Python scipy.optimized受限最小化SLSQP-无法与目标100%_Python_Math_Optimization_Machine Learning_Scipy

Python scipy.optimized受限最小化SLSQP-无法与目标100%

python math optimization machine-learning

Python scipy.optimized受限最小化SLSQP-无法与目标100%,python,math,optimization,machine-learning,scipy,Python,Math,Optimization,Machine Learning,Scipy,我试图得到最小的重量，接近平均重量的总和。我目前的问题是使用SLSQP解算器，我无法找到满足目标100%的正确权重。有没有其他解决方案可以用来解决我的问题？或者任何数学建议。请帮忙我现在的数学是 **min⁡(∑|x-mean(x)|)** **s.t.** Aw-b=0, w>=0 **bound** 0.2'<'x<0.5 问题：找到接近平均值的最小重量 Python代码： A = np.arra

我试图得到最小的重量，接近平均重量的总和。我目前的问题是使用SLSQP解算器，我无法找到满足目标100%的正确权重。有没有其他解决方案可以用来解决我的问题？或者任何数学建议。请帮忙

我现在的数学是

           **min⁡(∑|x-mean(x)|)**

           **s.t.** Aw-b=0, w>=0


           **bound** 0.2'<'x<0.5

问题：找到接近平均值的最小重量

Python代码：

A = np.array([[3582.000000, 3394.000000, 3356.000000, 3256.000000, 3415.000000,
        3505.000000, 3442.000000, 3381.000000, 3392.000000],
       [233445193.000000, 217344811.000000, 237746918.000000,
        219204892.000000, 225272825.000000, 242819442.000000,
        215258725.000000, 227681178.000000, 215189377.000000],
       [559090945.000000, 496500751.000000, 493639029.000000,
        461547877.000000, 501057960.000000, 505073223.000000,
        490458632.000000, 503102998.000000, 487026744.000000]])

b = np.array([8531, 1079115532, 429386951]) 

n=9
def fsolveMin(A,b,n):

    # The constraints: Ax = b

    def cons(x):
        return A.dot(x)-b

    cons = ({'type':'eq','fun':cons},

            {'type':'ineq','fun':lambda x:x[0]})

    # The minimizing constraints: the total absolute difference between the coefficients

    def fn(x):
        return np.sum(np.abs(x-np.mean(x)))       

    # Initialize the coefficients randomly
    z0 = abs(np.random.randn(len(A[1,:])))

    # Set up bound
   # bnds = [(0, None)]*n

    # Solve the problem

    sol = minimize(fn, x0 = z0, constraints = cons, method = 'SLSQP', options={'disp': True})



    #expected 35%
    print(sol.x)
    print(A.dot(sol.x))

    #print(fn(sol.x))

print(str(fsolveMin(A,b,n))+"\n\n")

首先引入一个带有约束的自由变量mu：

   mu = sum(i, x(i))/n

然后引入非负变量

y（i）

：

   -y(i) <= x(i) - mu <= y(i)

现在这是一个直接的LP（线性目标和线性约束），可以用任何LP解算器来解决。

为了让您了解像scipy的这样的低级工具会有多臃肿，我们必须模仿它们的标准形式：

其基本思想是：

为平均值添加一个辅助变量，如Erwin建议的
添加n个辅助变量来处理绝对值
（我使用n-extra-aux-vars作为
```
x-mean
```
的临时变量）

现在这个例子起作用了

对于您的数据，您必须小心边界。确保问题始终可行。问题本身非常不稳定，因为硬约束用于

Ax=b

；通常，您会在这里最小化一些范数/最小二乘（不再是LP；QP/SOCP），并将此误差添加到目标中）

可能需要在某个点将解算器从

method='simplex'

切换到

method='interior-point'

（仅自scipy 1.0起可用）

备选方案：

使用，公式更容易（提到的两种变体），并且您可以免费获得相当好的解算器（ECOS）

代码：

现在，对于您的原始数据，，事情变得艰难了
您需要：

缩放数据，因为变量的大小会影响解算器

根据您的任务，您可能需要分析缩放/反转的效果

使用内部点解算器

小心边界

例如：

A = np.array([[3582.000000, 3394.000000, 3356.000000, 3256.000000, 3415.000000, 3505.000000, 3442.000000, 3381.000000, 3392.000000], [233445193.000000, 217344811.000000, 237746918.000000, 219204892.000000, 225272825.000000, 242819442.000000, 215258725.000000, 227681178.000000, 215189377.000000], [559090945.000000, 496500751.000000, 493639029.000000, 461547877.000000, 501057960.000000, 505073223.000000, 490458632.000000, 503102998.000000, 487026744.000000]]) b = np.array([8531., 1079115532., 429386951.]) A /= 10000. # scaling b /= 10000. # scaling bounds = [(-50., 50.) for i in range(n)] + \ ... result = spo.linprog(c, A_ineq, b_ineq, A_eq, b_eq, bounds=bounds, method='interior-point')
输出：

