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如何只使用<代码>numpy限制最不发达国家广场。是否有任何办法将制约因素纳入<代码>numpy.linalg.lstsq(。或者有<代码>numpy+ 围起来,限制最不发达国家广场? 我知道,我可以很容易地利用<代码>cvxpy和scipy.optimize,但这种优化将成为numba的一部分。 JITed功能,这两个功能没有numba的支持。

EDIT: Here is a dummy example of a problem I want to solve

arr_true = np.vstack([np.random.normal(0.3, 2, size=20),
                      np.random.normal(1.4, 2, size=20),
                      np.random.normal(4.2, 2, size=20)]).transpose()
x_true = np.array([0.3, 0.35, 0.35])
y = arr_true @ x_true

def problem():
    noisy_arr = np.vstack([np.random.normal(0.3, 2, size=20),
                        np.random.normal(1.4, 2, size=20),
                        np.random.normal(4.2, 2, size=20)]).transpose()

    x = cvxpy.Variable(3)
    objective = cvxpy.Minimize(cvxpy.sum_squares(noisy_arr @ x - y))
    constraints = [0 <= x, x <= 1, cvxpy.sum(x) == 1]
    prob = cvxpy.Problem(objective, constraints)
    result = prob.solve()
    output = prob.value
    args = x.value

    return output, args

从根本上说,我的问题是,将1美元+x_2+x_3 = 1美元和4美元x_i、0 q x_i leq 1美元等值的Ax-y$降到最低。

Answer 1

Is it linear? Based on your numpy.linalg.lstsq suggestion I ll assume that for now. Also, I assume (to solve the simple case first) that you talk about equality constraints. Then you can easily solve it yourself by using Lagrange multipliers, and the solution can be found by solving a linear equation system.

If the problem is to minimize A * x - y over the constraints B * x = b, then solve

$egin{pmatrix}A^Tcdot A&B^T\B&end{pmatrix}cdotegin{pmatrix}x\lambdaend{pmatrix}=egin{pmatrix}A^Tcdot y\bend{pmatrix}$

As an example:

from numpy import array, zeros
from numpy.linalg import solve

A = array([[1, 2], [3, 4], [5, 6], [7, 8]])
y = array([3.2, 6.8, 11.1, 15.2])

B = array([[1, -1]])
b = array([0.1])

S = zeros((3, 3))
r = zeros(3)

S[:2, :2] = A.T @ A
S[2:, :2] = B
S[:2, 2:] = B.T
r[:2] = A.T @ y
r[2] = b

x = solve(S, r)[:2]
print(x)

记者: 也许可以列入插图,但简短的测试使我得以进入<条码>x。

[1.06262376 0.96262376]

This seems reasonable, and my constraint x1 - x2 = 0.1 is also fulfilled.

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