research: B4 时点干净，但实盘做的不是回测那批单（实盘口径 PF 0.73 vs 回测 4.21）
step63 因果回放：B4 的时点完全干净——100% 召回、100% 准时、零滞后， 一类那个把 PF 从 2.35 打到 0.92 的坑（§3.396）B4 没有。 但回放多产出 592 个全量口径里不存在的信号，PF 0.46 / t −5.03。 step68 按实盘口径复测（2001 根滚动窗口、每 500 根 init_stream、只做当根收盘， 逐行对齐 shadow_signal.py），并把三道滤网全测一遍： 无过滤 789 笔，额外占比 75%，PF 0.64 三道全开 195 笔，额外占比 74%，PF 0.73 滤网把成交量砍掉 75% 却几乎不改变额外信号占比——按同比例刷掉好的和坏的。 拆开看：回测里也有的 51 笔 PF 4.21/t+5.90，回测里没有的 144 笔 PF 0.26/t−6.21。 即回测报的 4.21 拿不到：那 51 笔要事后全量重算才能识别，实盘当下分不出来。 机制是重画——实时算出的中枢，数据变多后被修正掉。与 §3.396 同类： 一类错在时点，B4 错在存在性。 顺带闭掉一条待办：2001 根窗口与 2万→4万根增长窗口跑出完全相同的 789/197/592， 窗口左边界效应判为否。 限定：本轮是 5m/30m 而实盘跑 1m，需单独复验；同向过滤器被开了未来函数后门 （HTF 时间线用全量算），真实盘只会更差。 Co-authored-by: Cursor <cursoragent@cursor.com>
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"""按**实盘口径**重放 B4，并测三道过滤能否刷掉那批亏钱的额外信号。
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step63 的结论是一好一坏：
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好 时点干净 —— 100% 召回、100% 准时、零滞后（一类是 0% 准时、+15 根）
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坏 回放多出 592 个全量口径没有的信号，PF 0.46 / t −5.03，混合后 0.64 < 1
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但 step63 有两个口径问题，本脚本一并修掉：
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① 窗口不对。回测用全量 45000 根一次算完，step63 用 2 万涨到 4 万根的增长窗口，
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**而实盘用 2001 根滚动窗口、每 500 根 init_stream 拉回**（`shadow_signal.py`
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的 MAX_GROW）。三种口径的中枢结构都不一样。这也是 HANDOFF 里挂着的
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「回测用全量历史建中枢、实盘用 2000 根窗口」那条待办。
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顺带：窗口封顶后单步成本恒定，不再是 step63 那个平方级（143ms@2万根 ->
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292ms@4万根），所以本脚本快得多。
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② 没测过滤。step63 跑的是裸信号，而实盘有三道滤网。ATR 门控已单独测过——
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它刷掉 7.9% 的额外信号却刷掉 10.7% 的好信号，PF 纹丝不动。剩下两道要测。
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**一处刻意的简化，方向是保守的**：大级别分型时间线用全量历史算（真实盘的 HTF
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也会重画）。这等于**给同向过滤器开了未来函数的后门**。若连这样都刷不掉额外信号，
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结论只会更强。
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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" / "step68_live_window.feather"
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SL, SCALE_AT, RUNNER, RSTOP, MAXB = 2.0, 3.0, 8.0, 2.0, 48
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WIN, MAX_GROW, GATE_BP = 2001, 500, 8.0
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def _ladder(zones: pd.DataFrame) -> pd.DataFrame:
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z = zones.copy()
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pg, pdn = z["zg"].shift(), z["zd"].shift()
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z["z_above"], z["z_below"] = z["zd"] > pg, z["zg"] < pdn
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z["zone_i"] = np.arange(len(z))
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return z
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def replay(sym: str, ltf: str, htf: str, rows: int,
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steps: int) -> pd.DataFrame | None:
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from chanlun import TF_DF
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from chanlun.analysis.fast_bsp import (
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ensure_timestamp,
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find_fast_bsp3,
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zones_from_zs_list,
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)
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from lib.data import fetch_ohlcv
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from lib.fx_signal import extract_fx_signals, signals_to_frame
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from lib.nested_bsp import attach_htf_context, htf_fx_timeline
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try:
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df = fetch_ohlcv(f"{sym}/USDT:USDT", ltf, rows)
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if df is None or len(df) < WIN + steps + 100:
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return None
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df = df.iloc[-(WIN + steps):].reset_index(drop=True)
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full = TF_DF(df, 1, ltf)
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cdf = ensure_timestamp(full.dataframe)
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zs_full = full.cal_bi_zs_list_pure(full.bi_list)
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if not zs_full:
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return None
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sig_full = find_fast_bsp3(cdf, zones_from_zs_list(zs_full, cdf))
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full_keys = {(int(r.entry_idx), int(r.direction))
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for r in sig_full.itertuples()} if not sig_full.empty \
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else set()
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# 大级别分型时间线：全量算（见模块 docstring 的「刻意简化」）
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dh = fetch_ohlcv(f"{sym}/USDT:USDT", htf, rows)
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tl = None
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if dh is not None and len(dh) > 500:
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ch = TF_DF(dh, 1, htf)
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tl = htf_fx_timeline(
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signals_to_frame(extract_fx_signals(ch, ch.dataframe)),
