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[英]Scipy.optimize.minimize SLSQP with linear constraints fails
[英]Python's scipy.optimize.minimize with SLSQP fails with “Positive directional derivative for linesearch”
我有一個最小二乘最小化問題,受不等式約束的影響,我正在嘗試使用scipy.optimize.minimize解決。 對於不平等約束,似乎有兩種選擇:COBYLA和SLSQP。
我首先嘗試了SLSQP,因為它允許將函數的顯式偏導數最小化。 根據問題的嚴重程度,它會因錯誤而失敗:
Positive directional derivative for linesearch (Exit mode 8)
只要施加間隔或更普遍的不平等約束。
以前例如在這里已經觀察到了這一點 。 手動縮放要最小化的功能(以及相關的偏導數)似乎可以解決此問題,但是我無法通過更改選項中的ftol來實現相同的效果。
總的來說,這件事使我對常規程序的健壯性產生懷疑。 這是一個簡化的示例:
import numpy as np
import scipy.optimize as sp_optimize
def cost(x, A, y):
e = y - A.dot(x)
rss = np.sum(e ** 2)
return rss
def cost_deriv(x, A, y):
e = y - A.dot(x)
deriv0 = -2 * e.dot(A[:,0])
deriv1 = -2 * e.dot(A[:,1])
deriv = np.array([deriv0, deriv1])
return deriv
A = np.ones((10,2)); A[:,0] = np.linspace(-5,5, 10)
x_true = np.array([2, 2/20])
y = A.dot(x_true)
x_guess = x_true / 2
prm_bounds = ((0, 3), (0,1))
cons_SLSQP = ({'type': 'ineq', 'fun' : lambda x: np.array([x[0] - x[1]]),
'jac' : lambda x: np.array([1.0, -1.0])})
# works correctly
min_res_SLSQP = sp_optimize.minimize(cost, x_guess, args=(A, y), jac=cost_deriv, bounds=prm_bounds, method='SLSQP', constraints=cons_SLSQP, options={'disp': True})
print(min_res_SLSQP)
# fails
A = 100 * A
y = A.dot(x_true)
min_res_SLSQP = sp_optimize.minimize(cost, x_guess, args=(A, y), jac=cost_deriv, bounds=prm_bounds, method='SLSQP', constraints=cons_SLSQP, options={'disp': True})
print(min_res_SLSQP)
# works if bounds and inequality constraints removed
min_res_SLSQP = sp_optimize.minimize(cost, x_guess, args=(A, y), jac=cost_deriv,
method='SLSQP', options={'disp': True})
print(min_res_SLSQP)
應該如何設置以避免故障? 更一般而言,COBYLA是否會出現類似的問題? 對於這種類型的不等式約束最小二乘優化問題,COBYLA是更好的選擇嗎?
發現在成本函數中使用平方根可以提高性能。 但是,對於問題的非線性重新參數化(更簡單,但更接近我在實踐中需要做的事情),它再次失敗。 詳細信息如下:
import numpy as np
import scipy.optimize as sp_optimize
def cost(x, y, g):
e = ((y - x[1]) / x[0]) - g
rss = np.sqrt(np.sum(e ** 2))
return rss
def cost_deriv(x, y, g):
e = ((y- x[1]) / x[0]) - g
factor = 0.5 / np.sqrt(e.dot(e))
deriv0 = -2 * factor * e.dot(y - x[1]) / (x[0]**2)
deriv1 = -2 * factor * np.sum(e) / x[0]
deriv = np.array([deriv0, deriv1])
return deriv
x_true = np.array([1/300, .1])
N = 20
t = 20 * np.arange(N)
g = 100 * np.cos(2 * np.pi * 1e-3 * (t - t[-1] / 2))
y = g * x_true[0] + x_true[1]
x_guess = x_true / 2
prm_bounds = ((1e-4, 1e-2), (0, .4))
# check derivatives
delta = 1e-9
C0 = cost(x_guess, y, g)
C1 = cost(x_guess + np.array([delta, 0]), y, g)
approx_deriv0 = (C1 - C0) / delta
C1 = cost(x_guess + np.array([0, delta]), y, g)
approx_deriv1 = (C1 - C0) / delta
approx_deriv = np.array([approx_deriv0, approx_deriv1])
deriv = cost_deriv(x_guess, y, g)
# fails
min_res_SLSQP = sp_optimize.minimize(cost, x_guess, args=(y, g), jac=cost_deriv,
bounds=prm_bounds, method='SLSQP', options={'disp': True})
print(min_res_SLSQP)
代替最小化np.sum(e ** 2)
,最小化sqrt(np.sum(e ** 2))
或更好(就計算而言): np.linalg.norm(e)
!
