立方和cat色样条对图像的影响 [英] Impact of cubic and catmull splines on image
本文介绍了立方和cat色样条对图像的影响的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我尝试实现下面的一些功能
为此，我试图使用立方插值和 Catmull插值（分别检查两个比较最好的结果），我不明白是什么影响这些插值显示在图像上，我们如何可以获得这些点值，我们点击设置该曲线？
我从这些资源获得帮助
编辑
int main（int argc，const char ** argv）
{
Mat input = imread（E：\\img2。 jpg）;
for（int i = 0; i< input.rows; i ++）
{
for（int p = 0; p< input.cols; p ++）
{
// for（int t = 0; t
input.at< cv :: Vec3b>（i，p） [0] = 255 * correction（input.at （i，p）[0] /255.0,ctrl,N）; // B
input.at< cv :: Vec3b>（i，p）[1] = 255 * correction（input.at （i，p）[1] /255.0,ctrl ，N）。 // G
input.at< cv :: Vec3b>（i，p）[2] = 255 * correction（input.at （i，p）[2] /255.0,ctrl ，N）。 // R
//}
}
}
imshow（image，input）;
waitKey（）;
}
解决方案
总是在相同的x坐标
，并且沿整个范围线性分布，那么你可以这样做：
// ------------------------------------------- ----------------------------
const int N = 5; //控制点数（必须> = 4）
float ctrl [N] = //控制点使用线性函数启动的y值y = x
{// x值是索引* 1.0 /（N-1）
0.00，
0.25，
0.50，
0.75，
1.00，
}
// -------------------------------------------- -------------------------------
浮点修正（float col，float * ctrl，int n）
{
float di = 1.0 / float（n-1）;
int i0，i1，i2，i3;
float t，tt，ttt;
float a0，a1，a2，a3，d1，d2;
//找到开始控制点
col * = float（n-1）;
i1 = col col- = i1;
i0 = i1-1; if（i0 <0）i0 = 0;
i2 = i1 + 1; if（i2> = n）i2 = n-1;
i3 = i1 + 2; if（i3> = n）i3 = n-1;
//计算插值系数
d1 = 0.5 *（ctrl [i2] -ctrl [i0]）;
d2 = 0.5 *（ctrl [i3] -ctrl [i1]）;
a0 = ctrl [i1];
a1 = d1;
a2 =（3.0 *（ctrl [i2] -ctrl [i1]）） - （2.0 * d1）-d2;
a3 = d1 + d2 +（2.0 *（ - ctrl [i2] + ctrl [i1]））;
//现在插入新的colro强度
t = col; tt = t * t; ttt = tt * t;
t = a0 +（a1 * t）+（a2 * tt）+（a3 * ttt）
return t;
}
// --------------------------------------- ------------------------------------
- 它使用4点1D插值立方（从上面的注释中的链接）
-
要获得新颜色，只需执行以下操作：
new_col = correction（old_col，ctrl，N）
-
这是它的外观：
- 绿色箭头显示导出错误（始终只在整曲线的开始和结束点）
- 还有另外两个控制点
[注意事项] ]
- 颜色范围为< 0.0，1.0>
$ b b
[edit1]开始/结束派生固定一点
col，float * ctrl，int n）
{
float di = 1.0 / float（n-1）;
int i0，i1，i2，i3;
float t，tt，ttt;
float a0，a1，a2，a3，d1，d2;
//找到开始控制点
col * = float（n-1）;
i1 = col col- = i1;
i0 = i1-1;
i2 = i1 + 1; if（i2> = n）i2 = n-1;
i3 = i1 + 2;
//计算插值系数
if（i0> = 0）d1 = 0.5 *（ctrl [i2] -ctrl [i0]）; else d1 = ctrl [i2] -ctrl [i1];
if（i3< n）d2 = 0.5 *（ctrl [i3] -ctrl [i1]）; else d2 = ctrl [i2] -ctrl [i1];
a0 = ctrl [i1];
a1 = d1;
a2 =（3.0 *（ctrl [i2] -ctrl [i1]）） - （2.0 * d1）-d2;
a3 = d1 + d2 +（2.0 *（ - ctrl [i2] + ctrl [i1]））;
//现在插入新的colro强度
t = col; tt = t * t; ttt = tt * t;
t = a0 +（a1 * t）+（a2 * tt）+（a3 * ttt）
return t;
}
[edit2]
-
均来自此条件：
y（t）= a0 + a1 * t + a2 * t * t + a3 * t * t * t //直接价值
y' * t + 3 * a3 * t * t //第一个派生
-
y1，y2，y3
- 所以我选择了y（0）= y1和y（1）
- 现在我需要c1连续性，所以i添加y'（0）必须与y'（1）相同。
- 对于y'（1）我选择avg方向在点y0，y1，y2之间 在点y1，y2，y3之间的平均方向
- 这些对于下一个/上一个片段是相同的，因此足够
-
y（0）= y0 = a0 + a1 * 0 + a2 * 0 * 0 + a3 * 0 * 0 * 0
y（1）= y1 = a0 + a1 * 1 + a2 * 1 * 1 + a3 * 1 * 1 * 1
y'（0）= 0.5 *（y2-y0）= a1 + 2 * a2 * 0 + 3 * a3 * 0 * 0
y'（1）= 0.5 *（y3-y1）= a1 + 2 * a2 * 1 + 3 * a3 * 1 * 1
