"""Step 1b：诊断「多空双向短期均亏」的成因。

假设：sure_time 那根确认K线本身就是一段快速行情的末端，
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。
检验办法：看 entry bar 收盘价在其前后窗口里的分位位置。
"""
from __future__ import annotations

import sys
from pathlib import Path

import numpy as np
import pandas as pd

sys.path.insert(0, str(Path(__file__).resolve().parent))

from lib.bsp_eval import forward_returns, run_pipeline, to_events
from lib.data import fetch_ohlcv

pd.set_option("display.width", 220)


def main() -> None:
    symbol, tf = "BTC/USDT:USDT", "1h"
    df = fetch_ohlcv(symbol, tf, 50000)
    chan, bsp_list = run_pipeline(df, tf)
    events = to_events(bsp_list, df)
    fwd = forward_returns(events, df)

    closes = df["close"].to_numpy(dtype=float)
    n = len(df)
    look = 12  # 回看窗口

    rows = []
    for ev in events:
        i = ev.entry_idx
        if i - look < 0 or i + look >= n:
            continue
        past = closes[i - look : i + 1]
        # 入场价在过去 look 根里的分位：1.0=最高
        pct_past = float((past <= closes[i]).mean())
        # 顺信号方向的分位：做多时越靠近区间高点越像追高
        dir_pct = pct_past if ev.direction == 1 else 1.0 - pct_past
        # 确认K线自身的涨跌幅，及其相对近期波动的倍数
        bar_ret = (closes[i] - closes[i - 1]) / closes[i - 1]
        vol = np.std(np.diff(past) / past[:-1], ddof=1)
        rows.append(
            {
                "bsp_type": ev.bsp_type,
                "side": "LONG" if ev.direction == 1 else "SHORT",
                "dir_pct": dir_pct,
                "bar_ret_signed": ev.direction * bar_ret,
                "bar_ret_z": ev.direction * bar_ret / vol if vol > 0 else np.nan,
                # 从分型点到入场点，价格已经走掉多少（信号被吃掉的部分）
                "move_since_fx": ev.direction
                * (closes[i] - closes[i - ev.lag_bars])
                / closes[i - ev.lag_bars],
                "lag_bars": ev.lag_bars,
            }
        )

    diag = pd.DataFrame(rows)
    print(f"[样本] n={len(diag)}\n")

    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）")
    print(f"  全体均值 = {diag['dir_pct'].mean():.3f}   中位数 = {diag['dir_pct'].median():.3f}")
    print(diag.groupby("side")["dir_pct"].agg(["count", "mean", "median"]).to_string())
    print()

    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")
    print(f"  均值 = {diag['bar_ret_signed'].mean() * 100:+.3f}%   z均值 = {diag['bar_ret_z'].mean():+.2f}")
    print(diag.groupby("side")["bar_ret_signed"].agg(["mean", "median"]).map(lambda v: f"{v*100:+.3f}%").to_string())
    print()

    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")
    print(f"  均值 = {diag['move_since_fx'].mean() * 100:+.2f}%   中位数 = {diag['move_since_fx'].median() * 100:+.2f}%")
    print(
        diag.groupby("bsp_type")[["move_since_fx", "lag_bars"]]
        .agg({"move_since_fx": "mean", "lag_bars": "mean"})
        .assign(move_since_fx=lambda d: (d["move_since_fx"] * 100).map(lambda v: f"{v:+.2f}%"))
        .to_string()
    )
    print()

    # 对照：若能在分型点入场（理想但不可实现），收益如何
    ideal = []
    for ev in events:
        i = ev.entry_idx - ev.lag_bars  # 分型所在K线
        if i < 0:
            continue
        p0 = closes[i]
        for h in (1, 5, 10, 20, 40):
            j = i + h
            if j < n:
                ideal.append({"h": h, "ret": ev.direction * (closes[j] - p0) / p0})
    idf = pd.DataFrame(ideal)
    print("[对照] 假想在分型点入场（含未来函数，仅作上界参考）")
    print(
        idf.groupby("h")["ret"]
        .agg(n="count", mean="mean", winrate=lambda s: (s > 0).mean())
        .assign(mean=lambda d: (d["mean"] * 100).map(lambda v: f"{v:+.2f}%"),
                winrate=lambda d: (d["winrate"] * 100).map(lambda v: f"{v:.0f}%"))
        .to_string()
    )


if __name__ == "__main__":
    main()