jackyu66gitandCursor
7f393b93ed
删除根目录旧 Chan 模块、策略、配置、文档及 wyckoff 相关代码；更新缠论 pipeline 与笔中枢计算；补充 research 研究与 web 测试。 Co-authored-by: Cursor <cursoragent@cursor.com>
109 lines
4.2 KiB
Python
109 lines
4.2 KiB
Python
"""Step 1b：诊断「多空双向短期均亏」的成因。
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假设：sure_time 那根确认K线本身就是一段快速行情的末端，
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在它的收盘价入场等于追单，因此无论方向下一根都倾向回撤。
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检验办法：看 entry bar 收盘价在其前后窗口里的分位位置。
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"""
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from __future__ import annotations
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import sys
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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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sys.path.insert(0, str(Path(__file__).resolve().parent))
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from lib.bsp_eval import forward_returns, run_pipeline, to_events
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from lib.data import fetch_ohlcv
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pd.set_option("display.width", 220)
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def main() -> None:
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symbol, tf = "BTC/USDT:USDT", "1h"
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df = fetch_ohlcv(symbol, tf, 50000)
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chan, bsp_list = run_pipeline(df, tf)
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events = to_events(bsp_list, df)
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fwd = forward_returns(events, df)
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closes = df["close"].to_numpy(dtype=float)
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n = len(df)
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look = 12 # 回看窗口
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rows = []
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for ev in events:
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i = ev.entry_idx
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if i - look < 0 or i + look >= n:
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continue
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past = closes[i - look : i + 1]
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# 入场价在过去 look 根里的分位：1.0=最高
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pct_past = float((past <= closes[i]).mean())
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# 顺信号方向的分位：做多时越靠近区间高点越像追高
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dir_pct = pct_past if ev.direction == 1 else 1.0 - pct_past
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# 确认K线自身的涨跌幅，及其相对近期波动的倍数
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bar_ret = (closes[i] - closes[i - 1]) / closes[i - 1]
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vol = np.std(np.diff(past) / past[:-1], ddof=1)
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rows.append(
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{
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"bsp_type": ev.bsp_type,
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"side": "LONG" if ev.direction == 1 else "SHORT",
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"dir_pct": dir_pct,
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"bar_ret_signed": ev.direction * bar_ret,
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"bar_ret_z": ev.direction * bar_ret / vol if vol > 0 else np.nan,
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# 从分型点到入场点，价格已经走掉多少（信号被吃掉的部分）
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"move_since_fx": ev.direction
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* (closes[i] - closes[i - ev.lag_bars])
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/ closes[i - ev.lag_bars],
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"lag_bars": ev.lag_bars,
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}
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)
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diag = pd.DataFrame(rows)
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print(f"[样本] n={len(diag)}\n")
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print("[入场价在过去12根中的顺向分位] 0.5=中性, >0.5=追单（做多买在高位/做空卖在低位）")
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print(f" 全体均值 = {diag['dir_pct'].mean():.3f} 中位数 = {diag['dir_pct'].median():.3f}")
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print(diag.groupby("side")["dir_pct"].agg(["count", "mean", "median"]).to_string())
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print()
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print("[确认K线自身的顺向涨跌] 正=确认那根K线就在朝信号方向猛冲")
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print(f" 均值 = {diag['bar_ret_signed'].mean() * 100:+.3f}% z均值 = {diag['bar_ret_z'].mean():+.2f}")
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print(diag.groupby("side")["bar_ret_signed"].agg(["mean", "median"]).map(lambda v: f"{v*100:+.3f}%").to_string())
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print()
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print("[从分型点到确认点，价格已顺向走掉多少] 这是滞后吃掉的利润")
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print(f" 均值 = {diag['move_since_fx'].mean() * 100:+.2f}% 中位数 = {diag['move_since_fx'].median() * 100:+.2f}%")
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print(
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diag.groupby("bsp_type")[["move_since_fx", "lag_bars"]]
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.agg({"move_since_fx": "mean", "lag_bars": "mean"})
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.assign(move_since_fx=lambda d: (d["move_since_fx"] * 100).map(lambda v: f"{v:+.2f}%"))
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.to_string()
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)
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print()
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# 对照：若能在分型点入场（理想但不可实现），收益如何
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ideal = []
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for ev in events:
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i = ev.entry_idx - ev.lag_bars # 分型所在K线
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if i < 0:
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continue
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p0 = closes[i]
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for h in (1, 5, 10, 20, 40):
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j = i + h
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if j < n:
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ideal.append({"h": h, "ret": ev.direction * (closes[j] - p0) / p0})
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idf = pd.DataFrame(ideal)
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print("[对照] 假想在分型点入场（含未来函数，仅作上界参考）")
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print(
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idf.groupby("h")["ret"]
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.agg(n="count", mean="mean", winrate=lambda s: (s > 0).mean())
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.assign(mean=lambda d: (d["mean"] * 100).map(lambda v: f"{v:+.2f}%"),
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winrate=lambda d: (d["winrate"] * 100).map(lambda v: f"{v:.0f}%"))
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.to_string()
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)
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if __name__ == "__main__":
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main()
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