Original A [[ 4.17022005e-01 7.20324493e-01 1.14374817e-04] [ 3.02332573e-01 1.46755891e-01 9.23385948e-02]] hidden x: [ 0.18626021 0.34556073 0.39676747] target: [ 0.32663584 0.14366255] Reformulation y = mean A: [ 0.33333333 0.33333333 0.33333333 -1. 0. 0. 0. 0. 0. 0. ] y = mean b: [0] Ax=b A: [[ 4.17022005e-01 7.20324493e-01 1.14374817e-04 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00] [ 3.02332573e-01 1.46755891e-01 9.23385948e-02 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00]] Ax=b b: [ 0.32663584 0.14366255] z = x - mean A: [[-1. -0. -0. 1. 1. 0. 0. 0. 0. 0.] [-0. -1. -0. 1. 0. 1. 0. 0. 0. 0.] [-0. -0. -1. 1. 0. 0. 1. 0. 0. 0.]] z = x - mean b: [ 0. 0. 0.] solve... fun: 0.078779576294411263 message: 'Optimization terminated successfully.' nit: 10 slack: array([ 0.07877958, 0. , 0. , 0. , 0.07877958, 0. , 0.76273076, 0.68395118, 0.72334097]) status: 0 success: True x: array([ 0.23726924, 0.31604882, 0.27665903, 0.27665903, -0.03938979, 0.03938979, 0. , 0.03938979, 0.03938979, 0. ]) x: [ 0.23726924 0.31604882 0.27665903] residual Ax-b: [ 5.55111512e-17 0.00000000e+00] mean: 0.276659030053 x - mean: [-0.03938979 0.03938979 0. ] l1-norm(x - mean) / objective: 0.0787795762944

solve... con: array([ 3.98410760e-09, 1.18067724e-08, 8.12879938e-04, 1.75969041e-03, -3.93853838e-09, -3.96305566e-09, -4.10043555e-09, -3.94957667e-09, -3.88362764e-09, -3.89452381e-09, -3.95134592e-09, -3.92182287e-09, -3.85762178e-09]) fun: 52.742900697626389 message: 'Optimization terminated successfully.' nit: 8 slack: array([ 5.13245265e+01, 1.89309145e-08, 1.83429094e-09, 4.28687782e-09, 1.03726911e-08, 2.77000474e-09, 1.41837413e+00, 6.75769654e-09, 8.65285462e-10, 2.78501844e-09, 3.09591539e-09, 5.27429006e+01, 1.30944103e-08, 5.32994799e-09, 3.15369669e-08, 2.51943821e-09, 7.54848797e-09, 3.22510447e-09]) status: 0 success: True x: array([ -2.51938304e+01, 4.68432810e-01, 2.68398831e+01, 4.68432822e-01, 4.68432815e-01, 4.68432832e-01, -2.40754247e-01, 4.68432818e-01, 4.68432819e-01, 4.68432822e-01, -2.56622633e+01, -7.91749954e-09, 2.63714503e+01, 4.40376624e-09, -2.52137156e-09, 1.43834811e-08, -7.09187065e-01, 3.95395716e-10, 1.17990950e-09, 2.56622633e+01, 1.10134149e-08, 2.63714503e+01, 8.69064406e-09, 7.85131955e-09, 1.71534858e-08, 7.09187068e-01, 7.15309226e-09, 2.04519496e-09]) x: [-25.19383044 0.46843281 26.83988313 0.46843282 0.46843282 0.46843283 -0.24075425 0.46843282 0.46843282] residual Ax-b: [ -1.18067724e-08 -8.12879938e-04 -1.75969041e-03] mean: 0.468432821891 x - mean: [ -2.56622633e+01 -1.18805552e-08 2.63714503e+01 4.54189575e-10 -6.40499920e-09 1.04889573e-08 -7.09187069e-01 -3.52642715e-09 -2.67771227e-09] l1-norm(x - mean) / objective: 52.7429006758

