问题描述
从scipy教程中，我并没有真正了解optimize.minimize的工作方式.我想在以下方程组中最小化c3:
From scipy tutorial I don't really get how optimize.minimize works.I want to minimize c3 in following set of equations:
0 = cos(b1)+ cos(b2)- 0.0166
0 = sin(b1)+ sin(b2)+ 0.3077*c3 - 0.6278
0 = cos(b1)- cos(b2)+ 5.4155*c3 - 4.3547
间隔:
c3[0,1]
b1,b2[0,2*pi]
这是我的代码:
def fun(x):
return 4.9992-5.7233*x[0]-2*np.cos(x[2])-np.sin(x[2])-np.sin(x[1])
bnds = ((0,1),(0,2*np.pi),(0,2*np.pi))
i = optimize.minimize(fun, (0.05,np.pi*0.5,np.pi), method='SLSQP', bounds=bnds)
输出为
status: 0
success: True
njev: 6
nfev: 30
fun: -3.9601679766628886
x: array([ 1. , 1.57079633, 0.46367497])
message: 'Optimization terminated successfully.'
jac: array([ -5.72330004e+00, 0.00000000e+00, 6.11841679e-05,
0.00000000e+00])
nit: 6
L-BFGS-B中的结果相同.我的理解是，此处c3已变为1，这仍然可以，但我希望将其降低.如果我在函数上应用fsolve，它将找到c3 = 0.46的根.btw为什么我必须在代码中写x [0]，x [1]和x [2]而不是c3，b1，b2?是否有使用约束等更巧妙的方法?
The result is the same in L-BFGS-B My understanding is that here c3 has become 1 which is still ok, but I wanted it to be lower. If I apply fsolve on the function it finds a root for c3=0.46.btw why do I have to write x[0],x[1] and x[2] instead of c3,b1,b2 in the code?Is there a more clever way using constrains e.g.?
推荐答案
对于三个变量 b1 ， b2 和 c3 ，您具有三个超越方程em>.您需要做的是为您的变量求解这3个方程.由于方程式是超验的，因此可能没有解决方案，一个解决方案或许多解决方案.用 Mathematica 解决它们会得出:
You have three transcendent equations for the 3 variables b1, b2 and c3. What you need to do is to solve this 3 equations for your variables. As the equations are transcendent, it's possible that there is no solution, one solution or many solutions. Solving them with Mathematica gives:
In[30]:= eq1 = 0 == Cos[b1] + Cos[b2] - 0.0166;
eq2 = 0 == Sin[b1] + Sin[b2] + 0.3077*c3 - 0.6278;
eq3 = 0 == Cos[b1] - Cos[b2] + 5.4155*c3 - 4.3547;
In[42]:= Reduce[{eq1, eq2, eq3, b1 >= 0, b1 <= 2*Pi, b2 >= 0,
b2 <= 2*Pi, c3 >= 0, c3 <= 1}, {b1, b2, c3}]
During evaluation of In[42]:= Reduce::ratnz: Reduce was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result. >>
Out[42]= b1 == 0.214076 && b2 == 2.85985 && c3 == 0.446303
因此，实际上只有一种解决方案.现在，您还可以使用root从数字上找到方程组的该解决方案:
So there is in fact only one solution. Now you could also use root to find this solution of your system of equations numerically:
import scipy as sp
import scipy.optimize
def f(x):
b1, b2, c3 = x
e1 = sp.cos(b1) + sp.cos(b2) - 0.0166
e2 = sp.sin(b1) + sp.sin(b2) + 0.3077*c3 - 0.6278
e3 = sp.cos(b1) - sp.cos(b2) + 5.4155*c3 - 4.3547
return e1, e2, e3
res = sp.optimize.root(f, [0.5, 1.0, 1.0])
print('b1 = {}, b2 = {}, c3 = {}'.format(*res.x))
给予:
b1 = 0.214076256767, b2 = 2.85985240432, c3 = 0.446302998585
您对minimize所做的事情是最小化三个方程的总和，这不等同于最小化这组方程中的c3".
What you did with minimize is to minimize the sum of the three equations which is not equivalent to "minimize c3 in (the) set of equations".
minimize并非为您想要的. 最小化"的作用类似于最小化，找到最小化f(x) = x**2的x.答案显然是"x = 0".
minimize is not made for what you want. 'Minimize' does something like minimize find the x that minimizes f(x) = x**2. The answer would obviously be 'x=0'.
由于接口"minimize"的原因，您必须写"x [0]".该函数只是希望您以矢量形式给出要最小化的参数.
You have to write 'x[0]' because of the interface of 'minimize'. the function simply expects that you give the parameters you want to minimize in vector form.
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