"""Step 14：对 Step 13 结果的尽职调查。

Step 13 报表里的 +533966% 是每笔全仓连续复利的产物，且不同品种交易在时间上重叠，
不可当真。本步只问四个问题：

  1. 收益是不是被少数极端盈利撑起来的（去极值后还剩多少）
  2. 中位数收益是正是负
  3. 按固定风险仓位、按时间排序的真实权益曲线长什么样
  4. 成本敏感性 —— 手续费滑点翻倍还活不活
"""
from __future__ import annotations

import sys
import warnings
from pathlib import Path

import numpy as np
import pandas as pd

warnings.filterwarnings("ignore")
pd.set_option("display.width", 260)

TRADES = Path(__file__).parent / "out" / "step13_all_trades.csv"
FEE = 0.0008


def main() -> None:
    t = pd.read_csv(TRADES, parse_dates=["date"]).sort_values("date").reset_index(drop=True)
    r = t["ret"].to_numpy()
    print(f"[样本] {len(t)} 笔  {t['date'].min()} -> {t['date'].max()}\n")

    print("########## 1. 收益分布 ##########")
    qs = [0.01, 0.05, 0.25, 0.5, 0.75, 0.95, 0.99]
    d = pd.DataFrame({
        "分位": [f"P{int(q * 100)}" for q in qs],
        "收益": [f"{np.quantile(r, q) * 100:+.2f}%" for q in qs],
    })
    print(d.to_string(index=False))
    print(f"\n  均值 {r.mean() * 100:+.3f}%   中位数 {np.median(r) * 100:+.3f}%   "
          f"标准差 {r.std(ddof=1) * 100:.2f}%   偏度 {pd.Series(r).skew():.2f}")

    print("\n########## 2. 去极值稳健性 ##########")
    rows = []
    for k in (0, 1, 2, 5, 10):
        if k == 0:
            v = r
        else:
            hi = np.quantile(r, 1 - k / 100)
            v = r[r <= hi]
        win, loss = v[v > 0], v[v <= 0]
        pf = win.sum() / abs(loss.sum()) if len(loss) else np.inf
        sd = v.std(ddof=1)
        rows.append({
            "剔除最赚的": f"{k}%", "剩余笔数": len(v),
            "均收益": f"{v.mean() * 100:+.3f}%",
            "PF": f"{pf:.2f}",
            "t值": f"{v.mean() / (sd / np.sqrt(len(v))):+.2f}",
        })
    print(pd.DataFrame(rows).to_string(index=False))
    print("  剔掉最赚的 5% 后仍为正，才说明不是靠几笔暴利撑着。")

    print("\n########## 3. 真实权益曲线（固定 1% 风险，按时间顺序，不重叠加仓）##########")
    # 每笔风险敞口固定为账户 1%，收益按 ret/单笔风险幅度折算
    risk_per_trade = 0.01
    # 单笔风险幅度近似为止损距离，用 |最差收益| 的稳健估计代替
    stop_pct = np.abs(np.quantile(r, 0.05))
    scaled = r / max(stop_pct, 1e-6) * risk_per_trade
    scaled = np.clip(scaled, -0.1, 0.5)  # 防止单笔异常主导
    eq = np.cumprod(1 + scaled)
    dd = 1 - eq / np.maximum.accumulate(eq)
    yrs = (t["date"].max() - t["date"].min()).days / 365.25
    cagr = eq[-1] ** (1 / yrs) - 1 if yrs > 0 else np.nan
    sharpe = scaled.mean() / scaled.std(ddof=1) * np.sqrt(len(scaled) / yrs) if yrs > 0 else np.nan
    print(f"  总收益 {(eq[-1] - 1) * 100:+.1f}%   年化 {cagr * 100:+.1f}%   "
          f"最大回撤 {dd.max() * 100:.1f}%   Sharpe {sharpe:.2f}")
    print(f"  交易频率 {len(t) / yrs:.0f} 笔/年（9 个品种周期合计）")

    print("\n########## 4. 成本敏感性 ##########")
    rows = []
    for mult, name in [(0, "零成本"), (1, "0.08%基准"), (2, "0.16%"), (3, "0.24%"), (5, "0.40%")]:
        v = r + FEE - mult * FEE  # 还原毛收益再扣不同成本
        win, loss = v[v > 0], v[v <= 0]
        pf = win.sum() / abs(loss.sum()) if len(loss) else np.inf
        sd = v.std(ddof=1)
        rows.append({
            "成本": name, "均收益": f"{v.mean() * 100:+.3f}%", "PF": f"{pf:.2f}",
            "t值": f"{v.mean() / (sd / np.sqrt(len(v))):+.2f}",
        })
    print(pd.DataFrame(rows).to_string(index=False))

    print("\n########## 5. 分品种周期的稳定性（剔除极值后）##########")
    rows = []
    for (sym, tf), g in t.groupby(["symbol", "tf"]):
        if len(g) < 25:
            continue
        v = g["ret"].to_numpy()
        hi = np.quantile(v, 0.95)
        vt = v[v <= hi]
        win, loss = vt[vt > 0], vt[vt <= 0]
        rows.append({
            "品种周期": f"{sym} {tf}", "笔数": len(v),
            "原PF": f"{v[v > 0].sum() / abs(v[v <= 0].sum()):.2f}",
            "去极值PF": f"{win.sum() / abs(loss.sum()):.2f}" if len(loss) else "—",
            "中位收益": f"{np.median(v) * 100:+.3f}%",
        })
    print(pd.DataFrame(rows).to_string(index=False))

    print("\n########## 6. 出场原因分布 ##########")
    print(t.groupby("reason").agg(
        笔数=("ret", "size"),
        均收益=("ret", lambda x: f"{x.mean() * 100:+.3f}%"),
        占比=("ret", lambda x: f"{len(x) / len(t) * 100:.0f}%"),
    ).to_string())


if __name__ == "__main__":
    main()