research: 1m 出场口径重定、费率修正，与影子测量的判据常数
起因是用户看图指出「止盈没做好」，查下来 TP=3.0 确实把右尾截早了， 而且 1m 不该沿用 5m/15m/30m 的参数——成本固定在 bp、目标随 ATR 缩放， 1m 的 3 ATR 只有 0.39% 而 30m 是 1.87%，成本占比差 5 倍。 step41（5m/15m/30m）与 step42（1m）跑同一张全网格： SL × TP × MAX_BARS × 分批（3 ATR 减半 → 剩余目标 × 剩余半仓止损位）。 - 1m 最优 SL2 / 3ATR 减半 / 剩余止损保持 2.0 / 目标 8ATR / 48 根， 样本外 8/8 币、7/7 年全面提升，均R/R夏普/回撤/剔10%PF 四项全赢 - 分批要做，但**减仓后不要动止损**。止损位 0/0.5/1/1.5/2 ATR 严格单调， 越紧越差，三组初始 SL 全一致。保本损是全表最差的一档 - SL=1.0 在 1m 上是废的：剔10%PF 0.78~0.99、中位收益 −0.122% 费率此前写的 taker 3bp / maker 1bp 隐含「原始 taker 6bp」的错误前提， 实际是原始 taker 0.040% / maker 0.016%、返 50% 后 2.0 / 0.8bp。方向是保守的， 所以首轮跑出来的数字全部偏低。exit_model 已改，费率只在分析阶段套用， 不必重跑模拟。改完 1m 的均R +8%，5m/15m/30m 只动 2%——费率只对 1m 有杠杆。 顺带查证了用户的一个假设：余量逐年递减是不是跟波动率有关。成立，而且 r = +0.989。毛/ATR 七年在 2.26~2.67 之间没有趋势，衰减的是 ATR 本身 （2021 的 22.1bp 压到 2026 的 8.8bp）。**是波动率压缩，不是 alpha 衰减。** 由此引出 ATR 门控：低 ATR 桶的毛 R 其实最高（1.16 vs 高 ATR 桶的 0.94）， 断崖只在扣费之后出现。所以阈值是**费率的函数**（约 5 + 1.1×taker费）， 不是市场常数。当前费率下 ≥8bp，在 5m/15m/30m 上几乎不触发，可作全局规则。 lib/shadow_budget.py 放影子测量要对照的常数：逐币预算、门控阈值、 腿→maker/taker 映射、lag 阈值、判据。记录与报表归 research/live/， 分工的理由是这些数会变——今天预算就动了四次。 out/*.feather 转为 ignore：70MB+ 且重跑可得，摘要都在 HANDOFF。 Co-authored-by: Cursor <cursoragent@cursor.com>
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"""出场模拟与费率模型，step41（5m/15m/30m）与 step42（1m）共用。
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单独抽出来是因为两边必须用同一份实现——`fast_bsp3` 当初分散在两处的教训。
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## 费率模型（用户 2026-08-27 指出）
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原先所有数字都按「双边 taker」算（§5.3 的 6bp），这在两个方向上都错了：
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费率档位记高了（见下方 FEE_TAKER 注释），且没有区分出场性质——
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入场 信号在收盘出现，次根开盘市价单进场 → **taker + 滑点**
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止盈 挂在目标价的限价单被动成交 → **maker，无滑点**
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止损 stop-market，触发后市价成交 → **taker + 滑点**
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超时 到点市价平 → **taker + 滑点**
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止损做不成 maker：stop-limit 在急跌里可能不成交，那种情况下的损失远大于
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省下的 2bp。所以按出场原因分别计费，不是一刀切。
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分批离场不额外增加费用：手续费按名义额收，入场 1.0、出场 0.5+0.5，总额不变。
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但**第一批必然是止盈成交（maker）**，这正是分批在成本上占便宜的地方。
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## 出场配置
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整仓 `s{SL}_tp{TP}_m{MAXB}`
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分批 `s{SL}_so{目标}_k{剩余半仓止损}_m{MAXB}`
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到 3 ATR 平一半，剩余半仓止损挪到「开仓价下方 k 个 ATR」。
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k=0 即保本损，k 等于原 SL 即止损不动，k 更大则是主动放宽。
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## 实现
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每笔只前推两遍：第一遍记录各价位的首次触及根，第二遍从减仓根起记录剩余
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半仓各止损位的首次触及根。之后所有配置都是解析推导，不再重复走 K 线。
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同一根内止损与目标并存时一律判止损先到，宁可低估。
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"""
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from __future__ import annotations
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import numpy as np
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import pandas as pd
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# 单边费率。用户 2026-08-27 给的实际档位：返佣前 taker 0.040% / maker 0.016%，
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# API 返 50% → taker 0.020% / maker 0.008%。
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# 注意：此前这里写的是 0.030/0.010，隐含「原始 taker 6bp」的错误前提，
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# 所以 step41/42 首轮跑出来的所有数字都偏保守（taker 高估 50%）。
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FEE_TAKER = 0.00020
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FEE_MAKER = 0.00008
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SLIP = 0.00010
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TP, SL_, TIME = 0, 1, 2 # 出场原因编码
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def taker_notional(reason: np.ndarray, scaled: np.ndarray) -> np.ndarray:
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"""每笔走 taker 的名义额（入场 1.0，加上非止盈出场的部分）。滑点只发生在这上面。"""
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exit_taker = np.where(reason == TP, 0.0, 1.0)
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exit_taker = np.where(scaled == 1, 0.5 * exit_taker, exit_taker)
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return 1.0 + exit_taker
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def fee_of(reason: np.ndarray, scaled: np.ndarray,
