Question

如何扭曲图像？例如，每个角都有一个带有坐标的CGPoint - p1，p2，p3，p4。然后，我需要设置 - p4.x + = 50，p4.y + = 30。所以这个角（p4）应该在2D透视图中拉伸，图像应该扭曲。

alt text http://www.polar-b.com/scew.png

我尝试使用 CATransform3D ，但似乎无法以这种方式完成，因为它只是改变视角（旋转，更靠近/更远一侧）。也许 CGAffineTransform 可能有用吗？

如果您知道答案，请写一个示例代码。

提前致谢

Answer 1

CGAffineTransform无法实现。仿射变换总是可以分解为平移，旋转，剪切和缩放。它们都将平行四边形映射成平行四边形，而变换则不然。

对于您的转换，可以分两步完成。一个将方块转换为梯形。

p1-----p2       p1-----p2
 |     |   -->   |       \
p3-----p4       p3--------p4'

另一个垂直方向。一个天真的转换规则是

                   y - c
x' = (x - p1.x) * ———————— + p1.x
                  p1.y - c
y' = y

其中c是连接p1和p3以及p2和p4的线的交点的y坐标。

现在注意转换中的x * y因子。这表明这种变换不是线性的。因此，CATransform3D也不能将其作为2D变换执行。

然而，载体

[x, y, z, w=1]

将转换为实际的3D矢量

(x/w, y/w, z/w)

在投影之前，如果CA遵循通常的3D计算图形规则，那么您可以使用转换“作弊”

[ P . . Q ] [ x ]   [ x' ]
[ . R . S ] [ y ] = [ y' ]
[ . . 1 . ] [ z ]   [ z' ]
[ . T . U ] [ 1 ]   [ w' ]

使用适当的P，Q，R，S，T，U将4个点映射到预期位置。（6个独特的坐标和6个变量在大多数情况下应该只有1个解决方案。）

当你找到这6个常数时，你可以制作一个CATransform3D。请注意结构定义是

struct CATransform3D
   {
   CGFloat m11, m12, m13, m14;
   CGFloat m21, m22, m23, m24;
   CGFloat m31, m32, m33, m34;
   CGFloat m41, m42, m43, m44;
};
typedef struct CATransform3D CATransform3D;

因此，您可以直接更改矩阵元素，而不是依赖于CATransform3DMake函数。（由于使用行或列向量的惯例，您可能需要执行转置。）

获得将矩形（（X，Y），（W，H））转换为任何四边形（（x1a，y1a），（x2a，y2a）;（x3a，y3a），（x4a，y4a）的变换）），使用此功能（您可能需要转置）：

function compute_transform_matrix(X, Y, W, H, x1a, y1a, x2a, y2a, x3a, y3a, x4a, y4a) {
    var y21 = y2a - y1a, 
        y32 = y3a - y2a,
        y43 = y4a - y3a,
        y14 = y1a - y4a,
        y31 = y3a - y1a,
        y42 = y4a - y2a;

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

    var d = H*(-x4a*y21*y3a + x2a*y1a*y43 - x1a*y2a*y43 - x3a*y1a*y4a + x3a*y2a*y4a);
    var e = W*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42);
    var 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)));

    var g = H*(x3a*y21 - x4a*y21 + (-x1a + x2a)*y43);
    var h = W*(-x2a*y31 + x4a*y31 + (x1a - x3a)*y42);
    var 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));

    return [[a,b,0,c],[d,e,0,f],[0,0,1,0],[g,h,0,i]];
}

Answer 2

UIImage / CGImageRef上的3D变换

你应该能够自己计算每个像素的映射。不完美，但它可以解决问题......

此库http://github.com/hfossli/AGGeometryKit/

可用

有趣的文件是

https://github.com/hfossli/AGGeometryKit/blob/master/Source/AGTransformPixelMapper.m

https://github.com/hfossli/AGGeometryKit/blob/master/Source/CGImageRef%2BCATransform3D.m

https://github.com/hfossli/AGGeometryKit/blob/master/Source/UIImage%2BCATransform3D.m

UIView / UIImageView上的3D变换

https://stackoverflow.com/a/12820877/202451

然后您将完全控制四边形中的每个点。：）

Answer 3

struct CATransform3D
{
   CGFloat m11, m12, m13, m14;
   CGFloat m21, m22, m23, m24;
   CGFloat m31, m32, m33, m34;
   CGFloat m41, m42, m43, m44;
};

您必须调整m24和m14才能获得这样的形状。

Answer 4

我在Swift中尝试了很好的答案@KennyTM，并得到了一个错误“表达太复杂了，无法在合理的时间内解决”。

所以这是Swift的简化版本：

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

let a = -H*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42)
let b = W*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43)
let c0 = -H*W*x1a*(x4a*y32 - x3a*y42 + x2a*y43)
let cx = H*X*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42)
let cy = -W*Y*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43)
let c = c0 + cx + cy

let d = H*(-x4a*y21*y3a + x2a*y1a*y43 - x1a*y2a*y43 - x3a*y1a*y4a + x3a*y2a*y4a)
let e = W*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42)
let f0 = -W*H*(x4a*y1a*y32 - x3a*y1a*y42 + x2a*y1a*y43)
let fx = H*X*(x4a*y21*y3a - x2a*y1a*y43 - x3a*y21*y4a + x1a*y2a*y43)
let fy = -W*Y*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42)
let f = f0 + fx + fy;

let g = H*(x3a*y21 - x4a*y21 + (-x1a + x2a)*y43)
let h = W*(-x2a*y31 + x4a*y31 + (x1a - x3a)*y42)
let i0 = H*W*(x3a*y42 - x4a*y32 - x2a*y43)
let ix = H*X*(x4a*y21 - x3a*y21 + x1a*y43 - x2a*y43)
let iy = W*Y*(x2a*y31 - x4a*y31 - x1a*y42 + x3a*y42)
var i = i0 + ix + iy


let epsilon = CGFloat(0.0001);
if fabs(i) < epsilon {
    i = epsilon * (i > 0 ? 1 : -1);
}

return CATransform3D(m11: a/i, m12: d/i, m13: 0, m14: g/i, m21: b/i, m22: e/i, m23: 0, m24: h/i, m31: 0, m32: 0, m33: 1, m34: 0, m41: c/i, m42: f/i, m43: 0, m44: 1.0)

iPhone图像拉伸（歪斜）

4 个答案: