research: 因果回放推翻一类反手(step59)，线段顶点验证识别有效但不够(step60)
step59 逐根回放：init_stream 预热后逐根 append_bar，每根重算笔中枢与 bsp， 记录信号首现根并按它入场。信号身份用「类型+极值KLC时刻」而非 sure_time， 后者正是会被重画的字段。必须逐根，分段重建等于多给信息。 结果把 §3.395 推翻了：召回 100%、幻影 0，即存在性是因果的；但首现根比全量 sure_time 晚中位 15 根、P90 31 根，0% 能准时拿到。按真实首现根入场， B1 反手 PF 2.35→0.92、S1 2.75→1.30，t 0.20/0.64 完全不显著。 教训：sure_time 是引擎事后标注的确认时刻，不等于可执行时刻。任何拿它当 entry 的回测都要先过逐根回放。 step60 用线段终点当标准答案验证识别本身。对照组取所有同向笔端点——B1 按构造 就长在笔低点上，不设这个基准任何绝对命中率都无法解读。5m 上 B1 命中 30.3% 对基准 15.0%，提升 2.02 倍，S1 1.81 倍；B3/S3 恰为 0%，符合三类长在趋势 中段的预期，两者互为标签有效性旁证。 所以识别是对的，但精度只有 30%，且那 70% 噪声 PF 只有 0.08。更关键的是即使 用未来函数把精度提到 100%，命中组也只有 PF 0.70~0.83，仍不赚钱——因为入场价 已在结构底上方 2.90 ATR。一类线三层逐层否定后到此为止。 Co-authored-by: Cursor <cursoragent@cursor.com>
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jackyu66git
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"""一类买卖点到底有没有找到真正的趋势反转点？用线段顶点当标准答案。
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用户提出：线段的终点就是该级别的趋势反转点，虽然它是未来函数，但可以拿来当
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**标签**验证检测器，而不是拿来交易。这能把两件事分开：
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检测器对不对 B1 是否真的落在趋势反转底上
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能不能交易 step59 已证否（实盘首现比 sure_time 晚中位 15 根，PF 塌到 0.92）
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若检测器对而只是慢，那问题是延迟，还有救；若检测器本身就没找到反转点，
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这条线整个是死的。
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⚠️ **对照组是这个测试的全部意义**。B1 按构造就长在笔的低点上，而笔低点本来
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就有一定概率撞上线段底。所以要问的不是「B1 命中率多少」，而是
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**「在所有同向笔端点里，B1 这个标签把命中率提高了多少倍」**。
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没有这个基准，任何绝对数字都可以随便解读。
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（同样的坑 fast_bsp 文档头踩过：回抽极值命中笔端点 19.3%，看着不低，
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但随机基准是 22%，其实是负贡献。）
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"""
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from __future__ import annotations
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import argparse
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import sys
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import warnings
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from concurrent.futures import ProcessPoolExecutor, as_completed
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from pathlib import Path
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import numpy as np
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import pandas as pd
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warnings.filterwarnings("ignore")
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HERE = Path(__file__).resolve().parent
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sys.path.insert(0, str(HERE))
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sys.path.insert(0, str(HERE.parent))
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OUT = HERE / "out" / "step60_seg_truth.feather"
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SL, SCALE_AT, RUNNER, RSTOP, MAXB = 2.0, 3.0, 8.0, 2.0, 48
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TOL = [0, 1, 2, 3, 5]
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def collect(sym: str, tf: str, rows: int) -> pd.DataFrame | None:
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from chanlun import TF_DF
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from chanlun.core.ChanEnum import Chan_BSP_TYPE, Chan_SEG_DIR
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from lib.data import fetch_ohlcv
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from lib.exit_model import cfg_name, walk_exits
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try:
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df = fetch_ohlcv(f"{sym}/USDT:USDT", tf, rows)
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if df is None or len(df) < 5_000:
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return None
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chan = TF_DF(df, 1, tf, lean=False)
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cdf = chan.dataframe
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bz = chan.cal_bi_zs_list_pure(chan.bi_list)
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if not bz:
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return None
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bsp = chan.find_all_bsp(chan.bi_list, bz) or []
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dser = pd.to_datetime(cdf["date"])
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if dser.dt.tz is not None:
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dser = dser.dt.tz_localize(None)
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didx = pd.DatetimeIndex(dser)
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n = len(cdf)
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def to_i(ts) -> int:
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t = pd.Timestamp(ts)
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return int(didx.searchsorted(t.tz_localize(None) if t.tz else t))
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# ---- 标准答案：线段终点。下降线段终点=真底，上升线段终点=真顶 ----
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seg_bot, seg_top = [], []
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for sg in getattr(chan, "seg_list", []) or []:
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if sg.end_time is None:
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continue
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i = to_i(sg.end_time)
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if not (0 <= i < n):
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continue
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(seg_bot if sg.dir == Chan_SEG_DIR.DOWN else seg_top).append(i)
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if not seg_bot or not seg_top:
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return None
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truth = {1: np.array(sorted(seg_bot)), -1: np.array(sorted(seg_top))}
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def near(i: int, d: int, tol: int) -> bool:
