极小高维多元拟合问题

Question

我想使用 iminuit 执行多元拟合。当我使用二维时，一切工作正常，但如果我使用超过二维，我会收到“ValueError：太多值无法解压（预期为 2）”。

以下是 2D 拟合的工作代码：

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()

3D 拟合出现问题，如下所示：

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() 

有错误输出

---------------------------------------------------------------------------
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)

你能指出我做错了什么吗？任何帮助将不胜感激。

Answer 1

iminuit 中似乎有一个错误，它采用传入的 x 数组中的维数而不是第一个维度的大小。我已在 github 上提交了错误报告：https://github.com/scikit-hep/iminuit/issues/974