此修改:
x
解決方案 有了這一更改,所有情況都可以使用 , 即使使用數值微分也是如此(我太懶了,無法修改梯度,這需要反映這一點!)。
輸出示例(func-evals的數量給出num-diff):
Optimization terminated successfully. (Exit mode 0)
Current function value: 3.815547437029837e-06
Iterations: 16
Function evaluations: 88
Gradient evaluations: 16
fun: 3.815547437029837e-06
jac: array([-6.09663382, -2.48862544])
message: 'Optimization terminated successfully.'
nfev: 88
nit: 16
njev: 16
status: 0
success: True
x: array([ 2.00000037, 0.10000018])
Optimization terminated successfully. (Exit mode 0)
Current function value: 0.0002354577991007501
Iterations: 23
Function evaluations: 114
Gradient evaluations: 23
fun: 0.0002354577991007501
jac: array([ 435.97259208, 288.7483819 ])
message: 'Optimization terminated successfully.'
nfev: 114
nit: 23
njev: 23
status: 0
success: True
x: array([ 1.99999977, 0.10000014])
Optimization terminated successfully. (Exit mode 0)
Current function value: 0.0003392807206384532
Iterations: 21
Function evaluations: 112
Gradient evaluations: 21
fun: 0.0003392807206384532
jac: array([ 996.57340243, 51.19298764])
message: 'Optimization terminated successfully.'
nfev: 112
nit: 21
njev: 21
status: 0
success: True
x: array([ 2.00000008, 0.10000104])
盡管SLSQP可能存在一些問題,但鑒於廣泛的應用范圍,它仍然是最受測試和最強大的代碼之一!
我還希望SLSQP在這里比COBYLA更好,因為COBYLA很大程度上基於線性化。 (但僅作為猜測;考慮最小接口,很容易嘗試!)
替代
通常,用於凸二次規划的基於內部點的求解器將是此處的最佳方法。 但是為此,您需要保持科學。 (或者SOCP求解器可能會更好...我不確定)。
cvxpy帶來了一個很好的建模系統和一個很好的開源求解器( ECOS ;盡管從技術上講是一個圓錐形求解器->更通用,更不可靠;但是應該勝過SLSQP)。
使用cvxpy和ECOS,這看起來像:
import numpy as np
import cvxpy as cvx
""" Problem data """
A = np.ones((10,2)); A[:,0] = np.linspace(-5,5, 10)
x_true = np.array([2, 2/20])
y = A.dot(x_true)
x_guess = x_true / 2
prm_bounds = ((0, 3), (0,1))
# problematic case
A = 100 * A
y = A.dot(x_true)
""" Solve """
x = cvx.Variable(len(x_true))
constraints = [x[0] >= x[1]]
for ind, (lb, ub) in enumerate(prm_bounds): # ineffecient -> matrix-based expr better!
constraints.append(x[ind] >= lb)
constraints.append(x[ind] <= ub)
objective = cvx.Minimize(cvx.norm(A*x - y))
problem = cvx.Problem(objective, constraints)
problem.solve(solver=cvx.ECOS, verbose=False)
print(problem.status)
print(problem.value)
print(x.value.T)
# optimal
# -6.67593652593801e-10
# [[ 2. 0.1]]
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