jackyu66gitandCursor
2a59dcb9dc
承接上一条：既然同时开的必然同向，那关键是这批交易比孤立的好还是坏。 11 币 / 1161 笔 / 实盘口径： 真孤立（±5 分钟内无同伴） 763 笔 胜率 65.5% 毛R 0.832 PF 2.83 错开：5 分钟内但不同分钟 240 笔 胜率 82.1% 毛R 1.443 PF 7.31 同一分钟撞在一起 158 笔 胜率 63.3% 毛R 0.777 PF 2.20 必须把两者分开——结论相反，混在一起会得出错误判断。错开的是全样本最好的 一档，同分钟的反而略差于孤立组。机制上：错开 = 行情从某个币扩散开，后发是 对先发的确认；同分钟 = 全市场同时被一个冲击打中，即追高。 簇级复核（±5 分钟合一簇，排除重复计数）：多笔簇簇均毛R 1.180 vs 单笔簇 0.832，簇级 R 夏普 0.848 vs 0.459，结论不是重复计数撑起来的。多笔簇内 全赢 59.5%、全输 10.1%，簇内风险不可分散但偏度有利。 集中度上两类没差别（整簇同向 99.4%），差别纯在收益。所以「限制最多 N 个 并发仓位」把两类一视同仁是错的，它们期望收益差 1.9 倍。 注意：同分钟 vs 错开是看过数据后才划的切法，不是事先定的，208 天 158 个簇 容易切出噪声。当仓位规则用之前必须换一段时间验证。目前只有「扎堆整体更好」 是稳的（簇级也成立）。 Co-authored-by: Cursor <cursoragent@cursor.com>
242 lines
10 KiB
Python
242 lines
10 KiB
Python
"""Step 48：11 个币的开仓时刻到底挤不挤 —— 保证金要备多少、分散是不是真的。
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用户问：11 个币对开仓时间差距有多少，不是同时开吧。
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这个数决定三件事：
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1. 保证金峰值。若真是同时开 11 个，按每笔固定风险算的名义额会叠到很高
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2. "多币 = 分散"是否成立。§3.31 已知同时发生的信号 92.7% 同向——
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那种情况下多开的仓不是分散，是加杠杆
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3. 资金利用率。若大部分时间空仓，那"资金量不够"就不是真约束
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口径与实盘一致：深色（同向 ∧ 阶梯）∧ ATR≥8bp，持仓按回测实际出场根数
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（1m 上 1 根 = 1 分钟），不是一律按 48 根上限。
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"""
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from __future__ import annotations
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import argparse
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import os
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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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for v in ("OMP_NUM_THREADS", "OPENBLAS_NUM_THREADS", "MKL_NUM_THREADS"):
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os.environ.setdefault(v, "1")
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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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pd.set_option("display.width", 320)
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SL, SCALE_AT, RUNNER, RSTOP, MAXB = 2.0, 3.0, 8.0, 2.0, 48
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GATE_BP = 8.0
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OUT = HERE / "out" / "step48_signal_times.feather"
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def collect(sym: str, rows: int) -> pd.DataFrame | None:
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import warnings as _w
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_w.filterwarnings("ignore")
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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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from chanlun import TF_DF
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from chanlun.analysis.fast_bsp import (
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add_zone_ladder, attach_htf_agree, attach_zone_ladder,
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build_htf_zones, find_fast_bsp3, htf_fx_timeline,
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)
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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", "1m", rows)
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if df is None or len(df) < 50_000:
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return None
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chan = TF_DF(df, 1, "1m", lean=True)
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cdf = chan.dataframe
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zones = build_htf_zones(cdf, "1m", chan=chan)
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if zones.empty:
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return None
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zl = add_zone_ladder(zones.reset_index(drop=True))
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sig = find_fast_bsp3(cdf, zl)
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if sig.empty:
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return None
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dh = fetch_ohlcv(f"{sym}/USDT:USDT", "5m", 10 ** 9)
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ch = TF_DF(dh, 1, "5m", lean=True)
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sig = attach_zone_ladder(attach_htf_agree(sig, cdf, htf_fx_timeline(ch, ch.dataframe)), zl)
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res = walk_exits(cdf, sig, [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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idx = sig["entry_idx"].to_numpy().astype(int)
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atr = cdf["atr"].to_numpy(float)[idx]
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close = cdf["close"].to_numpy(float)[idx]
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return pd.DataFrame({
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"sym": sym,
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"date": cdf["date"].to_numpy()[idx],
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"dir": sig["direction"].to_numpy(),
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"atr_bp": atr / close * 1e4,
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"htf": sig["htf_agree"].to_numpy(),
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"lad": sig["ladder_ok"].to_numpy(),
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"hold": res[f"{cfg}_b"].to_numpy(),
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# 毛利与出场原因：扎堆的交易赚不赚钱要靠这几列，别只存时刻