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ch.dataframe)
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rec: dict[tuple, dict] = {}
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chan = None
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anchor = 0
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for i in range(WIN, len(df)):
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# 实盘的窗口纪律：2001 根起，长过 MAX_GROW 就 init_stream 拉回
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if chan is None or (i - anchor) >= MAX_GROW:
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w = df.iloc[i - WIN + 1:i + 1].copy()
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chan = TF_DF(w, 1, ltf)
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chan.init_stream(w, 1, ltf)
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anchor = i
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else:
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chan.append_bar(df.iloc[i])
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try:
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zl = chan.cal_bi_zs_list_pure(chan.bi_list)
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if not zl:
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continue
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sub = ensure_timestamp(chan.dataframe)
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z = _ladder(zones_from_zs_list(zl, sub))
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if z.empty:
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continue
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s = find_fast_bsp3(sub, z)
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if s is None or s.empty:
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continue
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last = len(sub) - 1
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s = s[s["entry_idx"].astype(int) == last] # 实盘只做当根
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if s.empty:
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continue
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if "zone_i" in s.columns:
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s = s.merge(z[["zone_i", "z_above", "z_below"]],
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on="zone_i", how="left")
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if tl is not None:
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s = attach_htf_context(s, sub, tl, "h1")
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except Exception: # noqa: BLE001
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continue
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for r in s.itertuples():
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k = (i, int(r.direction))
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if k in rec:
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continue
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push = getattr(r, "z_above" if r.direction == 1
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else "z_below", None)
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ag = getattr(r, "h1_agree", 0)
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rec[k] = {
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"sym": sym, "entry_idx": i, "direction": int(r.direction),
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"in_full": (i, int(r.direction)) in full_keys,
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"ladder_ok": int(bool(pd.notna(push) and bool(push))),
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"h1_agree": int(ag) if pd.notna(ag) else 0,
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}
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if not rec:
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return None
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r = pd.DataFrame(list(rec.values()))
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from lib.exit_model import cfg_name, walk_exits
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atr = cdf["atr"].to_numpy(float)
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cl = cdf["close"].to_numpy(float)
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r = r[(r.entry_idx < len(cdf) - 2)
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& np.isfinite(atr[r.entry_idx.values])
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& (atr[r.entry_idx.values] > 0)].reset_index(drop=True)
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if r.empty:
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return None
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res = walk_exits(cdf, pd.DataFrame({
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"entry_idx": r.entry_idx.values,
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"direction": r.direction.values}), [SL], [RUNNER], [MAXB],
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scale_at=SCALE_AT, runners=(RUNNER,), runner_stops=(RSTOP,))
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cfg = cfg_name(SL, RUNNER, MAXB, 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"):
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r[c] = res[f"{cfg}_{c}"].to_numpy()
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r["atr_pct"] = atr[r.entry_idx.values] / cl[r.entry_idx.values]
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r["gate_ok"] = (r.atr_pct * 1e4 >= GATE_BP).astype(int)
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r["pass_all"] = ((r.h1_agree == 1) & (r.ladder_ok == 1)
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& (r.gate_ok == 1)).astype(int)
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return r
