编写一个Harris角点检测图片的程序，自动输出角的大小

好的，这里有一个Python实现的Harris角点检测程序，使用OpenCV库进行图像处理：

import cv2
import numpy as np

# 读取图片
img = cv2.imread('image.jpg')

# 转换为灰度图
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)

# 计算导数
dx = cv2.Sobel(gray, cv2.CV_64F, 1, 0)
dy = cv2.Sobel(gray, cv2.CV_64F, 0, 1)

# 计算Ixx, Ixy, Iyy
Ixx = dx * dx
Ixy = dx * dy
Iyy = dy * dy

# 高斯滤波
kernel_size = 3
sigma = 1.0
ksize = int(2 * np.ceil(2 * sigma) + 1)
gaussian = cv2.getGaussianKernel(ksize, sigma)
gaussian = np.outer(gaussian, gaussian.transpose())

Ixx = cv2.filter2D(Ixx, -1, gaussian)
Ixy = cv2.filter2D(Ixy, -1, gaussian)
Iyy = cv2.filter2D(Iyy, -1, gaussian)

# 计算Harris响应
k = 0.04
det = Ixx * Iyy - Ixy * Ixy
trace = Ixx + Iyy
H = det - k * trace * trace

# 非极大值抑制
threshold = 0.01 * H.max()
kernel_size = 3
maxima = cv2.dilate(H, np.ones((kernel_size, kernel_size))) == H
H[~maxima] = 0

# 输出角点大小
corners = np.where(H > threshold)
for i in range(len(corners[0])):
    x = corners[0][i]
    y = corners[1][i]
    size = H[x][y]
    print("角点(%d, %d)大小为%.2f" % (x, y, size))

# 显示结果
cv2.imshow('Harris角点检测', img)
cv2.waitKey(0)

请将代码中的`image.jpg`替换为需要检测角点的图片路径。程序会自动输出检测到的角点位置和大小。