避开障碍物时路径优化错误

Question

我正在尝试使用平方距离方法找到避免起点和终点之间障碍物的最小路径。

为此，我在起点和终点之间定义了 n 个点 - 并计算优化路径和直线路径之间的平方距离。优化的路径必须与障碍物的距离最小。得到的优化路径是优化路径和直线路径之间的最小二乘距离。

我已经实现了如下代码，但在优化过程中，我收到以下错误：

无法将输入数组从形状 (27) 广播到形状 (27,3)

看起来 Scipy.minimize 将数组的形状从 3-D 数组更改为 1D 数组。您能否提出解决此问题的任何建议？

import numpy as np
import matplotlib.pyplot as plt
import random 
from mpl_toolkits.mplot3d import Axes3D
from scipy.optimize import minimize

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

## Setting Input Data:

startPoint = np.array([1,1,0])
endPoint = np.array([0,8,0])
obstacle = np.array([5,5,0])

## Get degree of freedom coordinates based on specified number of segments:
numberOfPoints = 10 
pipelineStraightVector = endPoint - startPoint 
normVector = pipelineStraightVector/np.linalg.norm(pipelineStraightVector)
stepSize = np.linalg.norm(pipelineStraightVector)/numberOfPoints
pointCoordinates = []
for n in range(numberOfPoints-1): 
  point = [normVector[0]*(n+1)*stepSize+startPoint[0],normVector[1]*(n+1)*stepSize+startPoint[1],normVector[2]*(n+1)*stepSize+startPoint[2]]
  pointCoordinates.append(point)
DOFCoordinates = np.array(pointCoordinates)

## Assign a random z value for the DOF coordinates - change later: 
for coordinate in range(len(DOFCoordinates)):
    DOFCoordinates[coordinate][2] = random.uniform(-1.0, -0.0)
    ##ax.scatter(DOFCoordinates[coordinate][0],DOFCoordinates[coordinate][1],DOFCoordinates[coordinate][2])

## function to calculate the squared residual:
def distance(a,b): 
  dist = ((a[0]-b[0])**2 + (a[1]-b[1])**2 + (a[2]-b[2])**2)
  return dist

## Get Straight Path Coordinates:
def straightPathCoordinates(DOF):
    allCoordinates = np.zeros((2+len(DOF),3))
    allCoordinates[0] = startPoint
    allCoordinates[1:len(DOF)+1]=DOF
    allCoordinates[1+len(DOF)]=endPoint
    return allCoordinates

pathPositions = straightPathCoordinates(DOFCoordinates)

## Set Degree of FreeDom Coordinates during optimization:
def setDOFCoordinates(DOF): 
    print 'DOF',DOF
    allCoordinates = np.zeros((2+len(DOF),3))
    allCoordinates[0] = startPoint
    allCoordinates[1:len(DOF)+1]=DOF
    allCoordinates[1+len(DOF)]=endPoint
    return allCoordinates

## Objective Function: Set Degree of FreeDom Coordinates and Get Square Distance between optimized and straight path coordinates:
def f(DOF):
   newCoordinates = setDOFCoordinates(DOF)
   print DOF
   sumDistance = 0.0
   for coordinate in range(len(pathPositions)):
        squaredDistance = distance(newCoordinates[coordinate],pathPositions[coordinate])
        sumDistance += squaredDistance
   return sumDistance 

## Constraints: all coordinates need to be away from an obstacle with a certain distance: 
constraint = []
minimumDistanceToObstacle = 0
for coordinate in range(len(DOFCoordinates)+2):
   cons = {'type': 'ineq', 'fun': lambda DOF:  minimumDistanceToObstacle-((obstacle[0] - setDOFCoordinates(DOF)[coordinate][0])**2 +(obstacle[1] - setDOFCoordinates(DOF)[coordinate][1])**2+(obstacle[2] - setDOFCoordinates(DOF)[coordinate][2])**2)}
   constraint.append(cons)

## Get Initial Guess:
starting_guess = DOFCoordinates

## Run the minimization:
objectiveFunction = lambda DOF: f(DOF)
result = minimize(objectiveFunction,starting_guess,constraints=constraint, method='COBYLA')
print result.x
print DOFCoordinates


ax.plot([startPoint[0],endPoint[0]],[startPoint[1],endPoint[1]],[startPoint[2],endPoint[2]])
ax.scatter(obstacle[0],obstacle[1],obstacle[2])

期望的结果是一组点及其在点 A 和点 B 之间的位置，这些点可以避开障碍物并返回最小距离。

Answer 1

这是因为要最小化的输入必须使用一维数组，

来自 scipy notes，

要最小化的目标函数。

 fun(x, *args) -> float

其中 x 是形状为 (n,) 的一维数组，args 是完全指定函数所需的固定参数的元组。

x0 : ndarray, 形状 (n,)

初步猜测。大小为 (n,) 的实元素数组， 其中“n”是自变量的数量。

这意味着您应该在输入中使用 starting_guess.ravel() 并更改 setDOFCoordinates 以使用一维数组。