如何裁剪和旋转图像到边框？ [英] How to crop and rotate an image to bounding box?

问题描述

• 我有一个包含手的数千个图像的数据集


• 我也有.mat文件包含边界框四个角的坐标
• 
  

• 然而，这些边界框的边缘与x& y轴。例如，
  

• I have a dataset of thousands of images containing hands
• I also have .mat files which contain the coordinates of 4 corners of the bounding box

• However, the edges of these bounding boxes are at an angle with the x & y axis. For example,

我想使用边界框坐标&然后旋转手，使其与x或y轴对齐。

I want to crop out the hands using the bounding box coordinates & then rotate the hands such that they are aligned with the x or y axis.

编辑：

手的表示如下：

但是，请记住矩形不是直的。

However, please keep in mind that the rectangle is NOT straight. So, I'll have to rotate it to straighten it out.

推荐答案

好的！

计算矩形的大小

 width = sqrt( sum( (b-a).^2 ) );
 height = sqrt( sum( (c-b).^2 ) );

第二步：

计算仿射从 a ... d 转换为直立图像

Second step:

Compute an affine transformation from a...d to an upright image

 Xin = [a(2) b(2) c(2) d(2)];
 Yin = [a(1) b(1) c(1) d(1)];
 Xout = [width 1 1 width];
 Yout = [1 1 height height];
 A = [Xin;Yin;ones(1,4)]';
 B = [Xout; Yout]';
 H = B \ A; % affine transformation

请注意，尽管我们允许fo H 是仿射的，角的选择（取决于 width 和 height code> H 不会扭曲裁剪的矩形。

Note that despite the fact that we allow fo H to be affine, the choise of corners (depending on width and height) will acertain that H will not distort the cropped rectangle.

可以使用 cp2tform ：

 H2 = cp2tform( [Xin;Yin]', [Xout;Yout]', 'nonreflectivesimilarity' );

第三步

获取相关图片部分

Third step

Use the transformation to get the relevant image part

 thumb = tformarray( img, maketform( 'affine', H' ), ... %//'
                     makeresampler( 'cubic', 'fill' ), ...
                     1:2, 1:2, ceil( [height width] ), [], 0 );

可选择使用 imtransform ：

optionally use imtransform:

 thumb = imtransform( img, H2, 'bicubic' );

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有关向量化的注意事项：

取决于如何存储角落的坐标（ a ... d ）前两个步骤可以很容易地向量化。

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A note regarding vectorization:

depends on how the coordinates of the corners are stored (a...d) the first two steps can be easily vectorize.

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