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fee_taker: float = FEE_TAKER, fee_maker: float = FEE_MAKER) -> np.ndarray:
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"""只算手续费，不含滑点。分批时第一半必然是止盈成交（maker）。"""
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exit_fee = np.where(reason == TP, fee_maker, fee_taker)
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exit_fee = np.where(scaled == 1, 0.5 * fee_maker + 0.5 * exit_fee, exit_fee)
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return fee_taker + exit_fee
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def cost_of(reason: np.ndarray, scaled: np.ndarray, slip: float = SLIP) -> np.ndarray:
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"""新口径：入场 taker、止盈 maker、止损/超时 taker，滑点只加在 taker 腿上。"""
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return fee_of(reason, scaled) + slip * taker_notional(reason, scaled)
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def all_taker_cost(reason: np.ndarray, scaled: np.ndarray) -> np.ndarray:
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"""旧口径：进出都当 taker，双边费 + 双边滑点。用来对照新旧差多少。"""
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return np.full(len(reason), 2.0 * (FEE_TAKER + SLIP))
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def flat_cost(reason: np.ndarray, scaled: np.ndarray) -> np.ndarray:
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"""研究口径的固定 5bp，用于和 step28/35/37 的历史数字对齐。"""
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return np.full(len(reason), 0.0005)
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def slip_budget(gross: np.ndarray, reason: np.ndarray, scaled: np.ndarray) -> float:
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"""盈亏平衡的单边滑点上限（bp）：毛收益扣掉手续费后，摊到走 taker 的名义额上。"""
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net_of_fee = gross.mean() - fee_of(reason, scaled).mean()
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return net_of_fee / taker_notional(reason, scaled).mean() * 10000
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def cfg_name(sl: float, target, maxb: int, rstop=None) -> str:
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"""整仓 s{SL}_tp{T}_m{B}；分批 s{SL}_so{T}_k{K}_m{B}。target=None 表示不设目标。"""
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t = f"{target:g}" if target is not None else "R"
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if rstop is None:
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return f"s{sl:g}_tp{t}_m{maxb}"
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return f"s{sl:g}_so{t}_k{rstop:g}_m{maxb}"
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def walk_exits(cdf: pd.DataFrame, sig: pd.DataFrame, sls, tps, maxbs,
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scale_at: float = 3.0, runners=(5.0, 6.0, 8.0, None),
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runner_stops=(0.0, 0.5, 1.0, 1.5, 2.0)) -> pd.DataFrame:
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"""前推每笔信号，解析出全部出场配置的结果。
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每个配置四列：`{cfg}_g` 毛收益率、`{cfg}_r` 出场原因、`{cfg}_c` 是否分批、
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`{cfg}_b` 持仓根数。另有 `s{SL}_mfe` 各初始止损下的最大有利偏移。
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"""
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high = cdf["high"].to_numpy(float)
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low = cdf["low"].to_numpy(float)
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open_ = cdf["open"].to_numpy(float)
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close = cdf["close"].to_numpy(float)
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atr = cdf["atr"].to_numpy(float)
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n = len(cdf)
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horizon = max(maxbs)
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ups = sorted(set(list(tps) + [scale_at] + [r for r in runners if r]))
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downs = sorted(set(list(sls) + list(runner_stops)))
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out = []
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for s, d in zip(sig["entry_idx"].astype(int), sig["direction"].astype(int)):
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e = s + 1
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if e >= n - 1:
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continue
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a = atr[s]
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if not np.isfinite(a) or a <= 0:
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continue
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entry = open_[e]