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except Exception as e: # noqa: BLE001
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print(f" {sym} 失败: {type(e).__name__}: {e}", flush=True)
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return None
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def perf(g: pd.DataFrame) -> dict | None:
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from lib.exit_model import fee_of, taker_notional
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if len(g) < 20:
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return None
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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), "胜率": f"{(net > 0).mean()*100:.1f}%",
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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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print("\n" + "=" * 92)
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print("【一】实盘窗口下还有多少额外信号")
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print("=" * 92)
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print(f"回放信号 {len(d)} · 其中全量口径也有 {int(d.in_full.sum())} · "
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f"**额外 {int((~d.in_full).sum())}**")
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print(f"（step63 的增长窗口口径：197 / 592）")
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print("\n" + "=" * 92)
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print("【二】三道过滤能不能刷掉额外信号 —— 这决定实盘是否在亏钱")
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print("=" * 92)
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rows = []
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for nm, m in [("① 无过滤", None), ("② 仅同向", d.h1_agree == 1),
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("③ 仅阶梯", d.ladder_ok == 1),
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("④ 仅ATR门控", d.gate_ok == 1),
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("⑤ 三道全开(实盘口径)", d.pass_all == 1)]:
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x = d if m is None else d[m]
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if x.empty:
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continue
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ex, bt = x[~x.in_full], x[x.in_full]
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row = {"过滤": nm, "留下": len(x),
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"额外占比": f"{(~x.in_full).mean()*100:.0f}%"}
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for lab, g in [("全部", x), ("额外", ex), ("回测口径", bt)]:
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s = perf(g)
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row[f"{lab}PF"] = "—" if s is None else s["PF"]
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row[f"{lab}n"] = len(g)
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rows.append(row)
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print(pd.DataFrame(rows).to_string(index=False))
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print("\n" + "=" * 92)
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print("【三】实盘口径（三道全开）的完整表现")
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print("=" * 92)
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rows = []
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for nm, g in [("实盘会做的全部", d[d.pass_all == 1]),
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(" 其中额外的", d[(d.pass_all == 1) & ~d.in_full]),
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(" 其中回测也有的", d[(d.pass_all == 1) & d.in_full])]:
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s = perf(g)
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if s:
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rows.append({"分组": nm, **s})
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print(pd.DataFrame(rows).to_string(index=False))
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print("""
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判读：「实盘会做的全部」PF > 1 -> 实盘安全，额外信号被滤网挡住了
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PF < 1 -> **实盘在做一批回测里不存在、且亏钱的信号**，要立刻处理""")
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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("--ltf", default="5m")
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ap.add_argument("--htf", default="30m")
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ap.add_argument("--rows", type=int, default=45_000)
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ap.add_argument("--steps", type=int, default=20_000)
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ap.add_argument("--workers", type=int, default=5)
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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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print(f"[实盘口径回放] {len(syms)} 币 × {args.steps} 根 · "
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f"{WIN} 根滚动窗口 / 每 {MAX_GROW} 根重建\n", flush=True)
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parts = []
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with ProcessPoolExecutor(max_workers=args.workers) as ex:
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fut = {ex.submit(replay, s, args.ltf, args.htf, args.rows,
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args.steps): s for s in syms}
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for i, f in enumerate(as_completed(fut), 1):
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r = f.result()
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print(f" [{i}/{len(syms)}] {fut[f]} "
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f"{0 if r is None else len(r)} 个信号", 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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