python疲劳驾驶困倦低头检测

admin • 2022-04-03 20:30 • 5G

python疲劳驾驶困倦低头检测

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