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https://github.com/InternLM/InternBootcamp.git
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78 lines
No EOL
2.6 KiB
Python
78 lines
No EOL
2.6 KiB
Python
import re
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import json
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import numpy as np
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from scipy.integrate import odeint
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from internbootcamp.bootcamp.base import Basebootcamp
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class LinearODEBootcamp(Basebootcamp):
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def __init__(
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self,
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k_range=(0.1, 1.0),
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x0_range=(0.5, 2.0),
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t_span=(0, 5),
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n_points=50,
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seed=None
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):
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self.k_range, self.x0_range = k_range, x0_range
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self.t0, self.t1 = t_span
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self.n_points = n_points
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if seed is not None:
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np.random.seed(seed)
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def case_generator(self):
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# 1. 随机采样参数 k 和初始值 x0
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k = float(np.random.uniform(*self.k_range))
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x0 = float(np.random.uniform(*self.x0_range))
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# 2. 构造时间序列并模拟 dx/dt = -k * x
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t = np.linspace(self.t0, self.t1, self.n_points).tolist()
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def model(x, t_val):
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return -k * x
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x = odeint(model, x0, t).flatten().tolist()
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return {"t": t, "x": x, "k": k}
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def prompt_func(self, identity) -> str:
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# 将 (t, x) 对格式化为提示
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points = ", ".join(f"({t:.2f}, {x:.2f})"
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for t, x in zip(identity["t"], identity["x"]))
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return (
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f"下面给出变量 x(t) 的观测数据点：\n{points}\n\n"
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"请找出其满足的微分方程，形式为：dx/dt = f(x)。\n"
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"只需返回 “dx/dt = <表达式>”。"
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)
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@staticmethod
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def extract_output(output: str) -> str:
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# 用正则提取“dx/dt = …”右侧的表达式
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m = re.search(r"dx/dt\s*=\s*([^\n\r]+)", output)
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return m.group(1).strip() if m else None
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@classmethod
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def _verify_correction(cls, solution: str, identity: dict) -> bool:
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# 解析 LLM 给出的系数 c，形如 “c*x”
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sol = solution.replace(" ", "")
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match = re.fullmatch(r"([\-0-9\.eE]+)\*x", sol)
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if not match:
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return False
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c = float(match.group(1))
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# 验证 c ≈ -k
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return abs(c + identity["k"]) < 1e-2
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if __name__ == "__main__":
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bootcamp = LinearODEBootcamp(seed=123)
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# 生成几个样例
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examples = [bootcamp.case_generator() for _ in range(3)]
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for identity in examples:
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# 构造“模型”返回答案，模拟 LLM 的输出
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coeff = -identity["k"]
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sol = f"{coeff:.4f}*x"
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# 调用 Basebootcamp 提供的 verify_score 接口进行验证
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score = bootcamp.verify_score(sol, identity, short_threshold=1e-2)
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# 打印结果
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print(json.dumps({
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"identity": identity,
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"solution": sol,
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"verify_score": score
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}, ensure_ascii=False, indent=2)) |