-
现在解决这个等式的系统（a0，a1，a2，a3 =？）
- 如果您需要不同的曲线属性，则只需创建不同的方程即可。
- li>
[edit3]用法
pic1 = pic0; //复制源图像到目的地pic是我的图像类...
for（y = 0; y
{
float i;
//读，转换，写像素
i = pic1.p [y] [x] .db [0]; i = 255.0 *校正（i / 255.0，红色控制点，5）; pic1.p [y] [x] .db [0] = i;
i = pic1.p [y] [x] .db [1]; i = 255.0 *校正（i / 255.0，绿色控制点，5）; pic1.p [y] [x] .db [1] = i;
i = pic1.p [y] [x] .db [2]; i = 255.0 *校正（i / 255.0，蓝色控制点，5）; pic1.p [y] [x] .db [2] = i;
}
- up是每个R，G，B的控制点
- 左下角是原始图片
- 右下角是正确的图片
I am trying to implement some function like below
For this I am trying to use Cubic interpolation and Catmull interpolation ( check both separately to compare the best result) , what i am not understanding is what impact these interpolation show on image and how we can get these points values where we clicked to set that curve ? and do we need to define the function these black points on the image separately ?
I am getting help from these resources
Edit
int main (int argc, const char** argv)
{
Mat input = imread ("E:\\img2.jpg");
for(int i=0 ; i<input.rows ; i++)
{
for (int p=0;p<input.cols;p++)
{
//for(int t=0; t<input.channels(); t++)
//{
input.at<cv::Vec3b>(i,p)[0] = 255*correction(input.at<cv::Vec3b>(i,p)[0]/255.0,ctrl,N); //B
input.at<cv::Vec3b>(i,p)[1] = 255*correction(input.at<cv::Vec3b>(i,p)[1]/255.0,ctrl,N); //G
input.at<cv::Vec3b>(i,p)[2] = 255*correction(input.at<cv::Vec3b>(i,p)[2]/255.0,ctrl,N); //R
//}
}
}
imshow("image" , input);
waitKey();
}
解决方案
So if your control points are always on the same x coordinate
and linearly dispersed along whole range then you can do it like this:
//---------------------------------------------------------------------------
const int N=5; // number of control points (must be >= 4)
float ctrl[N]= // control points y values initiated with linear function y=x
{ // x value is index*1.0/(N-1)
0.00,
0.25,
0.50,
0.75,
1.00,
};
//---------------------------------------------------------------------------
float correction(float col,float *ctrl,int n)
{
float di=1.0/float(n-1);
int i0,i1,i2,i3;
float t,tt,ttt;
float a0,a1,a2,a3,d1,d2;
// find start control point
col*=float(n-1);
i1=col; col-=i1;
i0=i1-1; if (i0< 0) i0=0;
i2=i1+1; if (i2>=n) i2=n-1;
i3=i1+2; if (i3>=n) i3=n-1;
// compute interpolation coefficients
d1=0.5*(ctrl[i2]-ctrl[i0]);
d2=0.5*(ctrl[i3]-ctrl[i1]);
a0=ctrl[i1];
a1=d1;
a2=(3.0*(ctrl[i2]-ctrl[i1]))-(2.0*d1)-d2;
a3=d1+d2+(2.0*(-ctrl[i2]+ctrl[i1]));
// now interpolate new colro intensity
t=col; tt=t*t; ttt=tt*t;
t=a0+(a1*t)+(a2*tt)+(a3*ttt);
return t;
}
//---------------------------------------------------------------------------
- it uses 4-point 1D interpolation cubic (from that link in my comment above)
to get new color just do this:
new_col = correction(old_col,ctrl,N);this is how it looks:
- the green arrows shows derivation error (always only on start and end point of whole curve)
- it can be corrected by adding 2 more control points
- one before and one after all others ...
