research: 趋势末端识别(step62)，ext_run 单调区分但 PF 仍不过 1
用户指出很多一二类实际在趋势中途被识别而非末期，若真在末期即使有延迟也该走出 行情。用 step60 的线段顶点当标签找实时可算的区分特征。 ext_run（极值越过中枢边界几个 ATR）单调有效：5m 上四分位的命中率是 12.0/22.1/37.0/43.8%，PF 0.12/0.11/0.22/0.46。短延伸那批就是趋势中途被识别的， 占一半且 PF 仅 0.11。最佳组合 ext_run≥P75 且 div≥中位：命中率 47.9%、PF 0.54， 相对基准 28.7%/0.22 精度接近翻倍。 两个反直觉结果：背驰越强反而越差（div Q1 命中 17.0%/PF 0.13，Q4 37.5%/0.28）， 是对 MACD 面积判据的直接证伪；趋势级数无区分力（命中率 28.5/30.6/28.6/25.5% 基本持平）。 另修正一个我先前的猜测：以为引擎漏了缠论「趋势 vs 盘整」前提，实测该条件在 2504 笔上恒为 True——B1 要求 enter_bi.dir == leave_bi.dir == DOWN，中枢向下进 向下出本身就定义了它嵌在下跌趋势里，引擎已隐含强制，过滤器无从添加。 zs_count 也不可用，它是全局中枢序号而非趋势内序号。 结论：识别可优化且幅度不小，但不是瓶颈，瓶颈是入场时点。 Co-authored-by: Cursor <cursoragent@cursor.com>
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"""怎么把「趋势末端的一类」从「趋势中途的一类」里挑出来。
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§3.397 的关键数字：一类里命中线段顶点（真反转）的只有 30%，那 30% 即使带着
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8~9 根滞后也有 PF 0.70~0.83；**没命中的 70% 是 PF 0.08**。
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所以亏损几乎全部来自被误识别在趋势中途的那批 —— 用户的判断。
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于是问题变成：有没有**实时可算**的特征能把两批分开。
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**首要候选来自缠论本身**：一类买点要求的是**趋势背驰**，而趋势的定义是
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「至少两个同向连续的中枢」。引擎的 `find_all_bsp` 对**任意**中枢都发信号，
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完全没查这个前提 —— 单个盘整中枢上的「背驰」只是盘整背驰，本就不该当一类用。
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`fast_bsp.add_zone_ladder` 早就实现了这个判定（B4 上把 PF 2.72 提到 3.41），
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一类这边却没接。
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测的特征全部只用信号时刻及之前的数据：
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ladder 本中枢相对前一中枢是否同向推进（下降趋势要求 zg < 前一个 zd）
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zs_count 该中枢在本段里的序号，越大趋势越成熟
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div 离开段 MACD 面积 / 进入段面积，越小背驰越强
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ext_run 极值越过中枢边界多少个 ATR，越大越延伸
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atr_z 极值处波动率（§3.398）
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评判分两层：**能否提高命中线段顶点的概率**（检测器精度），
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以及**能否提高实际收益**（可交易性）。前者好后者不好也没用。
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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" / "step62_trend_end.feather"
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SL, SCALE_AT, RUNNER, RSTOP, MAXB = 2.0, 3.0, 8.0, 2.0, 48
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BASE_WIN, TOL = 200, 2
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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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atr = cdf["atr"].to_numpy(float)
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cl = cdf["close"].to_numpy(float)
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base = (pd.Series(atr).rolling(BASE_WIN, min_periods=50)
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.median().shift(1).to_numpy())
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# 标准答案：线段终点（未来函数，只当标签用，不进入任何过滤器）
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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 0 <= i < n:
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(seg_bot if sg.dir == Chan_SEG_DIR.DOWN
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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) -> bool:
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a = truth[d]
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k = int(np.searchsorted(a, i))
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return any(0 <= j < len(a) and abs(int(a[j]) - i) <= TOL
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for j in (k - 1, k))
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# 中枢阶梯：按可用顺序排好，才谈得上「相对前一个」
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zs_seq = sorted(bz, key=lambda z: to_i(z.bi_list[0].start_time))
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pos = {id(z): k for k, z in enumerate(zs_seq)}
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want = {Chan_BSP_TYPE.B1: ("B1", 1), Chan_BSP_TYPE.S1: ("S1", -1)}
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rec = []
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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 or b.zs 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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a = atr[i_ext]
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if not np.isfinite(a) or a <= 0 or not np.isfinite(base[i_ext]):
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continue
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zs = b.zs
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k = pos.get(id(zs))
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# 趋势成熟度：本中枢往前数，连续同向推进的中枢有几个。
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# 单看「相对前一个是否同向」没有区分力 —— B1 要求
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# enter_bi.dir == leave_bi.dir == DOWN，即中枢向下进、向下出，
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# 这本身就定义了它嵌在下跌趋势里，连续纯中枢自然逐级下移，
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# 实测该条件在 2504 笔上恒为 True。**引擎已隐含强制了「趋势」前提。**
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# 有区分力的是「连了几级」，那才是趋势成熟度。
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ladder_n = 0
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if k is not None:
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j = k
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while j > 0:
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cur, prv = zs_seq[j], zs_seq[j - 1]
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ok = (float(cur.zg) < float(prv.zd) if d == 1
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else float(cur.zd) > float(prv.zg))
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if not ok:
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break
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ladder_n += 1
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j -= 1
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enter_bi = zs.bi_list[0].pre if zs.bi_list else None
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ea = abs(float(enter_bi.macd_hist)) if enter_bi is not None else np.nan
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la = abs(float(b.bi.macd_hist))
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edge = float(zs.zd) if d == 1 else float(zs.zg)
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ext = float(b.klc.low if d == 1 else b.klc.high)
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rec.append({