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end = min(e + horizon, n - 1)
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# 第一遍：各价位的首次触及根（与止损无关，纯价格事件）
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up_bar = {t: None for t in ups}
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dn_bar = {L: None for L in downs}
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mfe = {sl: 0.0 for sl in sls}
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alive = {sl: True for sl in sls}
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for j in range(e, end + 1):
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adv = (high[j] - entry) / a if d == 1 else (entry - low[j]) / a
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ret = (entry - low[j]) / a if d == 1 else (high[j] - entry) / a
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for L in downs:
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if dn_bar[L] is None and ret >= L:
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dn_bar[L] = j
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for sl in sls:
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if alive[sl]:
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if dn_bar[sl] is not None and dn_bar[sl] == j:
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alive[sl] = False # 同根内止损优先，不更新 MFE
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elif adv > mfe[sl]:
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mfe[sl] = adv
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for t in ups:
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if up_bar[t] is None and adv >= t:
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up_bar[t] = j
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for sl in sls:
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mfe[sl] = mfe[sl]
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# 第二遍：从减仓根起，剩余半仓各止损位的首次触及根
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j0 = up_bar[scale_at]
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dn_after = {L: None for L in runner_stops}
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if j0 is not None:
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for j in range(j0, end + 1):
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ret = (entry - low[j]) / a if d == 1 else (high[j] - entry) / a
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for L in runner_stops:
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if dn_after[L] is None and ret >= L:
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dn_after[L] = j
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if all(v is not None for v in dn_after.values()):
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break
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row = {"sig_idx": s, "direction": d, "atr_pct": a / entry}
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row.update({f"s{sl:g}_mfe": mfe[sl] for sl in sls})
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def resolve(target, stop_bar, stop_ret, cap, start):
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"""在 cap 根之前，目标与止损谁先到。返回 (毛收益率, 原因, 出场根)。"""
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tb = up_bar[target] if target is not None else None
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if tb is not None and tb <= cap and (stop_bar is None or tb < stop_bar):
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return target * a / entry, TP, tb
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if stop_bar is not None and stop_bar <= cap:
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return stop_ret, SL_, stop_bar
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k = min(cap, n - 1)
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return d * (close[k] - entry) / entry, TIME, k
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for sl in sls:
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dead = dn_bar[sl]
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for b in maxbs:
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cap = e + b
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for t in tps:
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g, r, xb = resolve(t, dead, -sl * a / entry, cap, e)
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c = cfg_name(sl, t, b)
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row[f"{c}_g"], row[f"{c}_r"], row[f"{c}_c"], row[f"{c}_b"] = g, r, 0, xb - e + 1
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# 分批：先看能否走到减仓点
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scaled_ok = (j0 is not None and j0 <= cap
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and (dead is None or j0 < dead))
