约束线性回归/二次规划python

Question

我有这样的数据集:

import numpy as np

a = np.array([1.2, 2.3, 4.2])
b = np.array([1, 5, 6])
c = np.array([5.4, 6.2, 1.9])

m = np.vstack([a,b,c])
y = np.array([5.3, 0.9, 5.6])

并希望适应约束线性回归

y = b1*a + b2*b + b3*c

其中所有b的总和为1且为正:b1 + b2 + b3 = 1

R中的类似问题在这里指定:

https://stats.stackexchange.com/questions/21565/how-do-i-fit-a-constrained-regression-in-r-so-that-c​​oefficients-total-1

我怎么能在python中这样做？

Answer 1

编辑: 这两种方法非常通用,可以用于中小规模的实例.要获得更有效的方法,请检查 chthonicdaemon 的答案(使用自定义预处理和scipy的optimize.nnls).

使用scipy

码

import numpy as np
from scipy.optimize import minimize

a = np.array([1.2, 2.3, 4.2])
b = np.array([1, 5, 6])
c = np.array([5.4, 6.2, 1.9])

m = np.vstack([a,b,c])
y = np.array([5.3, 0.9, 5.6])

def loss(x):
    return np.sum(np.square((np.dot(x, m) - y)))

cons = ({'type': 'eq',
         'fun' : lambda x: np.sum(x) - 1.0})

x0 = np.zeros(m.shape[0])
res = minimize(loss, x0, method='SLSQP', constraints=cons,
               bounds=[(0, np.inf) for i in range(m.shape[0])], options={'disp': True})

print(res.x)
print(np.dot(res.x, m.T))
print(np.sum(np.square(np.dot(res.x, m) - y)))

产量

Optimization terminated successfully.    (Exit mode 0)
        Current function value: 18.817792344
        Iterations: 5
        Function evaluations: 26
        Gradient evaluations: 5
[ 0.7760881  0.         0.2239119]
[ 1.87173571  2.11955951  4.61630834]
18.817792344

评估

• 很明显,模型能力/模型复杂性不足以获得良好的性能(高损失!)

使用由cvxpy建模的通用QP/SOCP优化

好处:

• cvxpy prooves,问题是凸出的
• 凸优化问题的收敛性得到保证(对于上述情况也可能如此)
• 一般来说:更准确
• 一般来说:在数值不稳定性方面更稳健(求解器只能解决SOCP;而不是像上面的SLSQP方法那样的非凸模型!)

码

import numpy as np
from cvxpy import *

a = np.array([1.2, 2.3, 4.2])
b = np.array([1, 5, 6])
c = np.array([5.4, 6.2, 1.9])

m = np.vstack([a,b,c])
y = np.array([5.3, 0.9, 5.6])

X = Variable(m.shape[0])
constraints = [X >= 0, sum_entries(X) == 1.0]

product = m.T * diag(X)
diff = sum_entries(product, axis=1) - y
problem = Problem(Minimize(norm(diff)), constraints)
problem.solve(verbose=True)

print(problem.value)
print(X.value)

产量

ECOS 2.0.4 - (C) embotech GmbH, Zurich Switzerland, 2012-15. Web: www.embotech.com/ECOS

It     pcost       dcost      gap   pres   dres    k/t    mu     step   sigma     IR    |   BT
 0  +0.000e+00  -0.000e+00  +2e+01  5e-01  1e-01  1e+00  4e+00    ---    ---    1  1  - |  -  - 
 1  +2.451e+00  +2.539e+00  +4e+00  1e-01  2e-02  2e-01  8e-01  0.8419  4e-02   2  2  2 |  0  0
 2  +4.301e+00  +4.306e+00  +2e-01  5e-03  7e-04  1e-02  4e-02  0.9619  1e-02   2  2  2 |  0  0
 3  +4.333e+00  +4.334e+00  +2e-02  4e-04  6e-05  1e-03  4e-03  0.9326  2e-02   2  1  2 |  0  0
 4  +4.338e+00  +4.338e+00  +5e-04  1e-05  2e-06  4e-05  1e-04  0.9698  1e-04   2  1  1 |  0  0
 5  +4.338e+00  +4.338e+00  +3e-05  8e-07  1e-07  3e-06  7e-06  0.9402  7e-03   2  1  1 |  0  0
 6  +4.338e+00  +4.338e+00  +7e-07  2e-08  2e-09  6e-08  2e-07  0.9796  1e-03   2  1  1 |  0  0
 7  +4.338e+00  +4.338e+00  +1e-07  3e-09  4e-10  1e-08  3e-08  0.8458  2e-02   2  1  1 |  0  0
 8  +4.338e+00  +4.338e+00  +7e-09  2e-10  2e-11  9e-10  2e-09  0.9839  5e-02   1  1  1 |  0  0

OPTIMAL (within feastol=1.7e-10, reltol=1.5e-09, abstol=6.5e-09).
Runtime: 0.000555 seconds.

4.337947939  # needs to be squared to be compared to scipy's output!
             #  as we are using l2-norm (outer sqrt) instead of sum-of-squares
             #  which is nicely converted to SOCP-form and easier to
             #  tackle by SOCP-based solvers like ECOS
             #  -> does not change the solution-vector x, only the obj-value
[[  7.76094262e-01]
 [  7.39698388e-10]
 [  2.23905737e-01]]

Answer 2

您可以通过一点数学和获得一个好的解决方案scipy.optimize.nnls：

首先我们做一下数学：

如果

y = b1 * a + b2 * b + b3 * c并且b1 + b2 + b3 = 1，则b3 = 1-b1-b2。

如果我们替代和简化，我们最终会得到

y-c = b1（a-c）+ b2（b-c）

现在，我们没有任何等式约束，并且nnls可以直接求解：

import scipy.optimize
A = np.vstack([a - c, b - c]).T
(b1, b2), norm = scipy.optimize.nnls(A, y - c)
b3 = 1 - b1 - b2

这将恢复使用cvxpy在其他答案中获得的解决方案。

b1 = 0.77608809648662802
b2 = 0.0
b3 = 0.22391190351337198
norm = 4.337947941595865

可以将这种方法推广到任意数量的尺寸，如下所示。假设我们有一个矩阵B，它由列中排列的原始问题的a，b，c构成。任何其他尺寸都将添加到此。

现在，我们可以做

A = B[:, :-1] - B[:, -1:]
bb, norm = scipy.optimize.nnls(A, y - B[:, -1])
bi = np.append(bb, 1 - sum(bb))