tzt
06700a5aa3
- difficulty：标记表编译拆分——英文标记改词边界正则（修真实误判：'int' 子串 命中 'print'/'point'、'log' 命中 'logic'、'list' 命中 'listen'），中文标记 保持子串语义；命中行为对合法用例不变（整词出现照常计数） - classifier：HuggingFaceClassifier 推理期异常回落内置 RuleClassifier（单次 推理异常不打垮路由）；build_classifier 的 hf 分支构造失败（ML 依赖缺失/ 模型加载失败）打印提示并回落规则分类器（外置规则照常合并） - 新增 tests/test_failsafe.py 5 项；全量 43 passed（38+5）
84 lines
3.1 KiB
Python
84 lines
3.1 KiB
Python
"""查询难度估计器（启发式，纯标准库）。
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依据：RouterArena 用 Bloom 分类法把问题分为 easy/medium/hard。
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这里用查询长度、指令动词、数学/推理标记做轻量估计。
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"""
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from __future__ import annotations
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import re
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from typing import Tuple
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# 触发 hard 的指令动词 / 推理标记
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_HARD_MARKERS = [
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"证明", "推导", "为什么", "如何", "对比", "比较", "分析", "评估", "设计", "优化",
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"复杂度", "时间复杂度", "空间复杂度", "原理", "机制", "优缺点", "区别", "推论", "定理",
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"proof", "prove", "derive", "explain why", "why", "how", "compare", "contrast",
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"analy", "evaluate", "design", "optimize", "refactor", "architect",
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"implement", "debug", "review", "plan", "synthesize",
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"int", "sum", "sqrt", "lim", "log", "derivative", "integral",
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]
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# 触发 medium 的标记
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_MEDIUM_MARKERS = [
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"用", "写", "计算", "求解", "生成", "翻译", "总结", "解释",
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"注意", "建议", "是否", "实现", "步骤",
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"write", "code", "function", "script", "calculate", "solve", "summarize",
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"translate", "fix", "explain", "describe", "list",
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]
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def _compile_markers(markers):
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"""标记表编译：英文标记用词边界正则，中文标记保持子串匹配。
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T-R3（采纳 llmrouter 词边界思想，super_hard 先于 hard 的同理）：
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纯子串匹配会让 "int" 误命中 "print"/"point"、"log" 误命中 "logic"、
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"list" 误命中 "listen"——英文必须整词命中才计数。
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"""
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substr, words = [], []
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for m in markers:
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(words if m.isascii() else substr).append(m)
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pattern = re.compile(r"\b(?:%s)\b" % "|".join(re.escape(w) for w in words)) \
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if words else None
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return substr, pattern
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_HARD_SUBSTR, _HARD_RE = _compile_markers(_HARD_MARKERS)
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_MEDIUM_SUBSTR, _MEDIUM_RE = _compile_markers(_MEDIUM_MARKERS)
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# 预编译正则（模块级一次，避免每次调用走 re 内部缓存查找）
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_RE_CODE_EXPR = re.compile(r"\b(def|class|function|import)\b")
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_RE_ARITH_EXPR = re.compile(r"[0-9]+\s*[+\-*/^=]\s*[0-9xya-z]")
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def _hit_count(substr, pattern, q: str) -> int:
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"""中文子串命中数 + 英文整词命中数。"""
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n = sum(1 for m in substr if m in q)
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if pattern is not None:
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n += len(pattern.findall(q))
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return n
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def estimate_difficulty(query: str) -> Tuple[str, float]:
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"""返回 (difficulty, score)，score 属于 [0,1]。"""
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q = query.lower()
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hard_hits = _hit_count(_HARD_SUBSTR, _HARD_RE, q)
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medium_hits = _hit_count(_MEDIUM_SUBSTR, _MEDIUM_RE, q)
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length = len(query)
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score = 0.0
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score += min(0.30, length / 600.0) # 长度贡献
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score += min(0.55, hard_hits * 0.25) # 推理标记贡献
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score += min(0.30, medium_hits * 0.08) # 一般指令贡献
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# 额外：代码 / 数学表达式（多步骤信号）
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if "```" in query or _RE_CODE_EXPR.search(q):
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score += 0.15
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if _RE_ARITH_EXPR.search(q):
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score += 0.15
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score = max(0.0, min(1.0, score))
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if score >= 0.55:
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return "hard", score
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if score >= 0.25:
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return "medium", score
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return "easy", score
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