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希望有人可以提供帮助，我正在尝试研究如何将图像从矩形转换为四边形，每个角都有给定的 x,y 屏幕坐标。

到目前为止，我已将图像放在 CALayer 上，但需要计算出 CATransform3D 以将矩形扭曲为所需的四边形。下面是我想要实现的一个示例（从 a 到 b）。

矩形到四边形图像的示例

如果我错了并且不能使用 CATransform3D 来完成，是否有任何其他方式可以通过示例来实现。

我认为 KennyTM 的答案与我在这里所需要的很接近。

iPhone 图像拉伸（倾斜）

我已经尝试过了，但运气不佳，他确实提到“您可能需要转置”，但如果是这种情况，我不知道该怎么做。

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3 回答 3

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CATransform 3D 绝对可以完成您尝试使用它的目的。我测试了您链接到的代码，它对我来说非常有效。请记住，像这样的变换矩阵仅定义为一个比例，因为它位于齐次坐标中。一旦你用他的方程生成矩阵，将每个元素除以右下角的元素。我能想到你需要转置的唯一原因是因为他给出的变换是行主要顺序的。如果您要填充列主要变换矩阵（我相信 CATransform3D 是），则需要在填充后对其进行转置。

这是我用来测试它的代码，它使用来自 openCV 的矩阵类并且在 c++ 中，但应该证明这一点

cv::Matx41d rect_tl(-10,-10,0,1);
cv::Matx41d rect_tr(10,-10,0,1);
cv::Matx41d rect_bl(-10,10,0,1);
cv::Matx41d rect_br(10,10,0,1);

cv::Matx41d quad_tl(2,2,0,1);
cv::Matx41d quad_tr(4,6,0,1);
cv::Matx41d quad_bl(2,-1,0,1);
cv::Matx41d quad_br(3,5,0,1);


double X = rect_tl(0);
double Y = rect_tl(0);
double W = 20;
double H = 20;

double x1a = quad_tl(0);
double y1a = quad_tl(1);

double x2a = quad_tr(0);
double y2a = quad_tr(1);

double x3a = quad_bl(0);
double y3a = quad_bl(1);

double x4a = quad_br(0);
double y4a = quad_br(1);



double y21 = y2a - y1a,
y32 = y3a - y2a,
y43 = y4a - y3a,
y14 = y1a - y4a,
y31 = y3a - y1a,
y42 = y4a - y2a;

double a = -H*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42);
double b = W*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43);
double c = H*X*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42) - H*W*x1a*(x4a*y32 - x3a*y42 + x2a*y43) - W*Y*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43);

double d = H*(-x4a*y21*y3a + x2a*y1a*y43 - x1a*y2a*y43 - x3a*y1a*y4a + x3a*y2a*y4a);
double e = W*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42);
double f = -(W*(x4a*(Y*y2a*y31 + H*y1a*y32) - x3a*(H + Y)*y1a*y42 + H*x2a*y1a*y43 + x2a*Y*(y1a - y3a)*y4a + x1a*Y*y3a*(-y2a + y4a)) - H*X*(x4a*y21*y3a - x2a*y1a*y43 + x3a*(y1a - y2a)*y4a + x1a*y2a*(-y3a + y4a)));

double g = H*(x3a*y21 - x4a*y21 + (-x1a + x2a)*y43);
double h = W*(-x2a*y31 + x4a*y31 + (x1a - x3a)*y42);
double i = W*Y*(x2a*y31 - x4a*y31 - x1a*y42 + x3a*y42) + H*(X*(-(x3a*y21) + x4a*y21 + x1a*y43 - x2a*y43) + W*(-(x3a*y2a) + x4a*y2a + x2a*y3a - x4a*y3a - x2a*y4a + x3a*y4a));

cv::Matx44d matrix(a,b,0,c
                   ,d,e,0,f
                   ,0,0,1,0
                   ,g,h,0,i);
matrix = matrix*(1/matrix(15));
//You may need a transpose here

cv::Matx41d test_tl = matrix*rect_tl;
test_tl *= (1/test_tl(3));
cv::Matx41d test_tr = matrix*rect_tr;
test_tr *= (1/test_tr(3));
cv::Matx41d test_bl = matrix*rect_bl;
test_bl *= (1/test_bl(3));
cv::Matx41d test_br = matrix*rect_br;
test_br *= (1/test_br(3));

执行后，底部的所有测试变量都与四边形对应项完美匹配。希望这可以解决问题。

于 2012-08-02T16:10:36.660 回答
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感谢 Hammers 的回答，我能够让它全部工作，需要转置并找到这个关于如何转置矩阵的精彩博客......