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a = truth[d]
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k = int(np.searchsorted(a, i))
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for j in (k - 1, k):
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if 0 <= j < len(a) and abs(int(a[j]) - i) <= tol:
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return True
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return False
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rec = []
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# ---- 对照组：所有笔端点。B1 本就长在笔低点上，基准必须同源 ----
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for bi in chan.bi_list:
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if not getattr(bi, "is_sure", False) or bi.end_klc is None:
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continue
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d = 1 if str(bi.dir).endswith("DOWN") else -1 # 下降笔终点=低点
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i = to_i(bi.end_klc.end_time)
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if not (0 <= i < n):
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continue
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rec.append({"kind": "笔端点", "dir": d, "i_ext": i, "i_sure": -1})
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want = {Chan_BSP_TYPE.B1: ("B1", 1), Chan_BSP_TYPE.S1: ("S1", -1),
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Chan_BSP_TYPE.B3: ("B3", 1), Chan_BSP_TYPE.S3: ("S3", -1)}
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for b in bsp:
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tag = want.get(b.type)
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if tag is None or b.sure_time is None:
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continue
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name, d = tag
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i_ext, i_sure = to_i(b.klc.end_time), to_i(b.sure_time)
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if not (0 <= i_ext < n and 0 <= i_sure < n):
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continue
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rec.append({"kind": name, "dir": d, "i_ext": i_ext,
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"i_sure": i_sure})
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r = pd.DataFrame(rec)
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for tol in TOL:
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r[f"hit{tol}"] = [near(i, d, tol)
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for i, d in zip(r.i_ext, r.dir)]
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# 命中线段顶点的那批一类，按原方向（抄底）做能不能赚
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atr = cdf["atr"].to_numpy(float)
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cl = cdf["close"].to_numpy(float)
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sig = r[(r.kind.isin(["B1", "S1"])) & (r.i_sure >= 0)
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& (r.i_sure < n - 2)].copy()
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sig = sig[np.isfinite(atr[sig.i_sure.values])
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& (atr[sig.i_sure.values] > 0)]
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if not sig.empty:
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cfg = cfg_name(SL, RUNNER, MAXB, RSTOP)
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res = walk_exits(cdf, pd.DataFrame({
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"entry_idx": sig.i_sure.values,
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"direction": sig.dir.values}), [SL], [RUNNER], [MAXB],
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scale_at=SCALE_AT, runners=(RUNNER,), runner_stops=(RSTOP,))
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if len(res) == len(sig):
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for c in ("g", "r", "c"):
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sig[c] = res[f"{cfg}_{c}"].to_numpy()
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sig["atr_pct"] = (atr[sig.i_sure.values]
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/ cl[sig.i_sure.values])
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r = r.merge(sig[["i_ext", "kind", "g", "r", "c", "atr_pct"]],
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on=["i_ext", "kind"], how="left")
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r["sym"], r["tf"] = sym, tf
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return r
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except Exception as e: # noqa: BLE001
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print(f" {sym} {tf} 失败: {type(e).__name__}: {e}", flush=True)
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return None
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def report(d: pd.DataFrame) -> None:
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from lib.exit_model import fee_of, taker_notional
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for tf, x in d.groupby("tf"):
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print("\n" + "#" * 92)
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print(f"########## {tf} · 线段顶点作为标准答案 ##########")
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print("\n【命中率】i_ext 落在同向线段终点 ±tol 根内的比例")
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rows = []
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for kind in ["笔端点", "B1", "S1", "B3", "S3"]:
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g = x[x.kind == kind]
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if len(g) < 30:
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continue
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row = {"信号": kind, "样本": len(g)}
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for tol in TOL:
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row[f"±{tol}根"] = f"{g[f'hit{tol}'].mean()*100:.1f}%"
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rows.append(row)
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t = pd.DataFrame(rows)
