使用iminuit进行高维多元拟合时出现ValueError的问题求助

问题描述 #

使用iminuit执行多元拟合，二维拟合正常，但维度超过二维时出现错误：ValueError: too many values to unpack (expected 2)。

可正常运行的二维拟合代码 #

from iminuit import Minuit
from iminuit.cost import LeastSquares
import numpy as np

def model1(xy, a, b, c):
    x, y = xy
    return a + (x ** b) + (y ** c)

np.random.seed(1)
data_x = np.linspace(0, 1, 10)
data_y = np.linspace(0, 1, 10)

data_terr = 0.1
data_t1 = model1([data_x,data_y], 2, 3, 10) + data_terr * np.random.randn(len(data_x))

least_squares = LeastSquares([data_x,data_y], data_t1, data_terr, model1)

m1 = Minuit(least_squares, a=1.0, b=1.0, c=1.0)
m1.migrad()

出错的三维拟合代码 #

from iminuit import Minuit
from iminuit.cost import LeastSquares
import numpy as np

def model2(xyz, a, b, c, d):
    x, y, z = xyz
    return a + (x ** b) + (y ** c) + (z ** d)

np.random.seed(1)
data_x = np.linspace(0, 1, 10)
data_y = np.linspace(0, 1, 10)
data_z = np.linspace(0, 1, 10)

data_terr = 0.1
data_t2 = model2([data_x,data_y, data_z], 2, 3, 10, 20) + data_terr * np.random.randn(len(data_x))

least_squares = LeastSquares([data_x, data_y, data_z], data_t2, data_terr, model2)

m2 = Minuit(least_squares, a=1.0, b=1.0, c=1.0, d=1.0)
m2.migrad() 

错误输出 #

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
Cell In[30], line 16
     13 least_squares = LeastSquares([data_x, data_y, data_z], data_t2, data_terr, model2)
     15 m2 = Minuit(least_squares, a=1.0, b=1.0, c=1.0, d=1.0)
---> 16 m2.migrad() 

File ~/anaconda3/envs/minuit_fitting/lib/python3.9/site-packages/iminuit/minuit.py:702, in Minuit.migrad(self, ncall, iterate)
    700 if self._precision is not None:
    701     migrad.precision = self._precision
--> 702 fm = migrad(ncall, self._tolerance)
    703 if fm.is_valid or fm.has_reached_call_limit:
    704     break

File ~/anaconda3/envs/minuit_fitting/lib/python3.9/site-packages/iminuit/cost.py:594, in Cost.__call__(self, *args)
    579 def __call__(self, *args: float) -> float:
    580     """
    581     Evaluate the cost function.
    582 
   (...)
    592     float
    593     """
--> 594     r = self._call(args)
    595     if self.verbose >= 1:
    596         print(args, "->", r)

File ~/anaconda3/envs/minuit_fitting/lib/python3.9/site-packages/iminuit/cost.py:1721, in LeastSquares._call(self, args)
   1719 def _call(self, args: Sequence[float]) -> float:
   1720     x = self._masked.T[0] if self._ndim == 1 else self._masked.T[: self._ndim]
-> 1721     y, yerror = self._masked.T[self._ndim :]
   1722     ym = self._model(x, *args)
   1723     ym = _normalize_model_output(ym)

ValueError: too many values to unpack (expected 2)

错误原因 #

LeastSquares构造函数对输入的自变量格式有明确要求：当处理多维自变量时，需要传入二维数组（形状为(n_samples, n_dims)），而非多个一维数组组成的列表。

二维代码能运行是巧合：当传入两个一维数组时，LeastSquares会将其识别为1个维度的自变量+观测值y，刚好能完成拆分；但当传入三个一维数组时，它会错误地将前三个数组都当作自变量维度，导致后续拆分y和yerror时，剩余的元素数量不符合预期，触发ValueError。

修正后的三维拟合代码 #

方案1：调整自变量为二维数组（推荐） #

将多维自变量合并为形状为(n_samples, n_dims)的二维数组，并修改模型函数适配该格式：

from iminuit import Minuit
from iminuit.cost import LeastSquares
import numpy as np

def model2(x, a, b, c, d):
    # x是形状为(10,3)的二维数组，每一行对应一个样本的x,y,z值
    x_vals = x[:, 0]
    y_vals = x[:, 1]
    z_vals = x[:, 2]
    return a + (x_vals ** b) + (y_vals ** c) + (z_vals ** d)

np.random.seed(1)
data_x = np.linspace(0, 1, 10)
data_y = np.linspace(0, 1, 10)
data_z = np.linspace(0, 1, 10)

# 合并为二维数组
data_xyz = np.column_stack([data_x, data_y, data_z])

data_terr = 0.1
# 生成模拟数据时，需传入二维数组的转置（匹配模型原输入格式）
data_t2 = model2(data_xyz.T, 2, 3, 10, 20) + data_terr * np.random.randn(len(data_x))

least_squares = LeastSquares(data_xyz, data_t2, data_terr, model2)

m2 = Minuit(least_squares, a=1.0, b=1.0, c=1.0, d=1.0)
m2.migrad() 

方案2：保持模型函数格式，调整自变量传入方式 #

如果想保留原模型函数的解包写法，可将自变量整理为(n_dims, n_samples)的数组，同时通过LeastSquares的ndim参数明确指定维度数：

from iminuit import Minuit
from iminuit.cost import LeastSquares
import numpy as np

def model2(xyz, a, b, c, d):
    x, y, z = xyz
    return a + (x ** b) + (y ** c) + (z ** d)

np.random.seed(1)
data_x = np.linspace(0, 1, 10)
data_y = np.linspace(0, 1, 10)
data_z = np.linspace(0, 1, 10)

# 整理为(n_dims, n_samples)的数组
data_xyz = np.array([data_x, data_y, data_z])

data_terr = 0.1
data_t2 = model2(data_xyz, 2, 3, 10, 20) + data_terr * np.random.randn(len(data_x))

# 明确指定ndim=3，让LeastSquares正确识别自变量维度
least_squares = LeastSquares(data_xyz, data_t2, data_terr, model2, ndim=3)

m2 = Minuit(least_squares, a=1.0, b=1.0, c=1.0, d=1.0)
m2.migrad() 

内容的提问来源于stack exchange，提问作者PitisCat