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"sym": sym, "tf": tf, "type": name, "dir": d,
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"i_ext": i_ext, "i_sure": i_sure,
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"lag_bars": i_sure - i_ext,
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"hit": near(i_ext, d),
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"ladder_n": int(ladder_n),
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"div": la / ea if (ea and np.isfinite(ea) and ea > 0) else np.nan,
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"ext_run": abs(ext - edge) / a,
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"atr_z": a / base[i_ext],
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})
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if not rec:
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return None
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r = pd.DataFrame(rec)
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r = r[(r.i_sure < n - 2) & np.isfinite(atr[r.i_sure.values])
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& (atr[r.i_sure.values] > 0)].reset_index(drop=True)
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if r.empty:
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return None
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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": r.i_sure.values, "direction": r.dir.values}),
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[SL], [RUNNER], [MAXB], scale_at=SCALE_AT, runners=(RUNNER,),
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runner_stops=(RSTOP,))
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if len(res) != len(r):
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return None
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for c in ("g", "r", "c", "b"):
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r[c] = res[f"{cfg}_{c}"].to_numpy()
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r["atr_pct"] = atr[r.i_sure.values] / cl[r.i_sure.values]
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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 perf(g: pd.DataFrame) -> dict:
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from lib.exit_model import fee_of, taker_notional
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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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return {
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"笔数": len(g), "命中率": f"{g.hit.mean()*100:.1f}%",
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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) / np.sqrt(len(g))), 2),
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}
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def report(d: pd.DataFrame) -> None:
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for tf, x in d.groupby("tf"):
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print("\n" + "#" * 96)
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print(f"########## {tf} · {len(x)} 笔一类 ##########")
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print("\n【一】单特征对「命中线段顶点」的区分力（命中率基准 "
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f"{x.hit.mean()*100:.1f}%）")
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rows = []
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for nm, col, qs in [("背驰div", "div", 4), ("延伸ext_run", "ext_run", 4),
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("波动atr_z", "atr_z", 4), ("趋势级数ladder_n", "", 0)]:
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if not col:
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q = pd.cut(x["ladder_n"], [-1, 1, 2, 3, 999],
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labels=["级数≤1", "=2", "=3", "≥4"])
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for k, g in x.groupby(q, observed=True):
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if len(g) >= 30:
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rows.append({"分组": str(k), **perf(g)})
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continue
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y = x.dropna(subset=[col])
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if len(y) < 100:
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continue
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q = pd.qcut(y[col], qs,
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labels=[f"{nm}Q{i+1}" for i in range(qs)],
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duplicates="drop")
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for k, g in y.groupby(q, observed=True):
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if len(g) >= 30:
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rows.append({"分组": str(k), **perf(g)})
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print(pd.DataFrame(rows).to_string(index=False))
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print("\n【二】叠加过滤：延伸是唯一单调的特征，看叠加还能不能推上去")
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rows = [{"过滤器": "无（现状）", **perf(x)}]
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e75 = x["ext_run"].quantile(.75)
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e50 = x["ext_run"].quantile(.50)
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for nm, g in [
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(f"ext_run≥P50({e50:.1f})", x[x["ext_run"] >= e50]),
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(f"ext_run≥P75({e75:.1f})", x[x["ext_run"] >= e75]),
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(f"ext_run≥P75 且 级数≥3",
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x[(x["ext_run"] >= e75) & (x["ladder_n"] >= 3)]),
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(f"ext_run≥P75 且 div≥中位",
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x[(x["ext_run"] >= e75) & (x["div"] >= x["div"].median())]),
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]:
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if len(g) >= 30:
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rows.append({"过滤器": nm, **perf(g)})
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print(pd.DataFrame(rows).to_string(index=False))
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print("\n判读：命中率若被显著抬高，说明特征确实在区分「趋势末端 vs 中途」。"
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"\n但 PF 才是能不能做的判据 —— 命中率上去而 PF 不过 1，"
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"说明滞后仍然吃掉了全部。")
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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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