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for rn in runners:
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for k in runner_stops:
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c = cfg_name(sl, rn, b, k)
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if not scaled_ok:
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g, r, xb = resolve(scale_at, dead, -sl * a / entry, cap, e)
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row[f"{c}_g"], row[f"{c}_r"] = g, r
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row[f"{c}_c"], row[f"{c}_b"] = 0, xb - e + 1
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else:
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rg, rr, rxb = resolve(rn, dn_after[k], -k * a / entry, cap, j0)
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row[f"{c}_g"] = 0.5 * (scale_at * a / entry) + 0.5 * rg
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row[f"{c}_r"], row[f"{c}_c"] = rr, 1
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row[f"{c}_b"] = rxb - e + 1
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out.append(row)
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return pd.DataFrame(out)
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def stat(g: pd.DataFrame, cfg: str, label: str, cost_fn=cost_of, minn: int = 25) -> dict:
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"""给一个配置出统计。`滑点余量bp` 是滑点的盈亏平衡上限。"""
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gross = g[f"{cfg}_g"].to_numpy()
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if len(gross) < minn:
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return {}
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reason, scaled = g[f"{cfg}_r"].to_numpy(), g[f"{cfg}_c"].to_numpy()
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r = gross - cost_fn(reason, scaled)
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w, o = r[r > 0], r[r <= 0]
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sd = r.std(ddof=1)
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t10 = r[r <= np.quantile(r, 0.90)]
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return {"口径": label, "笔数": len(r),
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"胜率": f"{(r > 0).mean() * 100:.1f}%",
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"毛bp": f"{gross.mean() * 10000:.2f}",
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"净均收益": f"{r.mean() * 100:+.3f}%",
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"中位": f"{np.median(r) * 100:+.3f}%",
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"PF": f"{w.sum() / abs(o.sum()):.2f}" if len(o) else "inf",
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"t值": f"{r.mean() / (sd / np.sqrt(len(r))):+.2f}",
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"剔10%PF": f"{t10[t10 > 0].sum() / abs(t10[t10 <= 0].sum()):.2f}" if (t10 <= 0).any() else "inf",
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"滑点余量bp": f"{slip_budget(gross, reason, scaled):.2f}",
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"止盈占比": f"{(reason == TP).mean() * 100:.0f}%",
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"均持仓": f"{g[f'{cfg}_b'].mean():.1f}"}
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def rstat(g: pd.DataFrame, cfg: str, label: str, sl: float, cost_fn=cost_of) -> dict:
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"""R 倍数口径：固定风险仓位下，不同初始止损之间唯一可比的量。
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SL 越宽，每笔风险越大、仓位越小，直接比百分比收益会把仓位差异算成策略优势。
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"""
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gross = g[f"{cfg}_g"].to_numpy()
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reason, scaled = g[f"{cfg}_r"].to_numpy(), g[f"{cfg}_c"].to_numpy()
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net = gross - cost_fn(reason, scaled)
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risk = sl * g["atr_pct"].to_numpy()
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r = net / risk
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return {"口径": label, "笔数": len(r), "风险%": f"{risk.mean() * 100:.2f}%",
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"净均收益": f"{net.mean() * 100:+.3f}%",
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"均R": f"{r.mean():+.3f}", "中位R": f"{np.median(r):+.3f}",
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"R夏普": f"{r.mean() / r.std(ddof=1):.3f}",
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"最差1%R": f"{np.percentile(r, 1):.2f}",
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"胜率": f"{(r > 0).mean() * 100:.1f}%"}
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