转置矩阵

我创建的结果工作方法如下......

- (CATransform3D)rectToQuad:(NSRect)rect quadTLX:(double)x1a quadTLY:(double)y1a quadTRX:(double)x2a quadTRY:(double)y2a quadBLX:(double)x3a quadBLY:(double)y3a quadBRX:(double)x4a quadBRY:(double)y4a
{
    double X = rect.origin.x;
    double Y = rect.origin.y;
    double W = rect.size.width;
    double H = rect.size.height;

    double y21 = y2a - y1a;
    double y32 = y3a - y2a;
    double y43 = y4a - y3a;
    double y14 = y1a - y4a;
    double y31 = y3a - y1a;
    double y42 = y4a - y2a;

    double a = -H*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42);
    double b = W*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43);
    double c = H*X*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42) - H*W*x1a*(x4a*y32 - x3a*y42 + x2a*y43) - W*Y*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43);

    double d = H*(-x4a*y21*y3a + x2a*y1a*y43 - x1a*y2a*y43 - x3a*y1a*y4a + x3a*y2a*y4a);
    double e = W*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42);
    double f = -(W*(x4a*(Y*y2a*y31 + H*y1a*y32) - x3a*(H + Y)*y1a*y42 + H*x2a*y1a*y43 + x2a*Y*(y1a - y3a)*y4a + x1a*Y*y3a*(-y2a + y4a)) - H*X*(x4a*y21*y3a - x2a*y1a*y43 + x3a*(y1a - y2a)*y4a + x1a*y2a*(-y3a + y4a)));

    double g = H*(x3a*y21 - x4a*y21 + (-x1a + x2a)*y43);
    double h = W*(-x2a*y31 + x4a*y31 + (x1a - x3a)*y42);
    double i = W*Y*(x2a*y31 - x4a*y31 - x1a*y42 + x3a*y42) + H*(X*(-(x3a*y21) + x4a*y21 + x1a*y43 - x2a*y43) + W*(-(x3a*y2a) + x4a*y2a + x2a*y3a - x4a*y3a - x2a*y4a + x3a*y4a));

    //Transposed matrix
    CATransform3D transform;
    transform.m11 = a / i;
    transform.m12 = d / i;
    transform.m13 = 0;
    transform.m14 = g / i;
    transform.m21 = b / i;
    transform.m22 = e / i;
    transform.m23 = 0;
    transform.m24 = h / i;
    transform.m31 = 0;
    transform.m32 = 0;
    transform.m33 = 1;
    transform.m34 = 0;
    transform.m41 = c / i;
    transform.m42 = f / i;
    transform.m43 = 0;
    transform.m44 = i / i;
    return transform;
}

对该方法的示例调用如下...

NSImage *image = // load a image

CALayer *layer = [CALayer layer];
[layer setContents:image];
[view setLayer:myLayer];
[view setFrame:NSMakeRect(0, 0, image.size.width, image.size.height)];

view.layer.transform = [self rectToQuad:view.frame quadTLX:0 quadTLY:0 quadTRX:image.size.width quadTRY:20 quadBLX:0 quadBLY:image.size.height quadBRX:image.size.width quadBRY:image.size.height + 90];
于 2012-08-03T09:34:03.540 回答
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感谢@Equinox2000 的帮助！

请注意，iOS CALayer 的默认锚点为 (0.5, 0.5)。如果您尝试应用所有值都相对于左上角坐标的变换，您需要将锚点更改为 (0.0, 0.0)。

于 2012-09-04T02:05:56.267 回答