问题:客户问「季季盈90天的起投金额是多少」会被引导到人工客服,而知识库里明明有答案。 实测根因不是阈值拍错了,而是专有名词在 embedding 空间里不占优势——该问句的向量 top1 只有 0.6291,够不到 0.75 硬门槛,只能靠与次优的差值勉强通过;而同一次查询用 title like "%季季盈%" 是唯一命中 PROD-007。既然客户已经说出了产品名,就不该再赌相似度。 做法(三条边界都是实测逼出来的,不是设想): 1. 只对产品集合做字面匹配。客户问「季季盈90天的起投金额是多少」与通用 FAQ 标题 「基金起投金额是多少?」有 7 个字连续重合;把 FAQ 纳入字面匹配会让它和真正的产品块 一起拿到满分、差距归零,反而又退化成"转人工"。 2. 字面命中只在向量结果不够确定时采用。客户问「基金赎回几天到账」时向量已给出正确答案 (FAQ-0016 得 0.8060),但手册章节标题「5.2 基金赎回流程」与问句也有 4 个字连续重合, 无条件采纳会把"操作步骤"顶掉客户真正问的"到账时间"。 3. 重叠门槛取 6 字而不是 4 字:"基金赎回"这类业务动作词正好 4 字,会骗过 4 字门槛; 产品名("南方季季盈90天")更长,6 字能同时保住产品名、挡住动作词。 未改动任何转人工判定阈值;VECTOR_CONFIDENT_SCORE 与 Agent 的 HIGH_SCORE 由单测锁定一致, 避免两处各自漂移出"谁都答不出来"的死角。 验证:季季盈类问法由"转人工"变为正确答出,基金赎回问法仍答 FAQ-0016; ruff / mypy(113 文件) / 453 unit+contract / 29 integration 全绿。
316 lines
14 KiB
Python
316 lines
14 KiB
Python
"""知识库检索:把客户问题向量化后在三个知识集合里检索(只读)。
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为什么不复用记忆那套 `VectorMemoryAdapter`:它只把命中折叠成 `(memory_uuid, score)`,
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会把知识块的标题与正文丢掉。而客服回答必须能把**原文与来源**一起交给客户——金融场景
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里「答案出自哪份文件的哪一条」本身就是交付物的一部分,丢了正文等于没法给来源引用。
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失败语义与基座一致:**任何一步失败都不抛异常给主链路**,而是返回 `degraded=True`
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的空结果,由调用方(客服 Agent)据此走「引导客户致电人工客服」的兜底路径。
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金融场景下"答不了"是可接受的结果,"答错"不是。
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"""
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from collections.abc import Awaitable, Callable, Sequence
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from dataclasses import dataclass, field
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from typing import Any, Protocol
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# 三个知识集合(方案 §2.4.4 / §4.1)
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FAQ_COLLECTION = "fin_faq_collection"
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PRODUCT_COLLECTION = "fin_product_collection"
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POLICY_COLLECTION = "fin_policy_collection"
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DEFAULT_COLLECTIONS: tuple[str, ...] = (FAQ_COLLECTION, PRODUCT_COLLECTION, POLICY_COLLECTION)
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# 检索输出字段:与灌库脚本写入的 schema 对齐
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OUTPUT_FIELDS = (
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"doc_id", "title", "content", "chapter", "section",
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"tags", "doc_no", "version", "source_file", "visibility",
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)
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# 关键字精确召回:客户问到产品名这类**专有名词**时,字面匹配比相似度更确定。
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#
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# 为什么需要:专有名词在 embedding 空间里不占优势。实测 160 条知识,客户问
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# 「季季盈90天的起投金额是多少」向量 top1 = PROD-007 仅 0.6291(够不到 0.75 的硬门槛,
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# 只能靠与次优的差值勉强通过);而同一次查询用 `title like "%季季盈%"` 是**唯一命中**
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# PROD-007。既然客户已经明确说出了产品名,就不该再让相似度去赌。
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KEYWORD_MATCH_SCORE = 1.0 # 字面命中的确定分,压过任何相似度分
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# 与标题的最长公共子串至少要这么长,才算"客户确切提到了它"。
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# 为什么不是 4:实测客户问「基金赎回几天到账」,与手册章节标题「5.2 基金赎回流程」
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# 正好有 4 个字连续重合——但"基金赎回"是业务动作词,不是专有名词。产品名
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# ("南方季季盈90天")通常比业务动作词长,取 6 字能同时保住产品名、挡住动作词。
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MIN_KEYWORD_OVERLAP = 6
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KEYWORD_SCAN_LIMIT = 500 # 一次最多扫描多少条标题;知识库到上千块后应改为倒排索引
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# 字面匹配只在向量结果**不够确定**时介入。
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#
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# 这条门禁是实测逼出来的:客户问「基金赎回几天到账」,向量已给出正确答案
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# (FAQ-0016「基金赎回到账需要多长时间?」得 0.8060),但产品手册里的章节标题
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# 「五、申购赎回操作指南」与问句也有 6 个字连续重合,无条件字面匹配会把它顶到第一,
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# 用操作步骤替换掉客户真正问的到账时间。所以字面匹配是**兜底**,不是优先。
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#
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# 门槛必须与客服 Agent 的高置信门槛一致,tests/unit 有断言锁定两者相等。
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VECTOR_CONFIDENT_SCORE = 0.75
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class VectorSearcher(Protocol):
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"""只依赖用到的两个方法,便于测试替身注入。"""
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def search(self, **kwargs: Any) -> Any: ...
