Python中支持Convex Optimization（凸规划）的模块为CVXOPT,其安装方式为：

卸载原Pyhon中的Numpy

安装CVXOPT的whl文件，链接为：https://www.lfd.uci.edu/~gohlke/pythonlibs/

安装Numpy+mkl的whl文件，链接为：https://www.lfd.uci.edu/~gohlke/pythonlibs/

之所以选择这种安装方式，是因为Python的whl和pip直接install的不兼容性。

CVXOPT的官方说明文档网址为：http://cvxopt.org/index.html， 现最新版本为1.1.9,由Martin Andersen, Joachim Dahl 和Lieven Vandenberghe共同开发完成，能够解决线性规划和二次型规划问题，其应用场景如SVM中的Hard Margin SVM.

CVXOPT使用举例如下：

线性规划问题

例1:

Python程序代码：

import numpy as npfrom cvxopt import matrix, solversA = matrix([[-1.0, -1.0, 0.0, 1.0], [1.0, -1.0, -1.0, -2.0]])b = matrix([1.0, -2.0, 0.0, 4.0])c = matrix([2.0, 1.0])sol = solvers.lp(c,A,b)print(sol['x'])print(np.dot(sol['x'].T, c))print(sol['primal objective'])

输出结果：

pcost dcost gap pres dres k/t 0: 2.6471e+00 -7.0588e-01 2e+01 8e-01 2e+00 1e+00 1: 3.0726e+00 2.8437e+00 1e+00 1e-01 2e-01 3e-01 2: 2.4891e+00 2.4808e+00 1e-01 1e-02 2e-02 5e-02 3: 2.4999e+00 2.4998e+00 1e-03 1e-04 2e-04 5e-04 4: 2.5000e+00 2.5000e+00 1e-05 1e-06 2e-06 5e-06 5: 2.5000e+00 2.5000e+00 1e-07 1e-08 2e-08 5e-08Optimal solution found.{'primal objective': 2.4999999895543072, 's': <4x1 matrix, tc='d'>, 'dual infeasibility': 2.257878974569382e-08, 'primal slack': 2.0388399547464153e-08, 'dual objective': 2.4999999817312535, 'residual as dual infeasibility certificate': None, 'dual slack': 3.529915972607509e-09, 'x': <2x1 matrix, tc='d'>, 'iterations': 5, 'gap': 1.3974945737723005e-07, 'residual as primal infeasibility certificate': None, 'z': <4x1 matrix, tc='d'>, 'y': <0x1 matrix, tc='d'>, 'status': 'optimal', 'primal infeasibility': 1.1368786228004961e-08, 'relative gap': 5.5899783359379607e-08}[ 5.00e-01][ 1.50e+00][[ 2.49999999]]

例2

Python程序代码

import numpy as npfrom cvxopt import matrix, solversA = matrix([[1.0, 0.0, -1.0], [0.0, 1.0, -1.0]])b = matrix([2.0, 2.0, -2.0])c = matrix([1.0, 2.0])d = matrix([-1.0, -2.0])sol1 = solvers.lp(c,A,b)min = np.dot(sol1['x'].T, c)sol2 = solvers.lp(d,A,b)max = -np.dot(sol2['x'].T, d)print('min=%s,max=%s'%(min[0][0], max[0][0]))

输出结果：

pcost dcost gap pres dres k/t 0: 4.0000e+00 -0.0000e+00 4e+00 0e+00 0e+00 1e+00 1: 2.7942e+00 1.9800e+00 8e-01 9e-17 7e-16 2e-01 2: 2.0095e+00 1.9875e+00 2e-02 4e-16 2e-16 7e-03 3: 2.0001e+00 1.9999e+00 2e-04 2e-16 6e-16 7e-05 4: 2.0000e+00 2.0000e+00 2e-06 6e-17 5e-16 7e-07 5: 2.0000e+00 2.0000e+00 2e-08 3e-16 7e-16 7e-09Optimal solution found. pcost dcost gap pres dres k/t 0: -4.0000e+00 -8.0000e+00 4e+00 0e+00 1e-16 1e+00 1: -5.2058e+00 -6.0200e+00 8e-01 1e-16 7e-16 2e-01 2: -5.9905e+00 -6.0125e+00 2e-02 1e-16 0e+00 7e-03 3: -5.9999e+00 -6.0001e+00 2e-04 1e-16 2e-16 7e-05 4: -6.0000e+00 -6.0000e+00 2e-06 1e-16 2e-16 7e-07Optimal solution found.min=2.00000000952,max=5.99999904803

二次型规划问题

其中P,q,G,h,A,b为输入矩阵，该问题求解采用QP算法。
例1：

Python程序代码：

from cvxopt import matrix, solversQ = 2*matrix([[2, .5], [.5, 1]])p = matrix([1.0, 1.0])G = matrix([[-1.0,0.0],[0.0,-1.0]])h = matrix([0.0,0.0])A = matrix([1.0, 1.0], (1,2))b = matrix(1.0)sol=solvers.qp(Q, p, G, h, A, b)print(sol['x'])print(sol['primal objective'])

输出结果：

pcost dcost gap pres dres 0: 1.8889e+00 7.7778e-01 1e+00 2e-16 2e+00 1: 1.8769e+00 1.8320e+00 4e-02 0e+00 6e-02 2: 1.8750e+00 1.8739e+00 1e-03 1e-16 5e-04 3: 1.8750e+00 1.8750e+00 1e-05 6e-17 5e-06 4: 1.8750e+00 1.8750e+00 1e-07 2e-16 5e-08Optimal solution found.[ 2.50e-01][ 7.50e-01]

例2：

Python程序代码：

from cvxopt import matrix, solversP = matrix([[1.0, 0.0], [0.0, 0.0]])q = matrix([3.0, 4.0])G = matrix([[-1.0, 0.0, -1.0, 2.0, 3.0], [0.0, -1.0, -3.0, 5.0, 4.0]])h = matrix([0.0, 0.0, -15.0, 100.0, 80.0])sol=solvers.qp(P, q, G, h)print(sol['x'])print(sol['primal objective'])

输出结果

pcost dcost gap pres dres 0: 1.0780e+02 -7.6366e+02 9e+02 0e+00 4e+01 1: 9.3245e+01 9.7637e+00 8e+01 6e-17 3e+00 2: 6.7311e+01 3.2553e+01 3e+01 6e-17 1e+00 3: 2.6071e+01 1.5068e+01 1e+01 2e-17 7e-01 4: 3.7092e+01 2.3152e+01 1e+01 5e-18 4e-01 5: 2.5352e+01 1.8652e+01 7e+00 7e-17 3e-16 6: 2.0062e+01 1.9974e+01 9e-02 2e-16 3e-16 7: 2.0001e+01 2.0000e+01 9e-04 8e-17 5e-16 8: 2.0000e+01 2.0000e+01 9e-06 1e-16 2e-16Optimal solution found.[ 7.13e-07][ 5.00e+00]20.00000617311241

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持。