o0o0o0o0o0o0o / image-processing-from-scratch Goto Github PK

Hi, I have a question about the function convlove()
"val = (filter * pad_mat[j * strides[1]:j * strides[1] + filter_size[0],
k * strides[0]:k * strides[0] + filter_size[1]]).sum()"
why you multiply "j" with stride[i]? In the "for j in range(0, mat_size[0], strides[1]):" , there is a strides[1]. I'm confused about it. Thank you!

Windows上使用VS2019运行分水岭算法代码时，出现如下错误：

找不出问题所做，请求帮助。

{
"resource": "/D:/Github/clone path/image-processing-from-scratch/sift/SIFT.py",
"owner": "python",
"code": "unsubscriptable-object",
"severity": 8,
"message": "Value 'info.shape' is unsubscriptable",
"source": "pylint",
"startLineNumber": 507,
"startColumn": 26,
"endLineNumber": 507,
"endColumn": 26
}

I tried using your codes for PCA&SVD and tested it on 0.jpg (21.9 kB) image provided, I found out that in spite of using PCA, the size of image 0_.jpg (73.1 kB) is not reduced.
Any leads will be helpful

up你好，b站来的。有一些SIFT代码问题求指教：
1、GuassianKernel( ) 求高斯卷积核的时候为啥不需要归一化咧？
2、LocateKeyPoint（）函数中，threshold = 0.5contrastThreshold/(n255SIFT_FIXPT_SCALE)，计算差分图像中的噪点阈值时候，这个公式与视频中所讲的0.50.4/n不同，是怎么来的啊？
3、LocateKeyPoint（）函数中，计算邻域极值的时候，我注意到这个邻域不是 33，而是34？这是为啥呢？对于i是i-1，但对于j就变成了j+2，不应该是j+1嘛？
eight_neiborhood_prev = img_prev[max(0, i - 1):min(i + 2, img_prev.shape[0]), max(0, j - 1):min(j + 2, img_prev.shape[1])] eight_neiborhood = img[max(0, i - 1):min(i + 2, img.shape[0]), max(0, j - 1):min(j + 2, img.shape[1])] eight_neiborhood_next = img_next[max(0, i - 1):min(i + 2, img_next.shape[0]), max(0, j - 1):min(j + 2, img_next.shape[1])]

Traceback (most recent call last):
File "/home/zsc/桌面/image-processing-from-scratch/sift/SIFT.py", line 557, in
drawLines(X1,X2,Y1,Y2,match[0][:,0],result)
File "/home/zsc/桌面/image-processing-from-scratch/sift/SIFT.py", line 503, in drawLines
img = Lines(img,info)
File "/home/zsc/桌面/image-processing-from-scratch/sift/SIFT.py", line 491, in Lines
result[i,j] = color
ValueError: cannot copy sequence with size 3 to array axis with dimension 4