"""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。

用户问：11 个币对开仓时间差距有多少，不是同时开吧。

这个数决定三件事：
  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
  2. "多币 = 分散"是否成立。§3.31 已知同时发生的信号 92.7% 同向——
     那种情况下多开的仓不是分散，是加杠杆
  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

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
（1m 上 1 根 = 1 分钟），不是一律按 48 根上限。
"""
from __future__ import annotations

import argparse
import os
import sys
import warnings
from concurrent.futures import ProcessPoolExecutor, as_completed
from pathlib import Path

import numpy as np
import pandas as pd

warnings.filterwarnings("ignore")
for v in ("OMP_NUM_THREADS", "OPENBLAS_NUM_THREADS", "MKL_NUM_THREADS"):
    os.environ.setdefault(v, "1")

HERE = Path(__file__).resolve().parent
sys.path.insert(0, str(HERE))
sys.path.insert(0, str(HERE.parent))
pd.set_option("display.width", 320)

SL, SCALE_AT, RUNNER, RSTOP, MAXB = 2.0, 3.0, 8.0, 2.0, 48
GATE_BP = 8.0
OUT = HERE / "out" / "step48_signal_times.feather"


def collect(sym: str, rows: int) -> pd.DataFrame | None:
    import warnings as _w
    _w.filterwarnings("ignore")
    sys.path.insert(0, str(HERE))
    sys.path.insert(0, str(HERE.parent))

    from chanlun import TF_DF
    from chanlun.analysis.fast_bsp import (
        add_zone_ladder, attach_htf_agree, attach_zone_ladder,
        build_htf_zones, find_fast_bsp3, htf_fx_timeline,
    )
    from lib.data import fetch_ohlcv
    from lib.exit_model import cfg_name, walk_exits

    try:
        df = fetch_ohlcv(f"{sym}/USDT:USDT", "1m", rows)
        if df is None or len(df) < 50_000:
            return None
        chan = TF_DF(df, 1, "1m", lean=True)
        cdf = chan.dataframe
        zones = build_htf_zones(cdf, "1m", chan=chan)
        if zones.empty:
            return None
        zl = add_zone_ladder(zones.reset_index(drop=True))
        sig = find_fast_bsp3(cdf, zl)
        if sig.empty:
            return None

        dh = fetch_ohlcv(f"{sym}/USDT:USDT", "5m", 10 ** 9)
        ch = TF_DF(dh, 1, "5m", lean=True)
        sig = attach_zone_ladder(attach_htf_agree(sig, cdf, htf_fx_timeline(ch, ch.dataframe)), zl)

        res = walk_exits(cdf, sig, [SL], [RUNNER], [MAXB],
                         scale_at=SCALE_AT, runners=(RUNNER,), runner_stops=(RSTOP,))
        cfg = cfg_name(SL, RUNNER, MAXB, RSTOP)
        idx = sig["entry_idx"].to_numpy().astype(int)
        atr = cdf["atr"].to_numpy(float)[idx]
        close = cdf["close"].to_numpy(float)[idx]

        vol = cdf["volume"]
        vr = vol / vol.rolling(60, min_periods=10).mean()
        vr60 = vr.to_numpy(float)
        # shift(1) 把信号根自己排除在外
        vpre10 = vr.rolling(10, min_periods=5).mean().shift(1).to_numpy(float)
        vpre30 = vr.rolling(30, min_periods=15).mean().shift(1).to_numpy(float)
        c_all = cdf["close"].to_numpy(float)
        a_all = cdf["atr"].to_numpy(float)
        d_all = sig["direction"].to_numpy().astype(int)

        def mom(k: int) -> np.ndarray:
            """顺方向动量（ATR 单位）。用全序列算好再取下标，避免逐笔切片。"""
            prev = np.concatenate([np.full(k, np.nan), c_all[:-k]])
            with np.errstate(invalid="ignore", divide="ignore"):
                m = (c_all - prev) / a_all
            out = np.full(len(idx), np.nan)
            out[:] = m[idx] * d_all
            return out

