python教程分享python疲劳驾驶困倦低头检测功能的实现

def get_head_pose(shape):  # 头部姿态估计      # （像素坐标集合）填写2d参考点      # 17左眉左上角/21左眉右角/22右眉左上角/26右眉右上角/36左眼左上角/39左眼右上角/42右眼左上角/      # 45右眼右上角/31鼻子左上角/35鼻子右上角/48左上角/54嘴右上角/57嘴中央下角/8下巴角      image_pts = np.float32([shape[17], shape[21], shape[22], shape[26], shape[36],                              shape[39], shape[42], shape[45], shape[31], shape[35],                              shape[48], shape[54], shape[57], shape[8]])      # solvepnp计算姿势——求解旋转和平移矩阵：      # rotation_vec表示旋转矩阵，translation_vec表示平移矩阵，cam_matrix与k矩阵对应，dist_coeffs与d矩阵对应。      _, rotation_vec, translation_vec = cv2.solvepnp(object_pts, image_pts, cam_matrix, dist_coeffs)      # projectpoints重新投影误差：原2d点和重投影2d点的距离（输入3d点、相机内参、相机畸变、r、t，输出重投影2d点）      reprojectdst, _ = cv2.projectpoints(reprojectsrc, rotation_vec, translation_vec, cam_matrix, dist_coeffs)      reprojectdst = tuple(map(tuple, reprojectdst.reshape(8, 2)))  # 以8行2列显示         # 计算欧拉角calc euler angle      rotation_mat, _ = cv2.rodrigues(rotation_vec)  # 罗德里格斯公式（将旋转矩阵转换为旋转向量）      pose_mat = cv2.hconcat((rotation_mat, translation_vec))  # 水平拼接，vconcat垂直拼接      # decomposeprojectionmatrix将投影矩阵分解为旋转矩阵和相机矩阵      _, _, _, _, _, _, euler_angle = cv2.decomposeprojectionmatrix(pose_mat)      pitch, yaw, roll = [math.radians(_) for _ in euler_angle]      pitch = math.degrees(math.asin(math.sin(pitch)))      roll = -math.degrees(math.asin(math.sin(roll)))      yaw = math.degrees(math.asin(math.sin(yaw)))      print('pitch:{}, yaw:{}, roll:{}'.format(pitch, yaw, roll))      return reprojectdst, euler_angle  # 投影误差，欧拉角  def eye_aspect_ratio(eye):      # 垂直眼标志（x，y）坐标      a = dist.euclidean(eye[1], eye[5])  # 计算两个集合之间的欧式距离      b = dist.euclidean(eye[2], eye[4])      # 计算水平之间的欧几里得距离      # 水平眼标志（x，y）坐标      c = dist.euclidean(eye[0], eye[3])      # 眼睛长宽比的计算      ear = (a + b) / (2.0 * c)      # 返回眼睛的长宽比      return ear  def mouth_aspect_ratio(mouth):  # 嘴部      a = np.linalg.norm(mouth[2] - mouth[9])  # 51, 59      b = np.linalg.norm(mouth[4] - mouth[7])  # 53, 57      c = np.linalg.norm(mouth[0] - mouth[6])  # 49, 55      mar = (a + b) / (2.0 * c)      return mar