OpenCV : 仿射變換

幾何變換 : 仿射變換

仿射變換

$\left[\begin{matrix}\tilde{x} \\ \tilde{y}\end{matrix}\right] = \left[\begin{matrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{matrix} \right]\left[\begin{matrix}x \\ y\end{matrix}\right]+ \left[\begin{matrix}a_{13} \\ a_{14}\end{matrix}\right]$

爲了計算方便(進行一些變換):齊次座標

$\left[\begin{matrix}\tilde{x} \\ \tilde{y}\\ 1\end{matrix}\right] = \left[\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22}& a_{23}\\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ 1\end{matrix}\right]= A\left[\begin{matrix}x \\ y \\ 1\end{matrix}\right]$

平移

$(\tilde{x},\tilde{y}) = (x+t_x, y+t_y)$
$\left[\begin{matrix}\tilde{x} \\ \tilde{y}\\ 1\end{matrix}\right] = \left[\begin{matrix}1 & 0 & t_x\\0 & 1 & t_y\\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ 1\end{matrix}\right]$
$\left[\begin{matrix}\tilde{x} \\ \tilde{y}\\ 1\end{matrix}\right] = \left[\begin{matrix}1 & 0 & t_x\\0 & 1 & t_y\end{matrix}\right] \left[\begin{matrix}x \\ y \\ 1\end{matrix}\right]$

放大和縮小

二維空間縮放:二維空間的座標$(x, y)$,以$(0, 0)$爲中心在$x$軸方向和$y$軸方向上縮放.以縮放$S$倍爲例:

• $(\tilde{x}, \tilde{y}) = (S_x*x, S_y*y)$
• $S_x = S_y$ : 等比例縮放

$\left[\begin{matrix}\tilde{x} \\ \tilde{y}\\ 1\end{matrix}\right] = \left[\begin{matrix}S_x & 0 & 0\\0 & S_y & 0\\ 0 & 0 & 1\end{matrix} \right]\left[\begin{matrix}x \\ y \\ 1\end{matrix}\right]$

二維空間:以任意座標$x_0, y_0)$爲中心進行縮放:$(\tilde{x}, \tilde{y}) = (x_0+S_x*(x-x_0), y_0+S_y*(y-y_0))$

$\left[\begin{matrix}\tilde{x} \\ \tilde{y}\\ 1\end{matrix}\right] = \left[\begin{matrix}1 & 0 & x_0\\0 & 1 & y_0\\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}S_x & 0 & 0\\0 & S_y & 0\\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}1 & 0 & -x_0\\0 & 1 & -y_0\\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ 1\end{matrix}\right]$

順時針旋轉$\alpha$

計算$(x, y)\to(\tilde{x}, \tilde{y})$, $p$爲$(x, y)$到原點的距離:
$cos\theta = \frac{x}{p}, sin\theta = \frac{y}{p}$
$cos(\theta + \alpha)$
$cos(\theta + \alpha) = cos\theta cos\alpha - sin\theta sin\alpha = \frac{x}{p}cos\alpha - \frac{y}{p}sin\alpha = \frac{\tilde{x}}{p}$
$sin(\theta + \alpha)$
$sin(\theta + \alpha) = sin\theta cos\alpha + cos\theta sin\alpha = \frac{y}{p}cos\alpha + \frac{x}{p}sin\alpha = \frac{\tilde{y}}{p}$
$\tilde{x}, \tilde{y}$

$\tilde{x} = xcos\alpha - ysin\alpha, \tilde{y} = xsin\alpha + ycos\alpha$

矩陣形式

$\left[\begin{matrix}\tilde{x} \\ \tilde{y}\\ 1\end{matrix}\right] = \left[\begin{matrix}cos\alpha & -sin\alpha & 0\\sin\alpha & cos\alpha & 0\\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ 1\end{matrix}\right], A = \left[\begin{matrix}cos\alpha & -sin\alpha & 0\\sin\alpha & cos\alpha & 0\\ 0 & 0 & 1\end{matrix}\right]$