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 +2.637e-17 -1.550e+06 +7e+08 1e-01 2e-04 1e+00 2e+07 --- --- 1 1 - | - - 1 -8.613e+04 -1.014e+05 +8e+06 1e-03 2e-06 2e+03 2e+05 0.9890 1e-04 2 1 1 | 0 0 2 -1.287e+03 -1.464e+03 +1e+05 1e-05 9e-08 4e+01 3e+03 0.9872 1e-04 3 1 1 | 0 0 3 +1.794e+02 +1.900e+02 +2e+03 2e-07 1e-07 1e+01 5e+01 0.9890 5e-03 5 3 4 | 0 0 4 -1.388e+00 -6.826e-01 +1e+02 1e-08 7e-08 9e-01 3e+00 0.9458 4e-03 7 6 6 | 0 0 5 +5.491e+00 +5.683e+00 +1e+01 1e-09 8e-09 2e-01 3e-01 0.9617 6e-02 1 1 1 | 0 0 6 +6.480e+00 +6.505e+00 +1e+00 2e-10 5e-10 3e-02 4e-02 0.8928 2e-02 1 1 1 | 0 0 7 +6.746e+00 +6.746e+00 +2e-02 3e-12 5e-10 5e-04 6e-04 0.9890 5e-03 1 0 0 | 0 0 8 +6.759e+00 +6.759e+00 +3e-04 2e-12 2e-10 6e-06 7e-06 0.9890 1e-04 1 0 0 | 0 0 9 +6.759e+00 +6.759e+00 +3e-06 2e-13 2e-10 6e-08 8e-08 0.9890 1e-04 2 0 0 | 0 0 10 +6.758e+00 +6.758e+00 +3e-08 5e-14 2e-10 7e-10 9e-10 0.9890 1e-04 1 0 0 | 0 0 OPTIMAL (within feastol=2.0e-10, reltol=4.7e-09, abstol=3.2e-08). Runtime: 0.002901 seconds. Objective: 6.757722879805085 x: [[-18.09169736 -5.55768047 11.12130645 11.48355878 -1.13982006 12.4290884 -3.00165819 -1.05158589 -2.4468432 ]] mean: 0.416074272576 Ax-b: [[ 2.17051777e-03] [ 1.90734863e-06] [ -5.72204590e-06]] x - mean: [[-18.50777164 -5.97375474 10.70523218 11.0674845 -1.55589434 12.01301413 -3.41773246 -1.46766016 -2.86291747]]

Objective: 785913288.2410747 x: [[ -5.57966858e-08 -4.74997454e-08 1.56066068e+00 1.68021234e-07 -3.55602958e-08 1.75340641e-06 -4.69609562e-08 -3.10216680e-08 -4.39482554e-08]] mean: 0.173406926909 Ax-b: [[ -3.29341696e+03] [ -7.08072860e+08] [ 3.41016903e+08]] x - mean: [[-0.17340698 -0.17340697 1.38725375 -0.17340676 -0.17340696 -0.17340517 -0.17340697 -0.17340696 -0.17340697]]
编辑
这里是一个基于SOCP的最小二乘（软约束）方法，我建议在数值稳定性方面使用它！这种方法可以而且应该根据您的需要进行调整。它是使用前面提到的建模工具（使用ECOS解算器）实现的
基本思想是：

代替：
最小（l1标准（x-平均值（x））标准轴=b

求解：
min（l2范数（Ax-b）+c*l1范数（x-均值（x））

其中
c
是非负权衡参数

小的
c
：
Ax=b
更重要

大
c
：
x-均值（x）
更重要

数据的示例代码和输出以及
-50、50
和
c=1e-3
的界限：

import numpy as np import cvxpy as cvx """ DATA """ A = np.array([[3582.000000, 3394.000000, 3356.000000, 3256.000000, 3415.000000, 3505.000000, 3442.000000, 3381.000000, 3392.000000], [233445193.000000, 217344811.000000, 237746918.000000, 219204892.000000, 225272825.000000, 242819442.000000, 215258725.000000, 227681178.000000, 215189377.000000], [559090945.000000, 496500751.000000, 493639029.000000, 461547877.000000, 501057960.000000, 505073223.000000, 490458632.000000, 503102998.000000, 487026744.000000]]) b = np.array([8531., 1079115532., 429386951.]) n = 9 # A /= 10000. scaling would be a good idea # b /= 10000. """ """ SOCP-based least-squares approach """ def solve(A, b, n, c=1e-1): x = cvx.Variable(n) y = cvx.Variable(1) # mean lower_bounds = np.zeros(n) - 50 # -50 upper_bounds = np.zeros(n) + 50 # 50 constraints = [] constraints.append(x >= lower_bounds) constraints.append(x <= upper_bounds) constraints.append(y == cvx.sum_entries(x) / n) objective = cvx.Minimize(cvx.norm(A*x-b, 2) + c * cvx.norm(x - y, 1)) problem = cvx.Problem(objective, constraints) problem.solve(solver=cvx.ECOS, verbose=True) print('Objective: ', problem.value) print('x: ', x.T.value) print('mean: ', y.value) print('Ax-b: ') print((A*x - b).value) print('x - mean: ', (x - y).T.value) solve(A, b, n)
这种方法将始终输出一个可行的解决方案（对于我们的任务），然后您可以决定观察到的残差是否适合您
正如您所观察到的，在所有公式中，0的下限都是致命的（看看数据中的大小差异！）
在这里，0的下限将为您提供一个残差较高的解决方案
例如：