        return pd.DataFrame({
            "sym": sym,
            "date": cdf["date"].to_numpy()[idx],
            "dir": sig["direction"].to_numpy(),
            "atr_bp": atr / close * 1e4,
            "htf": sig["htf_agree"].to_numpy(),
            "lad": sig["ladder_ok"].to_numpy(),
            "hold": res[f"{cfg}_b"].to_numpy(),
            # 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
            "g": res[f"{cfg}_g"].to_numpy(),
            "r": res[f"{cfg}_r"].to_numpy(),
            "c": res[f"{cfg}_c"].to_numpy(),
            "atr_pct": atr / close,
            # 信号根的相对成交量。当根已收盘，开仓时可知，是合规的可交易信息。
            # vr10 是引擎自带口径（前 10 根均量），vr60 换个基准做稳健性对照。
            "vr10": cdf["volume_ratio"].to_numpy(float)[idx],
            "vr60": vr60[idx],
            # 开仓**之前**那段的量与走势。vpre 已 shift(1)，不含信号根本身——
            # 信号根的量单独由 vr60 承担，两者混在一起就分不清是哪一段在起作用。
            "vpre10": vpre10[idx],
            "vpre30": vpre30[idx],
            # 顺方向动量，ATR 为单位。B4/S4 是回抽后转强，所以 mom10 多为负
            # （入场前那几根逆着你走）；负得多 = 回抽深。
            "mom10": mom(10),
            "mom60": mom(60),
        })
    except Exception as e:
        print(f"  {sym} 失败: {e!r}", flush=True)
        return None


def analyse(d: pd.DataFrame, label: str) -> None:
    d = d.sort_values("date").reset_index(drop=True)
    span = (d.date.max() - d.date.min()).total_seconds() / 86400
    print("\n" + "=" * 96)
    print(f"########## {label}：{d.sym.nunique()} 币 / {len(d)} 笔 / 跨 {span:.0f} 天 ##########")
    print(f"组合 {len(d) / span:.2f} 笔每天 —— 平均每 {span * 24 / len(d):.1f} 小时才 1 笔")

    g = d.date.diff().dt.total_seconds().dropna() / 60
    print("\n相邻两笔间隔（分钟）")
    qs = [(q, g.quantile(q)) for q in (0.05, 0.10, 0.25, 0.50, 0.75, 0.90)]
    print("  " + "  ".join(f"{int(q*100)}%:{v:.0f}" for q, v in qs))
    for lab, m in (("同一分钟", g == 0), ("≤5 分钟", g <= 5),
                   ("≤60 分钟", g <= 60), (">1 小时", g > 60)):
        print(f"  {lab:<9}{m.sum():>5} 次  {m.mean()*100:>5.1f}%")

    # 并发持仓：按回测实际持仓根数
    ev = []
    for t, h in zip(d.date, d.hold):
        ev.append((t, 1))
        ev.append((t + pd.Timedelta(minutes=float(h)), -1))
    ev.sort()
    cur, prev, dur = 0, None, {}
    for t, delta in ev:
        if prev is not None and t > prev:
            dur[cur] = dur.get(cur, 0) + (t - prev).total_seconds()
        cur += delta
        prev = t
    tot = sum(dur.values())
    print("\n同时持仓数的时间占比")
    for k in sorted(dur):
        if dur[k] / tot > 0.0005:
            print(f"  {k:>2} 个: {dur[k]/tot*100:>5.1f}%")
    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}%")

    # 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
    same = d.groupby(d.date)["dir"].agg(["count", "nunique"])
    multi = same[same["count"] > 1]
    if len(multi):
        agree = (multi["nunique"] == 1).mean()
        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 "
              f"{agree*100:.1f}% —— 这部分不是分散，是同一笔押注被拆成几个币")

    if "g" in d.columns:
        profit_by_cluster(d)


def profit_by_cluster(d: pd.DataFrame) -> None:
    """扎堆的交易赚不赚钱。

    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：
    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。
    """
    from lib.exit_model import fee_of, taker_notional

    d = d.sort_values("date").reset_index(drop=True)
    t = d.date.values.astype("datetime64[m]").astype(np.int64)
    n0 = np.searchsorted(t, t, "right") - np.searchsorted(t, t, "left")
    n5 = np.searchsorted(t, t + 5, "right") - np.searchsorted(t, t - 5, "left")
    net = d.g.values - fee_of(d.r.values, d.c.values)
    R = net / (SL * d.atr_pct.values)
    gR = d.g.values / (SL * d.atr_pct.values)
    tn = taker_notional(d.r.values, d.c.values)

