""" ___ """
from scipy.optimize import root
import numpy as np
LENGTH = 3
def process(x):
return x[0, 0] + x[0, 1] * 5
def draw(process, length):
""" """
X = np.matrix(np.random.normal(0, 10, (length, 2)))
y = np.matrix([process(x) for x in X])
y += np.random.normal(3, 1, len(y))
return y.T, X.T
def maximum_likelyhood(y, X):
def objective(b):
return (X.T * (y - X * b.T))
x0 = np.matrix([0, 0])
res = root(objective, x0=x0)
return res.x
y, X = draw(process, LENGTH)
X = X.transpose()
b = np.matrix([[0], [1]])
print maximum_likelyhood(y, X)
产生
Traceback (most recent call last):
File "ml.py", line 33, in <module>
maximum_likelyhood(y, X)
File "ml.py", line 26, in maximum_likelyhood
res = root(objective, x0=x0)
File "/usr/local/lib/python2.7/dist-packages/scipy/optimize/_root.py", line 168, in root
sol = _root_hybr(fun, x0, args=args, jac=jac, **options)
File "/usr/local/lib/python2.7/dist-packages/scipy/optimize/minpack.py", line 193, in _root_hybr
ml, mu, epsfcn, factor, diag)
ValueError: object too deep for desired array
我甚至无法解释问题是在进入目标的b中的问题 功能?还是在输出中?
答案 0 :(得分:11)
问题是fsolve和root不接受矩阵作为目标函数的返回值。
例如,这是上述问题的解决方案:
def maximum_likelyhood(y, X):
def objective(b):
b = np.matrix(b).T
return np.transpose(np.array((X.T * (y - X * b))))[0]
x0 = (1, 1)
res = root(objective, x0=x0)
return res.x