Python之CVXOPT模块

摘要：中支持凸规划的模块为其安装方式为卸载原中的安装的文件，链接为安装的文件，链接为之所以选择这种安装方式，是因为的和直接的不兼容性。

  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/~gohl...

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

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 np
from cvxopt import matrix, solvers

A = 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-08
Optimal 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 np
from cvxopt import matrix, solvers

A = 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-09
Optimal 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-07
Optimal 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, solvers

Q = 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-08
Optimal solution found.
[ 2.50e-01]
[ 7.50e-01]

例2：

Python程序代码：

from cvxopt import matrix, solvers

P = 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-16
Optimal solution found.
[ 7.13e-07]
[ 5.00e+00]

20.00000617311241