我想确定一个最小二乘法问题的系数，约束条件是系数之和为1，并且系数在0和1之间。最小化的函数用函数'predictory_combination'表示，目标是确定x。https:/docs.scipy.orgdocscipyreferenceoptimize.minimiz-trustconstr.html。)，从而引发ValueError "预期平方矩阵"。我应该怎么改？
from scipy.optimize import LinearConstraint
from scipy.optimize import minimize
import numpy as np
import pandas as pd
def combination_jacobian(x, vDataOut, YPred1, YPred2):
jac= np.zeros_like(x)
jac[0] = np.sum(-2*np.multiply(YPred1, vDataOut - x[0]*YPred1 - x[1]*YPred2))
jac[1] = np.sum(-2*np.multiply(YPred1, vDataOut - x[0]*YPred1 - x[1]*YPred2))
print(jac)
return jac
def combination_hessian(x, vDataOut, YPred1, YPred2):
hess = np.zeros((len(x), len(x)))
hess[0,0] = np.sum(YPred1**2)
hess[0,1] = np.sum(YPred1*YPred2)
hess[1,0] = np.sum(YPred1*YPred2)
hess[1,1] = np.sum(YPred2**2)
print(hess)
return hess
def forecast_combination(x, vDataOut, YPred1, YPred2):
return np.sum((vDataOut - x[0]*YPred1 - x[1]*YPred2)**2)
vDataOut = pd.DataFrame(np.ones((100, 1))+2).values
YPred1 = pd.DataFrame(np.ones((100, 1))+1).values
YPred2 = pd.DataFrame(np.ones((100, 1))+7).values
constrained_least_squares = minimize(fun = forecast_combination, x0 = np.array([0.5, 0.5]), method='trust-constr', args = (vDataOut, YPred1, YPred2), jac = combination_jacobian, hess = combination_hessian, constraints=[LinearConstraint([[1, 1]], [1], [1])],bounds=([0, 0], [1,1]), options={'factorization_method': None})
weights = constrained_least_squares.x
使用因子化方法'SVDFactorization'，结果是初始权重（不能是最小二乘权重），没有误差。
ValueError Traceback (most recent call last)
<ipython-input-48-cbfef6ae97b2> in <module>
----> 6 constrained_least_squares = minimize(fun = forecast_combination, x0 = np.array([0.5, 0.5]), method='trust-constr', args = (vDataOut, YPred1, YPred2), jac = combination_jacobian, hess = combination_hessian, constraints=[LinearConstraint([[1, 1]], [1], [1])],bounds=([0, 0], [1,1]), options={'factorization_method': None})
~\AppData\Local\Continuum\anaconda3\envs\thesis\lib\site-packages\scipy\optimize\_minimize.py in minimize(fun, x0, args, method, jac, hess, hessp, bounds, constraints, tol, callback, options)
620 return _minimize_trustregion_constr(fun, x0, args, jac, hess, hessp,
621 bounds, constraints,
--> 622 callback=callback, **options)
623 elif meth == 'dogleg':
624 return _minimize_dogleg(fun, x0, args, jac, hess,
~\AppData\Local\Continuum\anaconda3\envs\thesis\lib\site-packages\scipy\optimize\_trustregion_constr\minimize_trustregion_constr.py in _minimize_trustregion_constr(fun, x0, args, grad, hess, hessp, bounds, constraints, xtol, gtol, barrier_tol, sparse_jacobian, callback, maxiter, verbose, finite_diff_rel_step, initial_constr_penalty, initial_tr_radius, initial_barrier_parameter, initial_barrier_tolerance, factorization_method, disp)
503 stop_criteria, state,
504 initial_constr_penalty, initial_tr_radius,
--> 505 factorization_method)
506
507 elif method == 'tr_interior_point':
~\AppData\Local\Continuum\anaconda3\envs\thesis\lib\site-packages\scipy\optimize\_trustregion_constr\equality_constrained_sqp.py in equality_constrained_sqp(fun_and_constr, grad_and_jac, lagr_hess, x0, fun0, grad0, constr0, jac0, stop_criteria, state, initial_penalty, initial_trust_radius, factorization_method, trust_lb, trust_ub, scaling)
79 S = scaling(x)
80 # Get projections
---> 81 Z, LS, Y = projections(A, factorization_method)
82 # Compute least-square lagrange multipliers
83 v = -LS.dot(c)
~\AppData\Local\Continuum\anaconda3\envs\thesis\lib\site-packages\scipy\optimize\_trustregion_constr\projections.py in projections(A, method, orth_tol, max_refin, tol)
400 = svd_factorization_projections(A, m, n, orth_tol, max_refin, tol)
401
--> 402 Z = LinearOperator((n, n), null_space)
