Question

我正在尝试将imrotate应用于图像，但图像的角度未知，并随每张图像而变化。

如果我已将图像转换为二进制，那么可以找到最低y位置'1'和最左边'1'并使用它们之间的渐变/角度作为我图像旋转的角度？

使用这两个位置之间的角度并将其与x轴对齐？

当前进度 - 转换为二进制并使边缘更具可辨性：

% convert to binary
greyImage = rgb2gray(C); % greyscale
cannyImage = edge(greyImage, 'canny'); % canny edge detection

% fill the gaps in the shape
se = strel('disk',2);
bw = imclose(cannyImage, se);

filled = imfill(bw, 'holes');

imshow(filled);

[~,lowerMostCol] = max(cumsum(sum(filled,2)));
[~,leftMostRow] = max(sum(filled,1)==1);

Answer 1

方法＃1

使用a作为二进制图像，您可以执行以下操作 -

[~,lowermost] = max(cumsum(sum(a,2)));
lowermostpt = [lowermost,find(a(lowermost,:),1,'first')]

[~,rightmost] = max(cumsum(sum(a,1)));
rightmostpt = [find(a(:,rightmost),1,'first'),rightmost]

[~,topmost] = max(sum(a,2)==1);
topmostpt = [topmost,find(a(topmost,:),1,'first')]

[~,leftmost] = max(sum(a,1)==1);
leftmostpt = [find(a(:,leftmost),1,'first'),leftmost]

为了提高性能，最好将总结存储一次，然后再重复使用。

示例运行 -

a =
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     1     0     0     0     0     0
     0     0     0     0     0     0     0     0     1     1     1     0     0     0     0
     0     0     0     0     0     0     0     1     1     1     1     1     0     0     0
     0     0     0     0     0     0     1     1     1     1     1     1     1     0     0
     0     0     0     0     0     1     1     1     1     1     1     1     1     1     0
     0     0     0     0     0     0     1     1     1     1     1     1     1     0     0
     0     0     0     0     0     0     0     1     1     1     1     1     0     0     0
     0     0     0     0     0     0     0     0     1     1     1     0     0     0     0
     0     0     0     0     0     0     0     0     0     1     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
lowermostpt =
    12    10
rightmostpt =
     8    14
topmostpt =
     4    10
leftmostpt =
     8     6

方法＃2 使用图像处理工具箱中的bwboundaries -

idx = cell2mat(bwboundaries(a))

[~,p1] = min(idx(:,1))
topmostpt = idx(p1,:)
[~,p2] = max(idx(:,1))
lowermostpt = idx(p2,:)

[~,p3] = min(idx(:,2))
leftmostpt = idx(p3,:)
[~,p4] = max(idx(:,2))
rightmostpt = idx(p4,:)

找到最低和最右边的像素以找到图像的角度

1 个答案: