这是你鹅愚蠢的问题,但由于附近可能有比较懒惰的人,(用编号参数可读性)走这条线的MATLAB代码:
x = quadprog(C1, C2, C3, C4, C5, C6, C7, C8, ...)
在Python相当于将是:
import numpy
import cvxopt
n = C1.shape[1] # This is for readability only
P = C1
q = C2
G = numpy.vstack([C3, -numpy.eye(n), numpy.eye(n)])
h = numpy.hstack([C4, -C7, C8])
A = C5
b = C6
sol = cvxopt.solvers.qp(cvxopt.matrix(P), cvxopt.matrix(q), cvxopt.matrix(G), cvxopt.matrix(h), cvxopt.matrix(A), cvxopt.matrix(b))
x = sol['x']
假设你一直在使用numpy.arrays。如果您一直在使用基质,请检查this doc以找到G和h的此配方:
G = numpy.bmat('C3, -numpy.eye(n), numpy.eye(n)')
h = numpy.bmat('C4, -C7, C8')
来源
2013-11-25 03:22:33
FvD