    def row(m, lab):
        if m.sum() < 20:
            return {"分组": lab, "笔数": int(m.sum()), "备注": "样本不足"}
        nn, rr, gg, tt = net[m], R[m], gR[m], tn[m]
        w, o = nn[nn > 0].sum(), -nn[nn <= 0].sum()
        return {"分组": lab, "笔数": int(m.sum()), "占比": f"{m.mean()*100:.0f}%",
                "胜率": f"{(nn > 0).mean()*100:.1f}%", "毛R": round(gg.mean(), 3),
                "净均R": round(rr.mean(), 3),
                "R夏普": round(rr.mean() / rr.std(ddof=1), 3),
                "PF": round(w / o, 2) if o > 0 else np.inf,
                "余量bp": round(nn.mean() / tt.mean() * 1e4, 2)}

    print("\n扎堆的交易赚不赚钱")
    print(pd.DataFrame([
        row((n0 == 1) & (n5 == 1), "真孤立（±5 分钟内无同伴）"),
        row((n0 == 1) & (n5 > 1), "错开：5 分钟内有同伴但不同分钟"),
        row(n0 > 1, "同一分钟撞在一起"),
    ]).to_string(index=False))

    # 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
    clu = (d.date.diff().dt.total_seconds().fillna(9e9) > 300).cumsum()
    c = pd.DataFrame({"clu": clu, "R": R, "gR": gR, "net": net}).groupby("clu")
    agg = c.agg(n=("R", "size"), R=("R", "mean"), gR=("gR", "mean"))
    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()}、"
          f"多笔簇 {(agg.n>1).sum()}")
    for lab, m in (("单笔簇", agg.n == 1), ("多笔簇", agg.n > 1)):
        g = agg[m]
        if len(g) < 10:
            continue
        print(f"  {lab:<5} {len(g):>4} 簇   簇均毛R {g.gR.mean():.3f}   "
              f"簇均净R {g.R.mean():.3f}   簇级R夏普 {g.R.mean()/g.R.std(ddof=1):.3f}")
    win = c["net"].apply(lambda s: (s > 0).all())
    lose = c["net"].apply(lambda s: (s <= 0).all())
    multi_idx = agg.index[agg.n > 1]
    if len(multi_idx):
        aw, al = win[multi_idx].mean(), lose[multi_idx].mean()
        print(f"  多笔簇内：全赢 {aw*100:.1f}%、全输 {al*100:.1f}%、"
              f"有赢有输 {(1-aw-al)*100:.1f}% —— 簇内风险不可分散，但偏度有利")


def main() -> None:
    ap = argparse.ArgumentParser()
    ap.add_argument("--symbols", default="BTC,BNB,ETH,SOL,LINK,LTC,AVAX,XRP,DOGE,ADA,TRX")
    ap.add_argument("--rows", type=int, default=300_000)
    ap.add_argument("--workers", type=int, default=3)
    ap.add_argument("--reuse", action="store_true", help="直接读已存的 feather")
    args = ap.parse_args()

    if args.reuse and OUT.exists():
        d = pd.read_feather(OUT)
    else:
        syms = [s.strip() for s in args.symbols.split(",")]
        print(f"[开仓时刻分布] {len(syms)} 币 × {args.rows} 根 1m\n", flush=True)
        parts = []
        with ProcessPoolExecutor(max_workers=args.workers) as ex:
            fut = {ex.submit(collect, s, args.rows): s for s in syms}
            for i, f in enumerate(as_completed(fut), 1):
                r = f.result()
                print(f"  [{i}/{len(syms)}] {fut[f]} {0 if r is None else len(r)}", flush=True)
                if r is not None:
                    parts.append(r)
        if not parts:
            print("无结果")
            return
        d = pd.concat(parts, ignore_index=True)
        d.to_feather(OUT)

    d["date"] = pd.to_datetime(d["date"])
    dark = (d["htf"] == 1.0) & d["lad"]
    analyse(d[dark & (d.atr_bp >= GATE_BP)], "实盘口径：深色 ∧ ATR≥8bp")
    analyse(d[dark], "对照：深色但不加 ATR 门控")

    print("\n########## 逐币笔数（实盘口径）##########")
    dd = d[dark & (d.atr_bp >= GATE_BP)]
    t = dd.groupby("sym").agg(笔数=("date", "size"), 中位持仓分钟=("hold", "median")).sort_values("笔数", ascending=False)
    span = (dd.date.max() - dd.date.min()).total_seconds() / 86400
    t["每天笔数"] = (t["笔数"] / span).round(3)
    print(t.to_string())


if __name__ == "__main__":
    main()