我是靠谱客的博主 细心硬币,最近开发中收集的这篇文章主要介绍python_多点拟合曲线并计算曲率半径,觉得挺不错的,现在分享给大家,希望可以做个参考。

概述

1、摘要

本文主要讲解:python_多点拟合二次曲线,再选取两边和中点拟合圆,计算曲率半径
主要思路:

  1. 使用numpy中的np.polyfit(x, y, 2)和 np.poly1d(f1)两个函数拟合二次函数
  2. 选取拟合的二次函数中的两边和中点拟合圆
  3. 利用这三点计算曲率半径
  4. 根据这些数据画拟合图

2、数据介绍

数据可使用任何列表数据

3、相关技术

np.polyfit多项式拟合,第三个参数为x的几次幂
np.poly1d得到多项式系数
np.matmul两个numpy数组的矩阵相乘

曲率半径

最小二乘法

4、完整代码和步骤

代码输出如下:标题为曲率半径的数值
在这里插入图片描述
主运行程序入口

from math import sqrt

import numpy as np
import numpy.linalg as LA

import matplotlib.pyplot as plt
from my_utils.read_write import readCsv

src = 'D:document项目\'

def PJcurvature(x, y):
    """
        输入:三点的坐标
        输出:曲率和标准方向
    """
    t_a = LA.norm([x[1] - x[0], y[1] - y[0]])
    t_b = LA.norm([x[2] - x[1], y[2] - y[1]])

    M = np.array([
        [1, -t_a, t_a ** 2],
        [1, 0, 0],
        [1, t_b, t_b ** 2]
    ])

    a = np.matmul(LA.inv(M), x)
    b = np.matmul(LA.inv(M), y)

    kappa = 2 * (a[2] * b[1] - b[2] * a[1]) / (a[1] ** 2. + b[1] ** 2.) ** (1.5)
    return kappa, [b[1], -a[1]] / np.sqrt(a[1] ** 2. + b[1] ** 2.)


theta = [1.3, 1.3001, 1.3002]

x = 5 * np.cos(theta)
y = 5 * np.sin(theta)


# 半径 2326
def build_test():
    test = [9, 8, 8, 7, 7, 7, 8, 8, 8, 9, 9, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12,
            12, 11, 11, 10, 9, 8, 8, 8, 8, 9, 9, 10, 10, 11, 11, 10, 10, 9, 8, 8, 8, 9, 10, 10, 11, 11, 12, 12, 12, 11,
            10, 9, 9, 8, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 11, 11, 10, 10, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10,
            10, 10, 10, 10, 10, 10, 10, 10, 10, 9, 8, 8, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10,
            10, 10, 10, 10, 10, 10, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 8, 8, 7, 6, 6, 6, 6, 6, 6, 6, 6, 6,
            7, 7, 7, 7, 8, 8, 8, 8, 9, 8, 8, 8, 8, 9, 9, 9, 9, 9, 8, 8, 7, 7, 7, 7, 8, 8, 8, 7, 7, 6, 6, 6, 7, 7, 7, 8,
            8, 8, 8, 8, 7, 7, 7, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 9, 9, 8, 8, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 7, 7,
            6, 6, 6, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6, 6, 6, 6, 6, 6, 5, 5, 5, 5, 6, 6, 5, 5, 5, 5, 5, 5, 6,
            6, 5, 5, 5, 5, 5, 4, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 3, 2, 2,
            2, 2, 2, 2, 3, 4, 4, 4, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 6, 5, 4, 4, 3, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4,
            4, 3, 3, 2, 2, 2, 3, 3, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 5, 6, 6, 5, 5,
            4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1,
            1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 2, 1, 1, 1, 1, 0, 0, 0, 0,
            1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2,
            2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1]
    x = [i for i in range(len(test))]
    return [x, test]



def build_function(x, y):
    f1 = np.polyfit(x, y, 2)
    print('参数 :n', f1)
    p1 = np.poly1d(f1)
    print('y=', p1)
    yvals = p1(x)  # 拟合y值
    print('yvals is :n', yvals)
    length = len(x)
    nhx = [x[0], x[int(length / 2)], x[length - 1]]
    nhy = [yvals[0], yvals[int(length / 2)], yvals[length - 1]]
    kappa, norm = PJcurvature(nhx, nhy)
    r = int(1 / kappa)
    title = '曲率半径(' + str(r) + ')'
    plt.plot(x, y, color="red", linewidth=1)
    plt.scatter(nhx, nhy, color="blue", alpha=0.5)
    plt.plot(x, yvals, color="orange", linewidth=2)
    plt.title(str(r))
    plt.ylabel('Height')
    plt.xlabel('X')
    filename = 'D:document\' + title
    plt.savefig(filename)
    plt.show()
    return p1

data = build_test()
build_function(data[0], data[1])

5、学习链接

【Python】车道线拟合曲线的曲率半径计算公式及代码

最后

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