问题描述
我想要一个这样的路径连接点： enter image description here
根据我的尝试，这是我能够在 Python 中实现的结果： enter image description here
看起来相当简单。可能吗？
解决方法
points=[[2031.0974638138432,8871.788899127823],[1946.0939073523768,8687.702718346474],[1868.9610243990464,8542.951197953029],[2061.006139498597,8393.47953820238],[2163.3253106537886,8264.46188196409],[2541.119051912334,8232.994153653774],[2785.1108448732557,8292.782817554034],[2967.711185007007,8424.947266512696],8602.739911576653],[2709.552146487491,8752.211571327301],[2429.355105791343,8808.853442507185],[2150.732166552858,8744.34463924972],[2087.7665291581216,8531.937522878434],[2402.594716131818,8379.319070407408],[2638.7157524747163,8461.135134180222],[2446.670637375166,8541.377851316203],[2492.849155914053,8630.642922304622],[2444.788613747456,8669.072915834848],[2366.462005771966,8620.088463227232]]
starting=5
import math
from ortools.constraint_solver import routing_enums_pb2
from ortools.constraint_solver import pywrapcp
def create_data_model(points,starting):
"""Stores the data for the problem."""
data = {}
# Locations in block units
data['locations'] = points
data['num_vehicles'] = 1
data['depot'] = starting
return data
def compute_euclidean_distance_matrix(locations):
"""Creates callback to return distance between points."""
distances = {}
for from_counter,from_node in enumerate(locations):
distances[from_counter] = {}
for to_counter,to_node in enumerate(locations):
if from_counter == to_counter:
distances[from_counter][to_counter] = 0
else:
# Euclidean distance
distances[from_counter][to_counter] = (int(
math.hypot((from_node[0] - to_node[0]),(from_node[1] - to_node[1]))))
return distances
def print_solution(manager,routing,solution):
"""Prints solution on console."""
index = routing.Start(0)
plan_output = []
route_distance = 0
while not routing.IsEnd(index):
plan_output.append(manager.IndexToNode(index))
previous_index = index
index = solution.Value(routing.NextVar(index))
route_distance += routing.GetArcCostForVehicle(previous_index,index,0)
return plan_output
def main(points,starting=0):
"""Entry point of the program."""
# Instantiate the data problem.
data = create_data_model(points,starting)
# Create the routing index manager.
manager = pywrapcp.RoutingIndexManager(len(data['locations']),data['num_vehicles'],data['depot'])
# Create Routing Model.
routing = pywrapcp.RoutingModel(manager)
distance_matrix = compute_euclidean_distance_matrix(data['locations'])
def distance_callback(from_index,to_index):
"""Returns the distance between the two nodes."""
# Convert from routing variable Index to distance matrix NodeIndex.
from_node = manager.IndexToNode(from_index)
to_node = manager.IndexToNode(to_index)
return distance_matrix[from_node][to_node]
transit_callback_index = routing.RegisterTransitCallback(distance_callback)
# Define cost of each arc.
routing.SetArcCostEvaluatorOfAllVehicles(transit_callback_index)
# Setting first solution heuristic.
search_parameters = pywrapcp.DefaultRoutingSearchParameters()
search_parameters.first_solution_strategy = (
routing_enums_pb2.FirstSolutionStrategy.PATH_CHEAPEST_ARC)
# Solve the problem.
solution = routing.SolveWithParameters(search_parameters)
# Print solution on console.
if solution:
return print_solution(manager,solution)
if __name__ == '__main__':
permutation = main(points,starting=starting)
points2=[]
for i in permutation:
points2.append(points[i])
x,y = map(list,zip(*points2))
from matplotlib import pyplot as plt
plt.plot(x,y,'-o')
您是否每次都通过最短路径将线从点连接到点？看起来它是这样的。我建议您标记您放置的每个点，以便您可以按照放置所有内容的方式连接所有内容。
,这看起来像是一个旅行商问题。我在网上查了一下，发现了一个关于旅行商问题的图书馆。 https://pypi.org/project/python-tsp/。它应该有助于创建有效的非重叠路线。一旦它为您提供排列，请使用这些点并组织您的新点列表。
示例：
points2=[]
for i in permutation:
points2.append(points[i])
之后，您可以绘制 points2。