我有一张图片和一组四个点(描述一个四边形 Q)。我想变换此图像,使其适合四边形 Q。Photoshop 将此变换称为“扭曲”。但根据这个四边形的来源(图像在空间中移动的视角),它实际上是一个缩放、一个旋转和一个透视矩阵的组合。
我想知道使用 CATransform3D 4x4 矩阵是否可行。你有关于如何做到这一点的任何提示吗?我试图取这四个点并建立 16 个方程(从 A' = A x u 中得出)但它没有用:我不确定我应该使用什么作为 z、z'、w 和 w' 系数……
下图是我想做的:
这里有一些点的例子:
276.523, 236.438, 517.656, 208.945, 275.984, 331.285, 502.23, 292.344
261.441, 235.059, 515.09, 211.5, 263.555, 327.066, 500.734, 295
229.031, 161.277, 427.125, 192.562, 229.16, 226, 416.48, 256
最佳答案
我已经为在 iOS 上执行此操作创建了一个工具包:https://github.com/hfossli/AGGeometryKit/
确保你的 anchor 在左上角(CGPointZero
)。
+ (CATransform3D)rectToQuad:(CGRect)rect
quadTL:(CGPoint)topLeft
quadTR:(CGPoint)topRight
quadBL:(CGPoint)bottomLeft
quadBR:(CGPoint)bottomRight
{
return [self rectToQuad:rect quadTLX:topLeft.x quadTLY:topLeft.y quadTRX:topRight.x quadTRY:topRight.y quadBLX:bottomLeft.x quadBLY:bottomLeft.y quadBRX:bottomRight.x quadBRY:bottomRight.y];
}
+ (CATransform3D)rectToQuad:(CGRect)rect
quadTLX:(CGFloat)x1a
quadTLY:(CGFloat)y1a
quadTRX:(CGFloat)x2a
quadTRY:(CGFloat)y2a
quadBLX:(CGFloat)x3a
quadBLY:(CGFloat)y3a
quadBRX:(CGFloat)x4a
quadBRY:(CGFloat)y4a
{
CGFloat X = rect.origin.x;
CGFloat Y = rect.origin.y;
CGFloat W = rect.size.width;
CGFloat H = rect.size.height;
CGFloat y21 = y2a - y1a;
CGFloat y32 = y3a - y2a;
CGFloat y43 = y4a - y3a;
CGFloat y14 = y1a - y4a;
CGFloat y31 = y3a - y1a;
CGFloat y42 = y4a - y2a;
CGFloat a = -H*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42);
CGFloat b = W*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43);
CGFloat 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);
CGFloat d = H*(-x4a*y21*y3a + x2a*y1a*y43 - x1a*y2a*y43 - x3a*y1a*y4a + x3a*y2a*y4a);
CGFloat e = W*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42);
CGFloat 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)));
CGFloat g = H*(x3a*y21 - x4a*y21 + (-x1a + x2a)*y43);
CGFloat h = W*(-x2a*y31 + x4a*y31 + (x1a - x3a)*y42);
CGFloat 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));
const double kEpsilon = 0.0001;
if(fabs(i) < kEpsilon)
{
i = kEpsilon* (i > 0 ? 1.0 : -1.0);
}
CATransform3D transform = {a/i, d/i, 0, g/i, b/i, e/i, 0, h/i, 0, 0, 1, 0, c/i, f/i, 0, 1.0};
return transform;
}
此代码不归功于我。我所做的只是在互联网上搜索并将各种不完整的答案放在一起。
关于iphone - 使用 CATransform3D 将矩形图像转换为四边形,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9470493/