我想优化一个火箭轨迹。我想解决的问题在这个视频中有描述:#https://www.youtube.com/watch?v=9qsiCGpvwKA&t=332s。它是关于找到最小的时间让火箭从开始的零高度和零速度到最后的零高度和零速度。上面提到的视频中提供了一个解决方案，但我想自己解决这个问题，而不使用APMonitor的工具箱。
我设置了这段代码作为我的目标函数来优化，通过使用8个控制点来控制火箭的轨迹:
import matplotlib.pyplot as plt
from scipy.integrate import solve_ivp
import numpy as np
from scipy import interpolate
def rocketmotion(t, Y, x, z):
f = interpolate.interp1d(x, z)
if t>x[1]:
u=f(x[1])
else:
u=f(t)
s=Y[0]; v=Y[1]; m=Y[2]
ds_dt=v
dv_dt=(u-0.2*v**2)/m
dm_dt=-0.01*u**2
return [ds_dt, dv_dt, dm_dt]
def objectiveFunction(x, other_args):
n_Gridpoints=8
u=np.zeros(n_Gridpoints)
u[0]=x[1]; u[1]=x[2]; u[2]=x[3]; u[3]=x[4]; u[4]=x[5]; u[5]=x[6]; u[6]=x[7]; u[7]=x[8]
Y0=[0, 0, 1]
t_end=x[0]
time_steps=np.linspace(0, t_end, n_Gridpoints)
T=[]; S=[]; V=[]; M=[]
for i in range(n_Gridpoints-1):
x=[time_steps[i], time_steps[i+1]]
z=[u[i], u[i+1]]
if i==0:
Y = solve_ivp(rocketmotion,[0, time_steps[i+1]],Y0, args=[x, z])
else:
Y = solve_ivp(rocketmotion,[time_steps[i], time_steps[i+1]], [Y.y[0][-1], Y.y[1][-1], Y.y[2][-1]], args=[x, z])
T.extend(Y.t)
S.extend(Y.y[0])
V.extend(Y.y[1])
M.extend(Y.y[2])
execMode=other_args
if execMode==1:
return T[-1]+abs(10-S[-1])*10+abs(V[-1])*10
elif execMode==2:
print((T[-1]+abs(10-S[-1])*10+abs(V[-1])*10).ndim)
return T[-1]+abs(10-S[-1])*10+abs(V[-1])*10
else:
print((T[-1]+abs(10-S[-1])*10+abs(V[-1])*10).ndim)
print(T[-1]+abs(10-S[-1])*10+abs(V[-1])*10)
return T, S, V, M
我使用奖励塑造来确保最终条件(位置=10，速度=0)将被达到。现在我正在尝试使用SLSQP来优化给定的函数:
from scipy.optimize import minimize
bounds_time=(5, 15)
bounds_u=(-1.1, 1.1)
bounds=[bounds_time, bounds_u, bounds_u, bounds_u, bounds_u, bounds_u, bounds_u, bounds_u, bounds_u]
x0=[7.5, 1.1, 1.1, 0.7, 0.7, 0.7, 0.7, -1.0, -0.8]
execMode=2
other_args=[execMode]
solution = minimize(objectiveFunction,x0,method='SLSQP', bounds=bounds, args=other_args,
options={'maxiter': 1000, 'disp': True})
我得到错误消息:f0通过的维度大于1。此错误消息发生在SLSQP算法使用的函数approx_derivative中。我不明白这个错误信息。所以我想请求你的帮助。回溯是:
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-76-4be01f8fa5f5> in <module>
9 other_args=[execMode]
10
---> 11 solution = minimize(objectiveFunction,x0,method='SLSQP', bounds=bounds, args=other_args,
12 options={'maxiter': 1000, 'disp': True})
13
~miniconda3libsite-packagesscipyoptimize_minimize.py in minimize(fun, x0, args, method, jac, hess, hessp, bounds, constraints, tol, callback, options)
623 return _minimize_cobyla(fun, x0, args, constraints, **options)
624 elif meth == 'slsqp':
--> 625 return _minimize_slsqp(fun, x0, args, jac, bounds,
626 constraints, callback=callback, **options)
627 elif meth == 'trust-constr':
~miniconda3libsite-packagesscipyoptimizeslsqp.py in _minimize_slsqp(func, x0, args, jac, bounds, constraints, maxiter, ftol, iprint, disp, eps, callback, finite_diff_rel_step, **unknown_options)
367
368 # ScalarFunction provides function and gradient evaluation
--> 369 sf = _prepare_scalar_function(func, x, jac=jac, args=args, epsilon=eps,
370 finite_diff_rel_step=finite_diff_rel_step,
371 bounds=new_bounds)
~miniconda3libsite-packagesscipyoptimizeoptimize.py in _prepare_scalar_function(fun, x0, jac, args, bounds, epsilon, finite_diff_rel_step, hess)
259 # ScalarFunction caches. Reuse of fun(x) during grad
260 # calculation reduces overall function evaluations.
--> 261 sf = ScalarFunction(fun, x0, args, grad, hess,
262 finite_diff_rel_step, bounds, epsilon=epsilon)
263
~miniconda3libsite-packagesscipyoptimize_differentiable_functions.py in __init__(self, fun, x0, args, grad, hess, finite_diff_rel_step, finite_diff_bounds, epsilon)
93
94 self._update_grad_impl = update_grad
---> 95 self._update_grad()
96
97 # Hessian Evaluation
~miniconda3libsite-packagesscipyoptimize_differentiable_functions.py in _update_grad(self)
169 def _update_grad(self):
170 if not self.g_updated:
--> 171 self._update_grad_impl()
172 self.g_updated = True
173
~miniconda3libsite-packagesscipyoptimize_differentiable_functions.py in update_grad()
89 self._update_fun()
90 self.ngev += 1
---> 91 self.g = approx_derivative(fun_wrapped, self.x, f0=self.f,
92 **finite_diff_options)
93
~miniconda3libsite-packagesscipyoptimize_numdiff.py in approx_derivative(fun, x0, method, rel_step, abs_step, f0, bounds, sparsity, as_linear_operator, args, kwargs)
386 f0 = np.atleast_1d(f0)
387 if f0.ndim > 1:
--> 388 raise ValueError("`f0` passed has more than 1 dimension.")
389
390 if np.any((x0 < lb) | (x0 > ub)):
ValueError: `f0` passed has more than 1 dimension
MRE显示问题:
from scipy.optimize import minimize
x0=[7.5, 1.1, 1.1, 0.7, 0.7, 0.7, 0.7, -1.0, -0.8]
execMode=2
other_args=[execMode]
def objectiveFunction(x, other_args):
execMode=other_args
print(execMode)
return x
solution = minimize(objectiveFunction, x0, method='SLSQP', args=other_args)
并且在破碎前打印一堆[2]。你期望的是2(我从elif execMode==2:知道它)。因此，简单的解决方案是:
def objectiveFunction(x, other_args):
...
execMode=other_args[0]
...
如你所见，MRE准备好后，解决问题就容易多了。这就是为什么你被要求在stackoverflow上发布它。