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Embedder = Callable[[str], Awaitable[list[float]]]
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@dataclass(frozen=True)
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class KnowledgeHit:
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"""一条知识命中;`score` 为 COSINE 相似度(越大越相似)。"""
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doc_id: str
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title: str
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content: str
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score: float
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source_file: str = ""
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visibility: str = "public"
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doc_no: str = ""
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version: str = ""
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chapter: str = ""
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@property
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def reference_title(self) -> str:
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"""给客户看的来源标题:优先带内部文件编号,便于人工核对。"""
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return f"{self.title}({self.doc_no})" if self.doc_no else self.title
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@dataclass(frozen=True)
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class KnowledgeSearchOutcome:
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"""检索结果;`degraded=True` 表示检索链路故障,调用方必须走兜底而非当'没找到'。"""
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hits: tuple[KnowledgeHit, ...] = ()
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degraded: bool = False
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reason: str = ""
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searched_collections: tuple[str, ...] = field(default_factory=tuple)
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@property
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def best(self) -> KnowledgeHit | None:
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return self.hits[0] if self.hits else None
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@property
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def top_score(self) -> float:
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return self.hits[0].score if self.hits else 0.0
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class KnowledgeSearchService:
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def __init__(
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self,
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client: VectorSearcher | None,
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embedder: Embedder | None,
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*,
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collections: Sequence[str] = DEFAULT_COLLECTIONS,
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) -> None:
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self._client = client
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self._embedder = embedder
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self._collections = tuple(collections)
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@property
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def available(self) -> bool:
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"""向量库与向量化能力是否都在位;缺任一项都不做检索,直接走兜底。"""
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return self._client is not None and self._embedder is not None
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async def search(
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self,
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query: str,
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*,
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collections: Sequence[str] | None = None,
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top_k: int = 5,
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include_internal: bool = False,
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) -> KnowledgeSearchOutcome:
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"""检索知识库。
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`include_internal=False`(默认)时在 Milvus 侧就过滤掉 `visibility=internal` 的块,
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内部资料不进入面向客户的答案——这是检索层的硬隔离,不依赖提示词约束。
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"""
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text = query.strip()
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if not text:
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return KnowledgeSearchOutcome(reason="empty_query")
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if not self.available:
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return KnowledgeSearchOutcome(degraded=True, reason="vector_backend_unavailable")
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client, embedder = self._client, self._embedder
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if client is None or embedder is None:
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# 与 available 重复,但这里需要类型收窄(mypy 不跨属性判断 Optional)
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return KnowledgeSearchOutcome(degraded=True, reason="vector_backend_unavailable")
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try:
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vector = await embedder(text)
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except Exception:
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return KnowledgeSearchOutcome(degraded=True, reason="embedding_failed")
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if not vector:
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return KnowledgeSearchOutcome(degraded=True, reason="embedding_empty")
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targets = tuple(collections or self._collections)
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expression = None if include_internal else 'visibility == "public"'
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collected: list[KnowledgeHit] = []
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failures = 0
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for collection in targets:
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try:
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raw = client.search(
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collection_name=collection,
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data=[vector],
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limit=max(1, min(top_k, 20)),
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output_fields=list(OUTPUT_FIELDS),
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filter=expression,
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)
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except Exception:
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failures += 1
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continue
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collected.extend(self._parse(raw, collection))
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# 第二路召回:客户确切说出的产品名按字面取回。只在向量结果不够确定时介入,
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# 否则会把向量已经答对的题顶掉(见 VECTOR_CONFIDENT_SCORE 的说明)。
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best_vector_score = max((hit.score for hit in collected), default=0.0)
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if best_vector_score < VECTOR_CONFIDENT_SCORE:
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collected.extend(self._product_keyword_hits(client, targets, text, expression))
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if not collected and failures == len(targets) and targets:
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# 三个集合全查失败:是链路故障,不是"知识库里没有"
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return KnowledgeSearchOutcome(
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degraded=True, reason="search_failed", searched_collections=targets
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)
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collected.sort(key=lambda hit: hit.score, reverse=True)
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# 同一内容可能同时存在于产品手册与问答对里,按 doc_id 去重保留最高分
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deduped: list[KnowledgeHit] = []
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seen: set[str] = set()
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for hit in collected:
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if hit.doc_id in seen:
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continue
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seen.add(hit.doc_id)
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deduped.append(hit)
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return KnowledgeSearchOutcome(
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hits=tuple(deduped[: max(1, top_k)]),
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degraded=failures > 0,
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reason="partial_collection_failure" if failures else "",
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searched_collections=targets,
