python 单目测距

Python实现单目测距可以使用OpenCV库。OpenCV是一个开源计算机视觉库，提供了很多图像处理和计算机视觉功能，包括单目测距。下面是一个简单的单目测距的Python代码示例：

import cv2

# 读取图像
img = cv2.imread('image.jpg')

# 定义相机内参矩阵
fx = 1000.0  # x方向焦距
fy = 1000.0  # y方向焦距
cx = img.shape[1] / 2.0  # x方向光心坐标
cy = img.shape[0] / 2.0  # y方向光心坐标
camera_matrix = np.array([[fx, 0, cx],
                          [0, fy, cy],
                          [0, 0, 1]])

# 定义畸变系数
dist_coeffs = np.zeros((4, 1))

# 定义物体的实际尺寸
object_width = 0.2  # 物体的宽度，单位为米

# 定义物体在图像中的坐标点
obj_pts = np.zeros((4, 3), dtype=np.float32)
obj_pts[0] = (-object_width / 2, 0, 0)
obj_pts[1] = (object_width / 2, 0, 0)
obj_pts[2] = (object_width / 2, object_width, 0)
obj_pts[3] = (-object_width / 2, object_width, 0)

# 定义图像中的坐标点
img_pts = np.zeros((4, 2), dtype=np.float32)
img_pts[0] = (141, 131)
img_pts[1] = (393, 130)
img_pts[2] = (352, 291)
img_pts[3] = (182, 292)

# 计算旋转向量和平移向量
success, rotation_vector, translation_vector = cv2.solvePnP(obj_pts, img_pts, camera_matrix, dist_coeffs)

# 计算物体在相机坐标系中的坐标
object_pts = np.zeros((4, 3), dtype=np.float32)
object_pts[0] = (-object_width / 2, 0, 0)
object_pts[1] = (object_width / 2, 0, 0)
object_pts[2] = (object_width / 2, object_width, 0)
object_pts[3] = (-object_width / 2, object_width, 0)
object_pts = np.reshape(object_pts, (4, 1, 3))
image_pts, _ = cv2.projectPoints(object_pts, rotation_vector, translation_vector, camera_matrix, dist_coeffs)

# 计算物体在图像中的宽度
distance_in_pixels = abs(image_pts[0][0][0] - image_pts[1][0][0])
distance_in_meters = object_width * fx / distance_in_pixels

print("物体在图像中的宽度为：{} 像素".format(distance_in_pixels))
print("物体距离相机的距离为：{} 米".format(distance_in_meters))

这个例子中，我们首先读取了一张图像。然后定义了相机内参矩阵和畸变系数。接着，我们定义了物体的实际尺寸和在图像中的坐标点。通过solvePnP函数，我们可以计算出物体在相机坐标系中的坐标。最后，我们通过projectPoints函数计算出物体在图像中的坐标，并根据相机焦距和物体在图像中的宽度计算出物体距离相机的距离。