c=1e-7

界限=
0/15

输出：

Original A [[ 4.17022005e-01 7.20324493e-01 1.14374817e-04] [ 3.02332573e-01 1.46755891e-01 9.23385948e-02]] hidden x: [ 0.18626021 0.34556073 0.39676747] target: [ 0.32663584 0.14366255] Reformulation y = mean A: [ 0.33333333 0.33333333 0.33333333 -1. 0. 0. 0. 0. 0. 0. ] y = mean b: [0] Ax=b A: [[ 4.17022005e-01 7.20324493e-01 1.14374817e-04 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00] [ 3.02332573e-01 1.46755891e-01 9.23385948e-02 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00 0.00000000e+00]] Ax=b b: [ 0.32663584 0.14366255] z = x - mean A: [[-1. -0. -0. 1. 1. 0. 0. 0. 0. 0.] [-0. -1. -0. 1. 0. 1. 0. 0. 0. 0.] [-0. -0. -1. 1. 0. 0. 1. 0. 0. 0.]] z = x - mean b: [ 0. 0. 0.] solve... fun: 0.078779576294411263 message: 'Optimization terminated successfully.' nit: 10 slack: array([ 0.07877958, 0. , 0. , 0. , 0.07877958, 0. , 0.76273076, 0.68395118, 0.72334097]) status: 0 success: True x: array([ 0.23726924, 0.31604882, 0.27665903, 0.27665903, -0.03938979, 0.03938979, 0. , 0.03938979, 0.03938979, 0. ]) x: [ 0.23726924 0.31604882 0.27665903] residual Ax-b: [ 5.55111512e-17 0.00000000e+00] mean: 0.276659030053 x - mean: [-0.03938979 0.03938979 0. ] l1-norm(x - mean) / objective: 0.0787795762944

solve... con: array([ 3.98410760e-09, 1.18067724e-08, 8.12879938e-04, 1.75969041e-03, -3.93853838e-09, -3.96305566e-09, -4.10043555e-09, -3.94957667e-09, -3.88362764e-09, -3.89452381e-09, -3.95134592e-09, -3.92182287e-09, -3.85762178e-09]) fun: 52.742900697626389 message: 'Optimization terminated successfully.' nit: 8 slack: array([ 5.13245265e+01, 1.89309145e-08, 1.83429094e-09, 4.28687782e-09, 1.03726911e-08, 2.77000474e-09, 1.41837413e+00, 6.75769654e-09, 8.65285462e-10, 2.78501844e-09, 3.09591539e-09, 5.27429006e+01, 1.30944103e-08, 5.32994799e-09, 3.15369669e-08, 2.51943821e-09, 7.54848797e-09, 3.22510447e-09]) status: 0 success: True x: array([ -2.51938304e+01, 4.68432810e-01, 2.68398831e+01, 4.68432822e-01, 4.68432815e-01, 4.68432832e-01, -2.40754247e-01, 4.68432818e-01, 4.68432819e-01, 4.68432822e-01, -2.56622633e+01, -7.91749954e-09, 2.63714503e+01, 4.40376624e-09, -2.52137156e-09, 1.43834811e-08, -7.09187065e-01, 3.95395716e-10, 1.17990950e-09, 2.56622633e+01, 1.10134149e-08, 2.63714503e+01, 8.69064406e-09, 7.85131955e-09, 1.71534858e-08, 7.09187068e-01, 7.15309226e-09, 2.04519496e-09]) x: [-25.19383044 0.46843281 26.83988313 0.46843282 0.46843282 0.46843283 -0.24075425 0.46843282 0.46843282] residual Ax-b: [ -1.18067724e-08 -8.12879938e-04 -1.75969041e-03] mean: 0.468432821891 x - mean: [ -2.56622633e+01 -1.18805552e-08 2.63714503e+01 4.54189575e-10 -6.40499920e-09 1.04889573e-08 -7.09187069e-01 -3.52642715e-09 -2.67771227e-09] l1-norm(x - mean) / objective: 52.7429006758

