jackyu66git
78708ddeee
用户提出一二类长在极值点、该区域波动大、这可能是确认慢的原因。拆成三条 子命题分别测。 ① 成立：5m 上 B1 的 atr_z 中位 1.224、74% 高于基准，而所有笔端点的基准恰好 落在 50%（构造正确的旁证）。但二类是镜像——B2/S2 的 atr_z 中位 0.94~0.97、 仅四成高于基准，长在低波动区，因为它是首轮反转冲动之后的回抽。所以 「一二类都在极值点」对一类成立、对二类不成立。 ② 只沾边：corr(atr_z, lag_bars) 仅 +0.10，滞后中位在四个 atr_z 分档里是 8/7/8/9 根，几乎不动。滞后是结构性的，笔要等分型确认，与波动率基本无关。 ③ 回落属实但不是死因：入场后 48 根平均 ATR 是入场时的 0.896，目标确实按虚高 ATR 定、绝对价格高估约 10%。但一类只有 9.4% 走到 3ATR 减仓（三类 30.7%）， 81.3% 直接止损、超时仅 17.9%。若死因是目标够不着，超时占比该很高。所以一类 不是走不动，是入场后立刻反向——与 §3.397 的诊断一致。
253 lines
11 KiB
Python
253 lines
11 KiB
Python
"""极值点的波动率：它是不是既拖慢了确认，又让目标够不着？
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用户的观察：一二类买卖点长在极值点上，那个区域波动天然很大，这可能正是确认
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慢的原因。
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这个假设若成立会同时解释两件事，而且机制不同：
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确认慢 波动大 -> 分型/笔要更多根才稳定下来 -> sure_time 更晚
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赚不到 ATR 被造成极值的那根插针抬高 -> 止损 2×虚高ATR 其实宽松，
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但目标 3/8 ATR 变得够不着，因为入场后波动率会均值回复下来
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第二条尤其要紧：它意味着交易不是「被打掉」，而是「永远走不到目标」，
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与 §3.397 判定的「趋势继续」是**不同的失败模式**，应对办法也不同
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（该换波动率口径，而不是换方向）。
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三个量：
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atr_z 极值处 ATR / 该点之前 200 根的中位 ATR —— 是否真的偏高
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atr_fwd 入场后 48 根的平均 ATR / 入场时 ATR —— 是否均值回复
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与滞后相关 atr_z 高的信号，lag_bars 是否更长
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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" / "step61_vol.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 = 200
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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
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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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# 基准 ATR 只用**该点之前**的窗口，含当根会把要检验的那根插针算进去
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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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rec = []
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# 对照组：所有笔端点，与 §3.397 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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i = to_i(bi.end_klc.end_time)
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if not (0 <= i < n) or not np.isfinite(base[i]) or base[i] <= 0:
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continue
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rec.append({"kind": "笔端点", "dir": 0, "i_ext": i, "i_sure": -1,
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"atr_z": atr[i] / base[i], "lag_bars": -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.B2: ("B2", 1), Chan_BSP_TYPE.S2: ("S2", -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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if not np.isfinite(base[i_ext]) or base[i_ext] <= 0:
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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, "atr_z": atr[i_ext] / base[i_ext],
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"lag_bars": i_sure - i_ext})
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r = pd.DataFrame(rec)
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# 入场后的实现波动率：目标够不够得着，取决于入场**之后**的 ATR
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ok = r.i_sure >= 0
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fwd = np.full(len(r), np.nan)
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for k, i in zip(np.where(ok)[0], r.i_sure[ok].values):
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j = min(int(i) + 1 + MAXB, n)
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if j > int(i) + 1 and atr[int(i)] > 0:
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fwd[k] = np.nanmean(atr[int(i) + 1:j]) / atr[int(i)]
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r["atr_fwd"] = fwd
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sig = r[ok & (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", "b"):
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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", "b",
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"atr_pct"]], 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" + "#" * 96)
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print(f"########## {tf} · 极值点的波动率 ##########")
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print("\n【一】极值处 ATR 是不是真的偏高（atr_z = 当点ATR / 前200根中位ATR）")
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rows = []
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for kind in ["笔端点", "B1", "S1", "B2", "S2", "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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rows.append({
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"信号": kind, "样本": len(g),
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"atr_z中位": round(g.atr_z.median(), 3),
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"P75": round(g.atr_z.quantile(.75), 3),
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"P90": round(g.atr_z.quantile(.90), 3),
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"高于基准占比": f"{(g.atr_z > 1).mean()*100:.0f}%",
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})
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print(pd.DataFrame(rows).to_string(index=False))
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print("\n【二】波动大是不是确认更慢（按 atr_z 四分位看 lag_bars）")
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y = x[x.kind.isin(["B1", "S1"]) & (x.lag_bars >= 0)].copy()
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if len(y) >= 100:
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q = pd.qcut(y.atr_z, 4, labels=["Q1低", "Q2", "Q3", "Q4高"],
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duplicates="drop")
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print(pd.DataFrame([{
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"atr_z档": k, "笔数": len(g),
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"atr_z中位": round(g.atr_z.median(), 2),
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"滞后中位": round(g.lag_bars.median(), 1),
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"滞后均值": round(g.lag_bars.mean(), 1),
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} for k, g in y.groupby(q, observed=True)]).to_string(index=False))
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c = np.corrcoef(y.atr_z, y.lag_bars)[0, 1]
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print(f" 相关系数 corr(atr_z, lag_bars) = {c:+.3f}")
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print(" 正相关支持「波动大 -> 确认慢」；接近 0 则该假设不成立。")
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print("\n【三】入场后波动率是否回落（atr_fwd = 后48根平均ATR / 入场ATR）")
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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)].dropna(subset=["atr_fwd"])
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if len(g) < 30:
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continue
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rows.append({
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"信号": kind, "样本": len(g),
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"atr_fwd中位": round(g.atr_fwd.median(), 3),
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"回落占比(<1)": f"{(g.atr_fwd < 1).mean()*100:.0f}%",
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})
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print(pd.DataFrame(rows).to_string(index=False))
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print(" <1 = 入场后波动率比入场那刻低，则按入场ATR定的 3/8 ATR 目标"
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"\n 在绝对价格上被高估，会系统性够不着。")
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if "g" not in x.columns:
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continue
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print("\n【四】目标够不着的证据：出场原因分布（一类，按原方向）")
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z = x[x.kind.isin(["B1", "S1"])].dropna(subset=["g"])
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if len(z) >= 60:
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print((z.r.value_counts(normalize=True) * 100).round(1)
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.to_frame("占比%").to_string())
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print(f" 中位持仓 {z.b.median():.0f} 根 / 上限 {MAXB} 根")
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print("\n【五】按 atr_z 分档看一类表现（原方向）")
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q = pd.qcut(z.atr_z, 4, labels=["Q1低", "Q2", "Q3", "Q4高"],
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duplicates="drop")
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rows = []
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for k, g in z.groupby(q, observed=True):
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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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"atr_z档": k, "笔数": len(g),
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"atr_z中位": round(g.atr_z.median(), 2),
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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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})
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print(pd.DataFrame(rows).to_string(index=False))
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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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