93 lines
4.6 KiB
Python
93 lines
4.6 KiB
Python
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"""计划装配接入逻辑状态四态的离线单测（不连库）。
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钉住三件事：
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一，四态随每张卡一起产出，并按约定的形状发给下游（每路带截止日与一句话说明）。
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二，四态**不改判决**。候选卡的三门槛判决与逻辑四态是正交的两维，收敛规则是单调的：
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逻辑强化只能提前卡内序，永远不能把判决往上升一档；逻辑存疑只能改分流通道，
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不能把仅展示变成可执行。
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三，取数任一路读不到都不让计划断产，那一路记缺失。
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跑法：python3 test_plan_logic_state.py 或 pytest test_plan_logic_state.py
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"""
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import logic_state as ls
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import plan
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DS = "2026-09-03"
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def t(name, cond):
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assert cond, name
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print(" ok", name)
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def claim(date, direction="利好"):
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return {"disclosure_date": date, "direction": direction, "mechanism": "机制",
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"doc_title": "研报", "claim_id": "c" + date.replace("-", "")}
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def jrow(leaning="偏多", verified=1, migrated=0, stale=3, seg="固态电解质"):
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return {"leaning": leaning, "leaning_prev": None, "migrated": migrated,
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"stale_days": stale, "verified": verified, "review_date": "2026-09-03",
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"n_materials": 43, "subject_name": seg, "segment_name": seg,
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"n_bull": 3, "n_bear": 3}
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def main():
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print("四态随卡产出")
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evd = {"logic": [claim("2026-08-25")]}
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seg_of = {"SH600000": ["固态电解质"]}
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seg_view = {"固态电解质": jrow()}
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broker = {"SH600000": ls.signal(ls.PATH_BROKER, ls.SIG_FLAT, as_of=DS,
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coverage=6, why="预测中位数变化 +1%，在阈值之内")}
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st = plan._logic_state_of("SH600000", evd, seg_of, seg_view, broker, DS) # noqa: SLF001
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t("三路都在且无负面 -> 逻辑成立", st["state"] == ls.STATE_HOLD)
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t("四路都记了名，缺的那一路写明是公司事件",
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len(st["paths"]) == 4 and ls.PATH_EVENT in st["missing"])
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print("取数缺席时不断产")
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st = plan._logic_state_of("SH600000", {}, {}, {}, {}, DS) # noqa: SLF001
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t("四路全缺 -> 无法判断加证据不足，不抛异常",
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st["state"] == ls.STATE_UNKNOWN and st["why"] == ls.WHY_THIN)
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t("缺失名单写全了四路", len(st["missing"]) == 4)
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st = plan._logic_state_of("SH999999", evd, seg_of, seg_view, broker, DS) # noqa: SLF001
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t("这只票不在任何被指向环节上 -> 产业研判缺席，其余照算",
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ls.PATH_JUDGE in st["missing"] and ls.PATH_CLAIM in st["usable"])
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print("一票挂多个环节")
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st = plan._logic_state_of( # noqa: SLF001
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"SH600000", evd, {"SH600000": ["没评过的环节", "固态电解质"]}, seg_view, broker, DS)
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t("取第一个有行业观点的那个环节", ls.PATH_JUDGE in st["usable"])
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print("发给下游的形状")
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out = plan._state_out(st) # noqa: SLF001
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t("状态、子因、截止日都在", set(out) >= {"state", "why", "as_of", "usable", "missing",
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"reasons", "paths"})
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t("每一路都带自己的截止日与一句话说明",
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all(set(p) == {"path", "signal", "as_of", "why"} for p in out["paths"]))
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t("不发内部中间量（硬触发标记、出处原值不外泄）",
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all("hard" not in p and "refs" not in p for p in out["paths"]))
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t("没有状态时发 None，不硬拼一个空壳", plan._state_out(None) is None) # noqa: SLF001
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print("四态不改判决（收敛规则单调）")
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for state in (ls.STATE_STRONG, ls.STATE_HOLD, ls.STATE_UNKNOWN, ls.STATE_DOUBT):
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for verdict in ("候选", "关注", "仅展示"):
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r = ls.apply_to_card(verdict, state)
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assert r["verdict"] == verdict, (verdict, state, r)
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t("十二种组合逐个扫过，判决一次都没被四态改动", True)
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t("逻辑强化最多提前卡内序", ls.apply_to_card("关注", ls.STATE_STRONG)["rank_bonus"] == 1)
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t("逻辑存疑不把仅展示变成可执行",
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not ls.apply_to_card("仅展示", ls.STATE_DOUBT)["force_confirm"])
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t("候选加逻辑存疑是唯一需要新语义的一格：强制人工确认",
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ls.apply_to_card("候选", ls.STATE_DOUBT)["force_confirm"])
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print("计划日与数据日差一天")
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t("次日推算正确", plan._next_day("2026-09-03") == "2026-09-04") # noqa: SLF001
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t("认不出的日期原样返回，不抛异常", plan._next_day("不是日期") == "不是日期") # noqa: SLF001
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print("ALL OK — 四态随卡产出 / 缺席不断产 / 下游形状 / 判决不被改动 / 计划日推算 全部通过")
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if __name__ == "__main__":
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main()
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