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 +2.637e-17 -1.550e+06 +7e+08 1e-01 2e-04 1e+00 2e+07 --- --- 1 1 - | - - 1 -8.613e+04 -1.014e+05 +8e+06 1e-03 2e-06 2e+03 2e+05 0.9890 1e-04 2 1 1 | 0 0 2 -1.287e+03 -1.464e+03 +1e+05 1e-05 9e-08 4e+01 3e+03 0.9872 1e-04 3 1 1 | 0 0 3 +1.794e+02 +1.900e+02 +2e+03 2e-07 1e-07 1e+01 5e+01 0.9890 5e-03 5 3 4 | 0 0 4 -1.388e+00 -6.826e-01 +1e+02 1e-08 7e-08 9e-01 3e+00 0.9458 4e-03 7 6 6 | 0 0 5 +5.491e+00 +5.683e+00 +1e+01 1e-09 8e-09 2e-01 3e-01 0.9617 6e-02 1 1 1 | 0 0 6 +6.480e+00 +6.505e+00 +1e+00 2e-10 5e-10 3e-02 4e-02 0.8928 2e-02 1 1 1 | 0 0 7 +6.746e+00 +6.746e+00 +2e-02 3e-12 5e-10 5e-04 6e-04 0.9890 5e-03 1 0 0 | 0 0 8 +6.759e+00 +6.759e+00 +3e-04 2e-12 2e-10 6e-06 7e-06 0.9890 1e-04 1 0 0 | 0 0 9 +6.759e+00 +6.759e+00 +3e-06 2e-13 2e-10 6e-08 8e-08 0.9890 1e-04 2 0 0 | 0 0 10 +6.758e+00 +6.758e+00 +3e-08 5e-14 2e-10 7e-10 9e-10 0.9890 1e-04 1 0 0 | 0 0 OPTIMAL (within feastol=2.0e-10, reltol=4.7e-09, abstol=3.2e-08). Runtime: 0.002901 seconds. Objective: 6.757722879805085 x: [[-18.09169736 -5.55768047 11.12130645 11.48355878 -1.13982006 12.4290884 -3.00165819 -1.05158589 -2.4468432 ]] mean: 0.416074272576 Ax-b: [[ 2.17051777e-03] [ 1.90734863e-06] [ -5.72204590e-06]] x - mean: [[-18.50777164 -5.97375474 10.70523218 11.0674845 -1.55589434 12.01301413 -3.41773246 -1.46766016 -2.86291747]]

Objective: 785913288.2410747 x: [[ -5.57966858e-08 -4.74997454e-08 1.56066068e+00 1.68021234e-07 -3.55602958e-08 1.75340641e-06 -4.69609562e-08 -3.10216680e-08 -4.39482554e-08]] mean: 0.173406926909 Ax-b: [[ -3.29341696e+03] [ -7.08072860e+08] [ 3.41016903e+08]] x - mean: [[-0.17340698 -0.17340697 1.38725375 -0.17340676 -0.17340696 -0.17340517 -0.17340697 -0.17340696 -0.17340697]]

你使这成为一个困难的NLP（不可微非线性规划问题）。我将把它表述为LP（线性规划）；这样可以得到一个经验证的全局最优值。@Erwin.你如何在线性规划部分添加约束？我在回答中详细说明了我的建议。谢谢Sascha。我肯定已经接近了。我收到了以下错误消息：“优化失败。找不到可行的起点。”我提到了可能的问题和事情检查并执行。因此：（1）确保编辑边界后问题仍然可行（2）使用内部点解算器（它也检测不可行）（3）放弃lp硬约束方式并制定SOCP。谢谢sascha。我调整了我的边界并得到了我想要的答案。非常感谢！！嗨sascha，当我将边界设置为0到50时。它给我一个错误消息“算法成功终止并确定问题不可行”。你能帮我吗？是的…或者没有非否定ive x解决方案或者你遇到了数字问题。我都提到过。这是不可能从这里调试的，即使有时很难向人们解释为什么这和那不正确，如果他们不知道这些解决方案的内部结构。（我或多或少已经提到，您尝试优化的模型非常不稳定；Ax=b上的硬约束）