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)
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# ---- 关键字精确召回(字面匹配) ----
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@staticmethod
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def _overlap_length(left: str, right: str) -> int:
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"""两段文本的最长公共子串长度。
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用最长公共子串而不是分词:知识库标题是「南方科技有限公司 个人理财产品手册 ·
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2.1 南方季季盈90天」这种没有词边界的长串,任何分词器都得先养一份自定义词典,
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而词典会和手册一起过期。子串匹配不需要词典,手册改版也不会失效。
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"""
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if not left or not right:
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return 0
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previous = [0] * (len(right) + 1)
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best = 0
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for i in range(1, len(left) + 1):
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current = [0] * (len(right) + 1)
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for j in range(1, len(right) + 1):
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if left[i - 1] == right[j - 1]:
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current[j] = previous[j - 1] + 1
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if current[j] > best:
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best = current[j]
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previous = current
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return best
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def _product_keyword_hits(
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self, client: Any, targets: Sequence[str], query: str, expression: str | None
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) -> list[KnowledgeHit]:
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"""客户确切说出某个产品名时,按字面把它取出来(兜底用)。
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两处边界都是实测逼出来的:
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1. **只对产品集合做**。客户问「季季盈90天的起投金额是多少」,与通用 FAQ 标题
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「基金起投金额是多少?」的最长公共子串有 7 个字;若把 FAQ 也纳入字面匹配,
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它会和真正的产品块一起拿到满分、差距归零,反而又退化成"转人工"。
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2. **命中不能无条件优先**。产品手册的标题里不只有产品名,还有章节名:客户问
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「基金赎回几天到账」时,「五、申购赎回操作指南」那一块与问句也有 6 个字连续
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重合。所以本方法产出什么是一回事,是否采用由调用方按"向量是否已经足够确定"
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决定(见 VECTOR_CONFIDENT_SCORE)。
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失败一律返回空:关键字路径是**增益**,它坏了不能让整个检索变成故障。
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"""
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lookup = getattr(client, "query", None)
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if lookup is None or PRODUCT_COLLECTION not in targets:
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return [] # 客户端不支持标量查询(如测试替身),或本次没查产品集合
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try:
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rows = lookup(
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collection_name=PRODUCT_COLLECTION,
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filter=expression,
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output_fields=["doc_id", "title"],
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limit=KEYWORD_SCAN_LIMIT,
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)
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except Exception:
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return []
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matched_ids = [
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str(row.get("doc_id") or "")
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for row in (rows if isinstance(rows, list) else [])
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if isinstance(row, dict)
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and self._overlap_length(query, str(row.get("title") or "")) >= MIN_KEYWORD_OVERLAP
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]
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matched_ids = [doc_id for doc_id in matched_ids if doc_id]
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if not matched_ids:
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return []
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quoted = ", ".join(f'"{doc_id}"' for doc_id in matched_ids)
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try:
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details = lookup(
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collection_name=PRODUCT_COLLECTION,
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filter=f"doc_id in [{quoted}]",
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output_fields=list(OUTPUT_FIELDS),
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limit=len(matched_ids),
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)
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except Exception:
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return []
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hits: list[KnowledgeHit] = []
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for row in details if isinstance(details, list) else []:
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if not isinstance(row, dict):
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continue
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content = str(row.get("content") or "")
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if not content:
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continue
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hits.append(KnowledgeHit(
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doc_id=str(row.get("doc_id") or ""),
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title=str(row.get("title") or ""),
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content=content,
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score=KEYWORD_MATCH_SCORE,
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source_file=str(row.get("source_file") or ""),
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visibility=str(row.get("visibility") or "public"),
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doc_no=str(row.get("doc_no") or ""),
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version=str(row.get("version") or ""),
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chapter=str(row.get("chapter") or ""),
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))
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# 一个产品名可能命中多个块(产品概览、费率表各一块):全都保留,
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# 是不是"只有一个明确候选"交给上层的 gap 判定,这里不替它做选择。
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return hits
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@staticmethod
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def _parse(raw: Any, collection: str) -> list[KnowledgeHit]:
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"""把 pymilvus 的 `[[{id, distance, entity}]]` 折叠成命中列表(纯函数,不抛异常)。"""
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hits: list[KnowledgeHit] = []
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groups = raw if isinstance(raw, (list, tuple)) else [raw]
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for group in groups:
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rows = group if isinstance(group, (list, tuple)) else [group]
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for row in rows:
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entity = row.get("entity") if isinstance(row, dict) else None
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if not isinstance(entity, dict):
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continue
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content = str(entity.get("content") or "")
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if not content:
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continue # 没有正文的命中无法作为答案来源,直接丢弃而不是猜造
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hits.append(KnowledgeHit(
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doc_id=str(entity.get("doc_id") or ""),
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title=str(entity.get("title") or ""),
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content=content,
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score=float(row.get("distance") or 0.0),
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source_file=str(entity.get("source_file") or ""),
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visibility=str(entity.get("visibility") or "public"),
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doc_no=str(entity.get("doc_no") or ""),
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version=str(entity.get("version") or ""),
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chapter=str(entity.get("chapter") or ""),
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))
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return hits
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