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"g": res[f"{cfg}_g"].to_numpy(),
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"r": res[f"{cfg}_r"].to_numpy(),
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"c": res[f"{cfg}_c"].to_numpy(),
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"atr_pct": atr / close,
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})
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except Exception as e:
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print(f" {sym} 失败: {e!r}", flush=True)
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return None
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def analyse(d: pd.DataFrame, label: str) -> None:
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d = d.sort_values("date").reset_index(drop=True)
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span = (d.date.max() - d.date.min()).total_seconds() / 86400
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print("\n" + "=" * 96)
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print(f"########## {label}：{d.sym.nunique()} 币 / {len(d)} 笔 / 跨 {span:.0f} 天 ##########")
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print(f"组合 {len(d) / span:.2f} 笔每天 —— 平均每 {span * 24 / len(d):.1f} 小时才 1 笔")
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g = d.date.diff().dt.total_seconds().dropna() / 60
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print("\n相邻两笔间隔（分钟）")
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qs = [(q, g.quantile(q)) for q in (0.05, 0.10, 0.25, 0.50, 0.75, 0.90)]
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print(" " + " ".join(f"{int(q*100)}%:{v:.0f}" for q, v in qs))
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for lab, m in (("同一分钟", g == 0), ("≤5 分钟", g <= 5),
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("≤60 分钟", g <= 60), (">1 小时", g > 60)):
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print(f" {lab:<9}{m.sum():>5} 次 {m.mean()*100:>5.1f}%")
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# 并发持仓：按回测实际持仓根数
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ev = []
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for t, h in zip(d.date, d.hold):
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ev.append((t, 1))
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ev.append((t + pd.Timedelta(minutes=float(h)), -1))
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ev.sort()
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cur, prev, dur = 0, None, {}
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for t, delta in ev:
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if prev is not None and t > prev:
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dur[cur] = dur.get(cur, 0) + (t - prev).total_seconds()
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cur += delta
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prev = t
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tot = sum(dur.values())
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print("\n同时持仓数的时间占比")
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for k in sorted(dur):
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if dur[k] / tot > 0.0005:
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print(f" {k:>2} 个: {dur[k]/tot*100:>5.1f}%")
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print(f" 最大并发 {max(dur)} 个；有仓位的时间仅占 {(1-dur.get(0,0)/tot)*100:.1f}%")
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# 同时开仓时的方向一致性——这决定"多币"到底是分散还是加杠杆
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same = d.groupby(d.date)["dir"].agg(["count", "nunique"])
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multi = same[same["count"] > 1]
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if len(multi):
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agree = (multi["nunique"] == 1).mean()
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print(f"\n同一分钟出现多笔的时刻 {len(multi)} 个，其中方向完全一致的占 "
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f"{agree*100:.1f}% —— 这部分不是分散，是同一笔押注被拆成几个币")
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if "g" in d.columns:
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profit_by_cluster(d)
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def profit_by_cluster(d: pd.DataFrame) -> None:
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"""扎堆的交易赚不赚钱。
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必须把「同一分钟」和「错开几分钟」分开看——8 币预试中两者结论相反：
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错开的是全样本最好的一档，同分钟的反而略差于孤立。混在一起会得出错误结论。
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"""
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from lib.exit_model import fee_of, taker_notional
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d = d.sort_values("date").reset_index(drop=True)
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t = d.date.values.astype("datetime64[m]").astype(np.int64)
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n0 = np.searchsorted(t, t, "right") - np.searchsorted(t, t, "left")
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n5 = np.searchsorted(t, t + 5, "right") - np.searchsorted(t, t - 5, "left")
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net = d.g.values - fee_of(d.r.values, d.c.values)
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R = net / (SL * d.atr_pct.values)
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gR = d.g.values / (SL * d.atr_pct.values)
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tn = taker_notional(d.r.values, d.c.values)
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def row(m, lab):
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if m.sum() < 20:
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return {"分组": lab, "笔数": int(m.sum()), "备注": "样本不足"}
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nn, rr, gg, tt = net[m], R[m], gR[m], tn[m]
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w, o = nn[nn > 0].sum(), -nn[nn <= 0].sum()
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return {"分组": lab, "笔数": int(m.sum()), "占比": f"{m.mean()*100:.0f}%",
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"胜率": f"{(nn > 0).mean()*100:.1f}%", "毛R": round(gg.mean(), 3),
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"净均R": round(rr.mean(), 3),
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"R夏普": round(rr.mean() / rr.std(ddof=1), 3),
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"PF": round(w / o, 2) if o > 0 else np.inf,
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"余量bp": round(nn.mean() / tt.mean() * 1e4, 2)}
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print("\n扎堆的交易赚不赚钱")
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print(pd.DataFrame([
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row((n0 == 1) & (n5 == 1), "真孤立（±5 分钟内无同伴）"),
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row((n0 == 1) & (n5 > 1), "错开：5 分钟内有同伴但不同分钟"),
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row(n0 > 1, "同一分钟撞在一起"),
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]).to_string(index=False))
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# 簇级：同一波行情里的几笔高度相关，逐笔统计会把有效样本算多
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clu = (d.date.diff().dt.total_seconds().fillna(9e9) > 300).cumsum()
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c = pd.DataFrame({"clu": clu, "R": R, "gR": gR, "net": net}).groupby("clu")
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agg = c.agg(n=("R", "size"), R=("R", "mean"), gR=("gR", "mean"))
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print(f"\n簇级（±5 分钟合为一簇，避免重复计数）：单笔簇 {(agg.n==1).sum()}、"
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f"多笔簇 {(agg.n>1).sum()}")
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for lab, m in (("单笔簇", agg.n == 1), ("多笔簇", agg.n > 1)):
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g = agg[m]
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if len(g) < 10:
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continue
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print(f" {lab:<5} {len(g):>4} 簇 簇均毛R {g.gR.mean():.3f} "
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f"簇均净R {g.R.mean():.3f} 簇级R夏普 {g.R.mean()/g.R.std(ddof=1):.3f}")
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win = c["net"].apply(lambda s: (s > 0).all())
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lose = c["net"].apply(lambda s: (s <= 0).all())
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multi_idx = agg.index[agg.n > 1]
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if len(multi_idx):
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aw, al = win[multi_idx].mean(), lose[multi_idx].mean()
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print(f" 多笔簇内：全赢 {aw*100:.1f}%、全输 {al*100:.1f}%、"
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f"有赢有输 {(1-aw-al)*100:.1f}% —— 簇内风险不可分散，但偏度有利")
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def main() -> None:
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ap = argparse.ArgumentParser()
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ap.add_argument("--symbols", default="BTC,BNB,ETH,SOL,LINK,LTC,AVAX,XRP,DOGE,ADA,TRX")
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ap.add_argument("--rows", type=int, default=300_000)
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ap.add_argument("--workers", type=int, default=3)
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ap.add_argument("--reuse", action="store_true", help="直接读已存的 feather")
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args = ap.parse_args()
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if args.reuse and OUT.exists():
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d = pd.read_feather(OUT)
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else:
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syms = [s.strip() for s in args.symbols.split(",")]
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print(f"[开仓时刻分布] {len(syms)} 币 × {args.rows} 根 1m\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(collect, s, args.rows): 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]} {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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d.to_feather(OUT)
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d["date"] = pd.to_datetime(d["date"])
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dark = (d["htf"] == 1.0) & d["lad"]
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analyse(d[dark & (d.atr_bp >= GATE_BP)], "实盘口径：深色 ∧ ATR≥8bp")
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analyse(d[dark], "对照：深色但不加 ATR 门控")
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print("\n########## 逐币笔数（实盘口径）##########")
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dd = d[dark & (d.atr_bp >= GATE_BP)]
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t = dd.groupby("sym").agg(笔数=("date", "size"), 中位持仓分钟=("hold", "median")).sort_values("笔数", ascending=False)
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span = (dd.date.max() - dd.date.min()).total_seconds() / 86400
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t["每天笔数"] = (t["笔数"] / span).round(3)
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print(t.to_string())
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if __name__ == "__main__":
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main()
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