403 LS = LinearOperator((m, n), least_squares)
404 Y = LinearOperator((n, m), row_space)
~\AppData\Local\Continuum\anaconda3\envs\thesis\lib\site-packages\scipy\sparse\linalg\interface.py in __init__(self, shape, matvec, rmatvec, matmat, dtype, rmatmat)
516 self.__matmat_impl = matmat
517
--> 518 self._init_dtype()
519
520 def _matmat(self, X):
~\AppData\Local\Continuum\anaconda3\envs\thesis\lib\site-packages\scipy\sparse\linalg\interface.py in _init_dtype(self)
173 if self.dtype is None:
174 v = np.zeros(self.shape[-1])
--> 175 self.dtype = np.asarray(self.matvec(v)).dtype
176
177 def _matmat(self, X):
~\AppData\Local\Continuum\anaconda3\envs\thesis\lib\site-packages\scipy\sparse\linalg\interface.py in matvec(self, x)
227 raise ValueError('dimension mismatch')
228
--> 229 y = self._matvec(x)
230
231 if isinstance(x, np.matrix):
~\AppData\Local\Continuum\anaconda3\envs\thesis\lib\site-packages\scipy\sparse\linalg\interface.py in _matvec(self, x)
525
526 def _matvec(self, x):
--> 527 return self.__matvec_impl(x)
528
529 def _rmatvec(self, x):
~\AppData\Local\Continuum\anaconda3\envs\thesis\lib\site-packages\scipy\optimize\_trustregion_constr\projections.py in null_space(x)
191 # v = P inv(R) Q.T x
192 aux1 = Q.T.dot(x)
--> 193 aux2 = scipy.linalg.solve_triangular(R, aux1, lower=False)
194 v = np.zeros(m)
195 v[P] = aux2
~\AppData\Local\Continuum\anaconda3\envs\thesis\lib\site-packages\scipy\linalg\basic.py in solve_triangular(a, b, trans, lower, unit_diagonal, overwrite_b, debug, check_finite)
336 b1 = _asarray_validated(b, check_finite=check_finite)
337 if len(a1.shape) != 2 or a1.shape[0] != a1.shape[1]:
--> 338 raise ValueError('expected square matrix')
339 if a1.shape[0] != b1.shape[0]:
340 raise ValueError('incompatible dimensions')
ValueError: expected square matrix
我不知道具体问题是什么，但我想提出重要的变化。你的拟合是 x[0] 和 x[1]即2个参数。因为你在强迫 x[0]+x[1]=1.，那么问题就出在单参数上。你应该找的是 x[0] 只用 x[1]=x[0]. 这让你的任务简化了很多。
我改了 LinearConstraint([[1, 1]], [1], [1]) => LinearConstraint([[1, 1]], [0.99], [1.01]) 和错误消失了。
我将展示我解决这个问题的两种方法的代码：按照@rpoleski的建议，解决原问题和转换问题，优化1个参数。
正如我在上面的评论中提到的那样，边界的表述应该是不同的，最小化函数中的Jacobian和Hessian不是最小化的函数（'predict_combination'）的Jacobian和Hessian，而是约束条件的Jacobian和Hessian。
import numpy as np
import pandas as pd
from scipy.optimize import LinearConstraint
from scipy.optimize import minimize
from scipy.optimize import Bounds
def forecast_combination(x, vDataOut, YPred1, YPred2):
return np.sum((vDataOut - x[0]*YPred1 - x[1]*YPred2)**2)
vDataOut = pd.DataFrame(np.ones((100, 1))+2).values
YPred1 = pd.DataFrame(np.ones((100, 1))+1).values
YPred2 = pd.DataFrame(np.ones((100, 1))+7).values
constrained_least_squares = minimize(fun = forecast_combination, x0 = np.array([0.5, 0.5]), method='trust-constr', args = (vDataOut, YPred1, YPred2), constraints=[LinearConstraint([1, 1], [1], [1])],bounds=Bounds([0, 0], [1,1]))
weights = constrained_least_squares.x
print('Weights of forecast combination regression: ', weights)
当使用x[1]=1-x[0]时，也会得到同样的结果。
import numpy as np
import pandas as pd
from scipy.optimize import LinearConstraint
from scipy.optimize import minimize
from scipy.optimize import Bounds
def forecast_combination(x, vDataOut, YPred1, YPred2):
return np.sum((vDataOut - x[0]*YPred1 - (1 - x[0])*YPred2)**2)
vDataOut = pd.DataFrame(np.ones((100, 1))+2).values
YPred1 = pd.DataFrame(np.ones((100, 1))+1).values
YPred2 = pd.DataFrame(np.ones((100, 1))+7).values
constrained_least_squares = minimize(fun = forecast_combination, x0 = np.array([0.5]), method='trust-constr', args = (vDataOut, YPred1, YPred2), bounds=Bounds([0],[1]))
weights = constrained_least_squares.x
print('Weights of forecast combination regression: ', weights)