逆時針旋轉$\alpha$

$cos(\theta - \alpha)$
$cos(\theta - \alpha) = cos\theta cos\alpha + sin\theta sin\alpha = \frac{x}{p}cos\alpha + \frac{y}{p}sin\alpha = \frac{\tilde{x}}{p}$
$sin(\theta - \alpha)$
$sin(\theta - \alpha) = sin\theta cos\alpha - cos\theta sin\alpha = \frac{y}{p}cos\alpha - \frac{x}{p}sin\alpha = \frac{\tilde{y}}{p}$
$\tilde{x}, \tilde{y}$

$\tilde{x} = xcos\alpha + ysin\alpha, \tilde{y} = -xsin\alpha + ycos\alpha$

矩陣形式

$\left[\begin{matrix}\tilde{x} \\ \tilde{y}\\ 1\end{matrix}\right] = \left[\begin{matrix}cos\alpha & sin\alpha & 0\\-sin\alpha & cos\alpha & 0\\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ 1\end{matrix}\right], A = \left[\begin{matrix}cos\alpha & sin\alpha & 0\\-sin\alpha & cos\alpha & 0\\ 0 & 0 & 1\end{matrix}\right]$

從任意位置旋轉$\alpha$矩陣形式

$\left[\begin{matrix}\tilde{x} \\ \tilde{y}\\ 1\end{matrix}\right] = \left[\begin{matrix}1 & 0 & x_0\\0 & 1 & y_0\\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}cos\alpha & sin\alpha & 0\\-sin\alpha & cos\alpha & 0\\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}1 & 0 & -x_0\\0 & 1 & -y_0\\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ 1\end{matrix}\right]= A\left[\begin{matrix}x \\ y \\ 1\end{matrix}\right]$

計算放射矩陣

• getAffineTransform(src, dst)
  1. Point2f[]:保存放射前後的座標,數據類型CV_64F
  2. Mat_:保存放射前後的座標,數據類型CV_32F
• getRotationMatrix2D(center, angle, scale)
  1. center : 原座標
  2. angle : 角度
  3. scale : 縮放倍數
• resize(Mat src, Mat dst, Size dsize, double fx=0, double fy=0, int interpolation)
  1. src : 輸入圖像矩陣
  2. dst : 輸出圖像矩陣
  3. dsize : (寬,高)輸出圖像大小
  4. fx, fy : 水平垂直方向的縮放比例
  5. interpolation : INTE_NEAREST, INTE_LINEAR(默認)
• warpAffine(Mat src, Mat dst, Mat s, Size dsize)
  1. s : 放射變換矩陣
• rotate(Mat src, Mat dst, int rotateCode)
  1. src : 輸入矩陣
  2. dst : 輸出矩陣
  3. rotateCode : ROTATE_90_CLOCKWISE : 順時針旋轉90度,ROTATE_180 : 順時針旋轉180度,ROTATE_90_COUNTERCLOCKWISE : 順時針旋轉270度

#include<iostream>
#include<opencv2/core.hpp>
#include<opencv2/highgui.hpp>
#include<opencv2/imgproc.hpp>

using namespace cv;
using namespace std;

int main()
{
	Point2f src[] = { Point2f(0, 0), Point2f(20,0),Point2f(0,20) };
	Point2f dst[] = { Point2f(0,0),Point2f(20,0),Point2f(0,20) };
	Mat A = getAffineTransform(src, dst);
	cout << A << endl;
	return 0;
}

#include<iostream>
#include<opencv2/core.hpp>
#include<opencv2/highgui.hpp>
#include<opencv2/imgproc.hpp>

using namespace cv;
using namespace std;

int main()
{
	Mat src = (Mat_<float>(3, 2) << 0, 0, 20, 0, 0, 20);
	Mat dst = (Mat_<float>(3, 2) << 0, 0, 10, 0, 0, 10);
	Mat A = getAffineTransform(src, dst);
	cout << A << endl;
	return 0;
}

插值算法

把圖像看作是一個二維的圖像函數$f_I$
$z = f_I(x, y) = I(x,y), 0 <= x < W, 0< = y <H, x \in N, y \in N$

• $N$ : 整數集合
• $I(x, y)$ : 矩陣$I$的$x$行$y$列的值
• $f_I$ : 圖像函數
• $f_I(x, y)$ : 圖像座標$(x, y)$處的值

