本文整理汇总了Python中openopt.NLP.c方法的典型用法代码示例。如果您正苦于以下问题：Python NLP.c方法的具体用法？Python NLP.c怎么用？Python NLP.c使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类openopt.NLP的用法示例。
在下文中一共展示了NLP.c方法的4个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: NLP
# 需要导入模块: from openopt import NLP [as 别名]
# 或者: from openopt.NLP import c [as 别名]
from openopt import NLP
from numpy import cos, arange, ones, asarray, abs, zeros
N = 30
M = 5
ff = lambda x: ((x-M)**2).sum()
p = NLP(ff, cos(arange(N)))
def df(x):
r = 2*(x-M)
r[0] += 15 #incorrect derivative
r[8] += 80 #incorrect derivative
return r
p.df = df
p.c = lambda x: [2* x[0] **4-32, x[1]**2+x[2]**2 - 8]
def dc(x):
r = zeros((2, p.n))
r[0,0] = 2 * 4 * x[0]**3
r[1,1] = 2 * x[1]
r[1,2] = 2 * x[2] + 15 #incorrect derivative
return r
p.dc = dc
p.h = lambda x: (1e1*(x[-1]-1)**4, (x[-2]-1.5)**4)
def dh(x):
r = zeros((2, p.n))
r[0,-1] = 1e1*4*(x[-1]-1)**3
r[1,-2] = 4*(x[-2]-1.5)**3 + 15 #incorrect derivative
return r示例2: array
# 需要导入模块: from openopt import NLP [as 别名]
# 或者: from openopt.NLP import c [as 别名]
# so p.b = array([1.5, -1.6, -825]) or p.b = (1.5, -1.6, -825) are valid as well
# Aeq x = beq
# x20+x21 = 2.5
p.Aeq = zeros(N)
p.Aeq[20:22] = 1
p.beq = 2.5
# non-linear inequality constraints c(x) <= 0
# 2*x0^4 <= 1/32
# x1^2+x2^2 <= 1/8
# x25^2 +x25*x35 + x35^2<= 2.5
p.c = lambda x: [2* x[0] **4-1./32, x[1]**2+x[2]**2 - 1./8, x[25]**2 + x[35]**2 + x[25]*x[35] -2.5]
# other valid c:
# p.c = [lambda x: c1(x), lambda x : c2(x), lambda x : c3(x)]
# p.c = (lambda x: c1(x), lambda x : c2(x), lambda x : c3(x))
# p.c = lambda x: numpy.array(c1(x), c2(x), c3(x))
# def c(x):
# return c1(x), c2(x), c3(x)
# p.c = c
# dc(x)/dx: non-lin ineq constraints gradients (optional):
def DC(x):
r = zeros((3, N))
r[0,0] = 2 * 4 * x[0]**3
r[1,1] = 2 * x[1]
r[1,2] = 2 * x[2]示例3: criterium
# 需要导入模块: from openopt import NLP [as 别名]
# 或者: from openopt.NLP import c [as 别名]
# Note that in ALGENCAN gradtol means norm of projected gradient of the Augmented Lagrangian
# so it should be something like 1e-3...1e-5
p.gradtol = 1e-5#5 # gradient stop criterium (default for NLP is 1e-6)
#print 'maxiter', p.maxiter
#print 'maxfun', p.maxfun
p.maxIter=50
# p.maxfun=100
#p.df_iter = 50
p.maxTime = 4000
h_args=(h,k,l,fq,fqerr,x,z,cosmat_list,coslist,flist)
if 0:
#p.h=[pos_sum,neg_sum]
p.h=[pos_sum,neg_sum]
p.c=[chisq]
# p.h=[pos_sum,neg_sum]
p.args.h=h_args
p.args.c=h_args
p.dh=[pos_sum_grad,neg_sum_grad]
p.df=chisq_grad
if 1:
#p.h=[pos_sum,neg_sum,chisq]
p.c=[chisq]
p.h=[pos_sum,neg_sum]
p.args.h=h_args
p.args.c=h_args
p.dh=[pos_sum_grad,neg_sum_grad]
p.dc=chisq_grad
#p.dh=[pos_sum_grad,neg_sum_grad,neg_sum_grad]示例4: well
# 需要导入模块: from openopt import NLP [as 别名]
# 或者: from openopt.NLP import c [as 别名]
Note! For oofun handling user parameters is performed
in the same way:
my_oofun.args = (...)
they will be passed to derivative function as well (if you have supplied it)
"""
from openopt import NLP
from numpy import asfarray
f = lambda x, a: (x**2).sum() + a * x[0]**4
x0 = [8, 15, 80]
p = NLP(f, x0)
#using c(x)<=0 constraints
p.c = lambda x, b, c: (x[0]-4)**2 - 1 + b*x[1]**4 + c*x[2]**4
#using h(x)=0 constraints
p.h = lambda x, d: (x[2]-4)**2 + d*x[2]**4 - 15
p.args.f = 4 # i.e. here we use a=4
# so it's the same to "a = 4; p.args.f = a" or just "p.args.f = a = 4"
p.args.c = (1,2)
p.args.h = 15
# Note 1: using tuple p.args.h = (15,) is valid as well
# Note 2: if all your funcs use same args, you can just use
# p.args = (your args)