持不变
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+ 2. 如果\后跟两个小写字母，保持不变
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+ 3. 其他情况，在\后添加空格
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+
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+ Args:
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+ input_string (str): 输入的LaTeX公式
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+
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+ Returns:
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+ str: 处理后的LaTeX公式
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+ """
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+
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+ def replace_func(match):
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+ # 获取\后面的字符
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+ next_char = match.group(1)
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+
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+ # 如果是特殊字符或空格，保持不变
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+ if next_char in "#$%&~_^|\\{} \t\n\r\v\f":
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+ return match.group(0)
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+
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+ # 如果是字母，检查下一个字符
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+ if 'a' <= next_char <= 'z' or 'A' <= next_char <= 'Z':
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+ pos = match.start() + 2 # \x后的位置
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+ if pos < len(input_string) and ('a' <= input_string[pos] <= 'z' or 'A' <= input_string[pos] <= 'Z'):
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+ # 下一个字符也是字母，保持不变
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+ return match.group(0)
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+
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+ # 其他情况，在\后添加空格
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+ return '\\' + ' ' + next_char
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+
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+ # 匹配\后面跟一个字符的情况
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+ pattern = r'\\(.)'
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+
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+ return re.sub(pattern, replace_func, input_string)
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+
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+
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def latex_rm_whitespace(s: str):
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"""Remove unnecessary whitespace from LaTeX code.
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"""
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+ # logger.debug(f"latex_orig: {s}")
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+
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+ s = fix_unbalanced_braces(s)
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+
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+ # left right不匹配的情况（只考虑了不在同一个组里挪到同一个组里的逻辑，没有考虑right比left多的情况）
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+ # 还有加一个\left或\right后至少要跟随一个符号，如果没符号就补.
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+ s = fix_latex_left_right(s)
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+ # s = fix_left_right_pairs(s)
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+
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+ # 用正则删除\left,\left.,\right,\right.
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+ # s = re.sub(r'\\left\.?|\\right\.?', '', s)
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+
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+
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+ # 替换\up命令
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+ s = re.sub(r'\\up([a-zA-Z]+)',
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+ lambda m: m.group(0) if m.group(1) in ["arrow", "downarrow", "lus", "silon",] else f"\\{m.group(1)}", s)
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+
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+ # 替换\underbar为underline
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+ s = re.sub(r'\\underbar', r'\\underline', s)
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+
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+ # 删除\lefteqn
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+ s = re.sub(r'\\lefteqn', r'', s)
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+
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+ # 删除\boldmath
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+ s = re.sub(r'\\boldmath', r'', s)
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+
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+ # \Bar换成\hat
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+ s = re.sub(r'\\Bar', r'\\hat', s)
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+
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+ # \后缺失空格的补空格
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+ s = process_latex(s)
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+
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+ # \qquad后补空格
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+ s = re.sub(r'\\qquad(?!\s)', r'\\qquad ', s)
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# 先保存 "\ " 序列，防止被错误处理
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s = re.sub(r'\\ ', r'\\SPACE', s)
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myhloli