[
{
instruction: '向东步行84米左转',
action: '左转',
polyline: [
[121.585851, 31.198207],
[121.586424, 31.198312],
[121.586424, 31.198312],
[121.586727, 31.198364],
],
},
{
instruction: '向北步行27米右转',
action: '右转',
polyline: [
[121.586727, 31.198364],
[121.58668, 31.198611],
],
},
{
instruction: '沿晨晖路向东步行26米左转',
action: '左转',
polyline: [
[121.586675, 31.198611],
[121.586784, 31.198633],
[121.586944, 31.198659],
],
},
{
instruction: '沿松涛路向北步行333米右转',
action: '右转',
polyline: [
[121.586944, 31.198659],
[121.586671, 31.200204],
[121.586671, 31.200204],
[121.586589, 31.20066],
[121.586589, 31.20066],
[121.586506, 31.201137],
[121.586506, 31.201137],
[121.586437, 31.201502],
[121.586437, 31.201502],
[121.586411, 31.201615],
[121.586389, 31.201641],
],
},
{
instruction: '沿祖冲之路向东步行97米到达目的地',
action: '',
polyline: [
[121.586385, 31.201641],
[121.586814, 31.201719],
[121.586814, 31.201719],
[121.587248, 31.201793],
[121.587248, 31.201793],
[121.5874, 31.201819],
],
},
]
像上面的一个导航规划的路线,给定一个坐标,如何判断当前处于哪个阶段,有没有好的算法啊
判断点在哪个线段上就行了吧. 点在首尾之间,点在线上.
如果点不一定完全在线上. 那就点在首尾之间,点到线的距离最短.
你好,我并没有太了解你的意思,你可以看看这个是不是你想要的。
import numpy as np
def distance_point_to_segment(px, py, ax, ay, bx, by):
"""计算点(px, py)到线段(ax, ay)和(bx, by)的最短距离"""
# 线段的长度
ab = np.array([bx - ax, by - ay])
ap = np.array([px - ax, py - ay])
bp = np.array([px - bx, py - by])
# 线段的长度的平方
ab_squared = np.dot(ab, ab)
# 计算投影比例
if ab_squared == 0:
return np.linalg.norm(ap)
t = np.clip(np.dot(ap, ab) / ab_squared, 0, 1) # 计算t值并限制范围
projection = np.array([ax, ay]) + t * ab # 求投影点
return np.linalg.norm(np.array([px, py]) - projection) # 返回距离
def find_current_stage(coordinate, route):
"""返回当前阶段"""
current_x, current_y = coordinate
closest_stage = None
min_distance = float('inf') # 用于追踪最小距离
for index, stage in enumerate(route):
start = stage['polyline'][0]
end = stage['polyline'][-1]
# 计算当前坐标到该阶段的线段的距离
dist = distance_point_to_segment(current_x, current_y, start[0], start[1], end[0], end[1])
# 输出调试信息
print(f"阶段 {index + 1} 的距离: {dist:.2f} 米")
# 如果距离小于最小距离和判断阈值,更新当前阶段
if dist < min_distance and dist < 10: # 修改阈值为10米
min_distance = dist
closest_stage = (index + 1, stage['instruction'])
return closest_stage if closest_stage else (None, None)
# 示例坐标
current_coordinate = (121.585851, 31.198207)
route = [
{
'instruction': '向东步行84米左转',
'action': '左转',
'polyline': [
[121.585851, 31.198207],
[121.586424, 31.198312],
[121.586424, 31.198312],
[121.586727, 31.198364],
],
},
{
'instruction': '向北步行27米右转',
'action': '右转',
'polyline': [
[121.586727, 31.198364],
[121.58668, 31.198611],
],
},
{
'instruction': '沿晨晖路向东步行26米左转',
'action': '左转',
'polyline': [
[121.586675, 31.198611],
[121.586784, 31.198633],
[121.586944, 31.198659],
],
},
{
'instruction': '沿松涛路向北步行333米右转',
'action': '右转',
'polyline': [
[121.586944, 31.198659],
[121.586671, 31.200204],
[121.586671, 31.200204],
[121.586589, 31.20066],
[121.586589, 31.20066],
[121.586506, 31.201137],
[121.586506, 31.201137],
[121.586437, 31.201502],
[121.586437, 31.201502],
[121.586411, 31.201615],
[121.586389, 31.201641],
],
},
{
'instruction': '沿祖冲之路向东步行97米到达目的地',
'action': '',
'polyline': [
[121.586385, 31.201641],
[121.586814, 31.201719],
[121.586814, 31.201719],
[121.587248, 31.201793],
[121.587248, 31.201793],
[121.5874, 31.201819],
],
},
]
current_stage, instruction = find_current_stage(current_coordinate, route)
print(f"当前阶段: {current_stage}, 指令: {instruction}")
将经纬度转换为平面坐标后,可以判断上一次位置到当前位置是以什么方式接近目标位置的,最好写一个计算算法,比如从X方向靠近还是Y方向靠近,如果从X方向靠近,则判断该方位目标位置,但是只适合数据量小的,如果是数据量比较大的,则需要考虑聚类(数据可能不是一条线,有误差情况下可能呈现出发散状态),通过一次次聚合来进行判断路径(中心点)
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