如果有一些稀疏的轨迹,如何将这些轨迹平滑插值处理呢?
方法1:线性插值
方法2:三次样条插值
方法3:贝塞尔曲线插值
方法4:拉格朗日插值

线性插值:在两两相邻的点之间差值,两点间所有插值点都在一条直线上。
贝塞尔曲线:贝塞尔曲线不会经过所有的坐标点,会根据坐标点的排列趋势去拟合出一条相对平滑的从第1个点到最后一个点之间的曲线。
三次样条插值:插值函数会经过所有的坐标点,拟合函数平滑。
拉格朗日插值:点太多,会出现不稳定的结果。见下图:

前三种插值算法都有特定的使用场景,按需使用就好了。
import time import numpy as np from scipy.interpolate import interp1d from scipy.special import comb def linear_interpolation(route, num_points): # 1. 线性插值 # 将经纬度分开 lons = np.array([point[0] for point in route]) lats = np.array([point[1] for point in route]) # 创建插值函数 distance = np.cumsum(np.sqrt(np.ediff1d(lons, to_begin=0) ** 2 + np.ediff1d(lats, to_begin=0) ** 2)) distance /= distance[-1] # 创建插值函数 lon_interp = interp1d(distance, lons, kind='linear') lat_interp = interp1d(distance, lats, kind='linear') # 生成新的距离点 new_distance = np.linspace(0, 1, num_points) # 插值 new_lons = lon_interp(new_distance) new_lats = lat_interp(new_distance) return list(zip(new_lons, new_lats)) def cubic_spline_interpolation(route, num_points): # 2. 三次样条插值 lons = np.array([point[0] for point in route]) lats = np.array([point[1] for point in route]) distance = np.cumsum(np.sqrt(np.ediff1d(lons, to_begin=0) ** 2 + np.ediff1d(lats, to_begin=0) ** 2)) distance /= distance[-1] lon_interp = interp1d(distance, lons, kind='cubic') lat_interp = interp1d(distance, lats, kind='cubic') new_distance = np.linspace(0, 1, num_points) new_lons = lon_interp(new_distance) new_lats = lat_interp(new_distance) return list(zip(new_lons, new_lats)) def bernstein_poly(i, n, t): return comb(n, i) * (t ** i) * ((1 - t) ** (n - i)) def bezier_curve(route, num_points=100): # 3. 贝塞尔曲线插值 n = len(route) - 1 t = np.linspace(0, 1, num_points) lons = np.array([point[0] for point in route]) lats = np.array([point[1] for point in route]) new_lons = np.zeros(num_points) new_lats = np.zeros(num_points) for i in range(n + 1): new_lons += bernstein_poly(i, n, t) * lons[i] new_lats += bernstein_poly(i, n, t) * lats[i] return list(zip(new_lons, new_lats)) import matplotlib.pyplot as plt def plot_routes(original, linear, cubic, bezier): plt.figure(figsize=(12, 8)) # 原始轨迹 orig_lons, orig_lats = zip(*original) plt.plot(orig_lons, orig_lats, 'ro-', label='Original', alpha=0.5) # 线性插值 lin_lons, lin_lats = zip(*linear) plt.plot(lin_lons, lin_lats, 'b-', label='Linear', alpha=0.7) # 三次样条插值 cub_lons, cub_lats = zip(*cubic) plt.plot(cub_lons, cub_lats, 'g-', label='Cubic Spline', alpha=0.7) # 贝塞尔曲线 bez_lons, bez_lats = zip(*bezier) plt.plot(bez_lons, bez_lats, 'm-', label='Bezier', alpha=0.7) plt.legend() plt.xlabel('Longitude') plt.ylabel('Latitude') plt.title('Trajectory Interpolation Comparison') plt.grid(True) plt.show() if __name__ == '__main__': # 原始轨迹数据 route = [ [122.123456, 31.123456], [122.234567, 31.234567], [122.345678, 31.345678], [122.456789, 31.456789], [122.567890, 31.567890], [122.678901, 31.578901], [122.789012, 31.789012], [122.890123, 31.890123], [122.901234, 31.901234], ] start_time = time.time() # 线性插值 linear_route = linear_interpolation(route, 1000) print("线性插值结果 (前5个点):", linear_route[:5]) print("线性插值用时:", time.time() - start_time, "秒") start_time = time.time() # 三次样条插值 cubic_route = cubic_spline_interpolation(route, 1000) print("三次样条插值结果 (前5个点):", cubic_route[:5]) print("三次样条插值用时:", time.time() - start_time, "秒") start_time = time.time() # 贝塞尔曲线插值 bezier_route = bezier_curve(route, 1000) print("贝塞尔曲线插值结果 (前5个点):", bezier_route[:5]) print("贝塞尔曲线插值用时:", time.time() - start_time, "秒") # 绘制比较图 plot_routes(route, linear_route, cubic_route, bezier_route)
# 4、拉格朗日插值算法 import time from scipy.interpolate import lagrange import numpy as np def lagrange_interp(x, y, x_new): """ Lagrange interpolation :param x: x coordinates of data points :param y: y coordinates of data points :param x_new: x coordinates of new interpolated points :return: y coordinates of new interpolated points """ f = lagrange(x, y) y_new = f(x_new) return y_new if __name__ == '__main__': # 原始数据 route = [ [122.123456, 31.123456], [122.234567, 31.234567], [122.345678, 31.345678], [122.456789, 31.456789], [122.567890, 31.567890], [122.678901, 31.678901], [122.789012, 31.789012], [122.890123, 31.890123], [122.901234, 31.901234], ] x_list = [i[0] for i in route] y_list = [i[1] for i in route] # 新数据 x_new = np.arange(122.123456, 122.990123, 0.01) y_new = lagrange_interp(x_list, y_list, x_new) # 绘图 import matplotlib.pyplot as plt plt.plot(x_list, y_list, 'o', label='original data') plt.plot(x_new, y_new, label='interpolated data') plt.legend() plt.show()