$(x, y) \to (\tilde{x}, \tilde{y})$的空間變換關係如下:
$\left[\begin{matrix}\tilde{x}\\ \tilde{y}\\ 1\end{matrix}\right] = A\left[\begin{matrix}x\\ y\\ 1\end{matrix}\right], A =\left[\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\0&0&1\end{matrix}\right]$

圖像矩陣的放大$f_O(2x, 2y) = f_I(x, y)$:假設:$(4,4) \longleftarrow (2,2)$,即$f_O(4,4) = f_I(2,2)$,同理$f_O(3,3)=f_I(1,5, 1.5)$,但是$f_I$在$(1.5,1.5)$處沒有函數值,因爲$f_I$只在第一象限的整數座標處有函數值,但是$(1.5, 1.5)$周圍四個臨近的座標有函數值$(1,1),(2,1),(1,2),(2,2)$,可以根據這四個位置的函數值估算$f_I(1.5, 1.5)$的值.

最鄰近插值(輸出圖像會出現鋸齒狀)

最鄰近插值同上,也是找到目標座標四個相鄰的座標,從中找到一個距離目標最近的一個,假設爲$(\hat{x},\hat{y})$
$f_I(x, y) = f_I(\hat{x}, \hat{y})$

線性插值

已知座標$(x_0,y_0),(x_1, y_1)$,要計算得到兩點所在直線上$x$處$y$的值:

1. $x$已知求$y$?
   $\frac{y-y_0}{x-x_0} = \frac{y_1 - y_0}{x_1 - x_0}$
   $y = \frac{y_1 - y_0}{x_1 - x_0}(x-x_0)+y_0$
   已知$y$計算$x$同上

雙線性插值

顧名思義,雙線性插值就是線性插值的double,$p_1,p_2,p_3,p_4$均已知,求$O$

1. 先計算$a$, $b$
   $y_a = \frac{y_{p_2}-y_{p_1}}{x_{p_2}-x_{p_1}}(x_a-x_{p_1})+y_{p_1} \tag{1}$
   $y_b = \frac{y_{p_4}-y_{p_3}}{x_{p_4}-x_{p_3}}(x_b-x_{p_3})+y_{p_3} \tag{2}$

2. $a, b$ 已知求$O$
   $y_O = \frac{y_a -y_b}{x_a - x_b}(x_O - x_b) + y_b \tag{3}$

圖像的縮放

#include<iostream>
#include<opencv2/core.hpp>
#include<opencv2/highgui.hpp>
#include<opencv2/imgproc.hpp>

using namespace cv;
using namespace std;

int main()
{
	Mat I = imread("./People/wave.jpg", 1);
	if (!I.data)
		return -1;
	// 放大放射矩陣
	Mat s = (Mat_<float>(2, 3) << 2, 0, 0, 0, 2, 0);
	Mat dst1;
	// 利用放射矩陣縮放
	warpAffine(I, dst1, s, Size(I.cols * 2, I.rows * 2));
	Mat dst2;
	// resize 放大矩陣2倍
	resize(I, dst2, Size(I.cols * 2, I.rows * 2), 2, 2);
	Mat dst3;
	resize(I, dst3, Size(I.cols * 10, I.rows * 10), 10, 10);
	imshow("original img", I);
	imshow("warpAffine", dst1);
	imshow("resize x 2", dst2);
	imshow("resize x 10", dst3);
	waitKey(0);
}

圖像的旋轉

#include<iostream>
#include<opencv2/core.hpp>
#include<opencv2/highgui.hpp>
#include<opencv2/imgproc.hpp>

using namespace cv;
using namespace std;

int main()
{
	Mat I = imread("./People/wave.jpg", 1);
	if (!I.data)
		return -1;
	Mat Ix10;
	resize(I, Ix10, Size(I.cols * 10, I.rows * 10), 10, 10);
	Mat I_90, I_180, I_270;
    // 順時針旋轉90, 180, 270
	rotate(Ix10, I_90, ROTATE_90_CLOCKWISE);
	rotate(Ix10, I_180, ROTATE_180);
	rotate(Ix10, I_270, ROTATE_90_CLOCKWISE);
	imshow("順時針旋轉90", I_90);
	imshow("順時針旋轉180", I_180);
	imshow("順時針旋轉270", I_270);
	waitKey(0);
}