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print(t.to_string(index=False))
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base = x[x.kind == "笔端点"]
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print("\n【提升倍数】相对「所有同向笔端点」这个基准。"
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"≈1 就是没有信息量")
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rows = []
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for kind in ["B1", "S1", "B3", "S3"]:
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g = x[x.kind == kind]
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if len(g) < 30:
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continue
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row = {"信号": kind}
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for tol in TOL:
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b = base[base.dir.isin(g.dir.unique())][f"hit{tol}"].mean()
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row[f"±{tol}根"] = (round(g[f"hit{tol}"].mean() / b, 2)
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if b > 0 else np.nan)
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rows.append(row)
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print(pd.DataFrame(rows).to_string(index=False))
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if "g" not in x.columns:
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continue
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print("\n【命中 vs 未命中】一类按原方向（抄底/摸顶）做的表现，"
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"tol=±2 根")
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y = x[x.kind.isin(["B1", "S1"])].dropna(subset=["g"]).copy()
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if len(y) < 60:
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continue
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rows = []
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for hit, g in y.groupby(y.hit2):
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net = g.g.values - fee_of(g.r.values, g.c.values)
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gR = g.g.values / (SL * g.atr_pct.values)
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tn = taker_notional(g.r.values, g.c.values)
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w, o = net[net > 0].sum(), -net[net <= 0].sum()
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rows.append({
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"命中线段顶点": "是" if hit else "否", "笔数": len(g),
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"胜率": f"{(net > 0).mean()*100:.1f}%",
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"毛R": round(gR.mean(), 3),
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"PF": round(w / o, 2) if o > 0 else np.inf,
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"余量bp": round(net.mean() / tn.mean() * 1e4, 2),
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"t值": round(gR.mean() / (gR.std(ddof=1)
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/ np.sqrt(len(g))), 2),
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})
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print(pd.DataFrame(rows).to_string(index=False))
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print("\n判读：若「命中」那组按原方向做显著为正，说明检测器是对的、"
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"只是掺了太多噪声，\n值得找实时可判的过滤器；若两组都为负，"
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"说明即使真站在线段底上，\n这个入场时点也已经太晚了。")
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def main() -> None:
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ap = argparse.ArgumentParser()
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ap.add_argument("--symbols", default="BTC,ETH,SOL,LINK,DOGE")
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ap.add_argument("--tfs", default="5m,15m")
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ap.add_argument("--rows", type=int, default=200_000)
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ap.add_argument("--workers", type=int, default=3)
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ap.add_argument("--reuse", action="store_true")
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args = ap.parse_args()
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if args.reuse and OUT.exists():
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report(pd.read_feather(OUT))
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return
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syms = [s.strip() for s in args.symbols.split(",")]
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tfs = [t.strip() for t in args.tfs.split(",")]
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parts = []
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with ProcessPoolExecutor(max_workers=args.workers) as ex:
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fut = {ex.submit(collect, s, t, args.rows): (s, t)
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for s in syms for t in tfs}
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for i, f in enumerate(as_completed(fut), 1):
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r = f.result()
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s, t = fut[f]
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print(f" [{i}/{len(fut)}] {s} {t} "
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f"{0 if r is None else len(r)}", flush=True)
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if r is not None:
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parts.append(r)
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if not parts:
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print("无结果")
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return
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d = pd.concat(parts, ignore_index=True)
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OUT.parent.mkdir(exist_ok=True)
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d.to_feather(OUT)
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report(d)
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if __name__ == "__main__":
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main()
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