[Notes]
- color range is < 0.0 , 1.0 >
- if you need other then just multiply the result and divide the input ...
[edit1] the start/end derivations fixed a little
float correction(float col,float *ctrl,int n)
{
float di=1.0/float(n-1);
int i0,i1,i2,i3;
float t,tt,ttt;
float a0,a1,a2,a3,d1,d2;
// find start control point
col*=float(n-1);
i1=col; col-=i1;
i0=i1-1;
i2=i1+1; if (i2>=n) i2=n-1;
i3=i1+2;
// compute interpolation coefficients
if (i0>=0) d1=0.5*(ctrl[i2]-ctrl[i0]); else d1=ctrl[i2]-ctrl[i1];
if (i3< n) d2=0.5*(ctrl[i3]-ctrl[i1]); else d2=ctrl[i2]-ctrl[i1];
a0=ctrl[i1];
a1=d1;
a2=(3.0*(ctrl[i2]-ctrl[i1]))-(2.0*d1)-d2;
a3=d1+d2+(2.0*(-ctrl[i2]+ctrl[i1]));
// now interpolate new colro intensity
t=col; tt=t*t; ttt=tt*t;
t=a0+(a1*t)+(a2*tt)+(a3*ttt);
return t;
}
[edit2] just some clarification on the coefficients
the are all derived from this conditions:
y(t) = a0 + a1*t + a2*t*t + a3*t*t*t // direct value y'(t) = a1 + 2*a2*t + 3*a3*t*t // first derivationnow you have points y0,y1,y2,y3
- so I chose that y(0)=y1 and y(1)=y2
- which gives c0 continuity (value is the same in the joint points between curves)
- now I need c1 continuity so i add y'(0) must be the same as y'(1) from previous curve
- for y'(0) I choose avg direction between points y0,y1,y2
- for y'(1) I choose avg direction between points y1,y2,y3
- these are the same for the next/previous segments so it is enough
now put it together
y(0) = y0 = a0 + a1*0 + a2*0*0 + a3*0*0*0 y(1) = y1 = a0 + a1*1 + a2*1*1 + a3*1*1*1 y'(0) = 0.5*(y2-y0) = a1 + 2*a2*0 + 3*a3*0*0 y'(1) = 0.5*(y3-y1) = a1 + 2*a2*1 + 3*a3*1*1now solve this system of equtions (a0,a1,a2,a3 = ?)
- and will get what I have in source code above
- if you need different properties of the curve then just make different equations ...
[edit3] usage
pic1=pic0; // copy source image to destination pic is mine image class ...
for (y=0;y<pic1.ys;y++) // go through all pixels
for (x=0;x<pic1.xs;x++)
{
float i;
// read, convert, write pixel
i=pic1.p[y][x].db[0]; i=255.0*correction(i/255.0,red control points,5); pic1.p[y][x].db[0]=i;
i=pic1.p[y][x].db[1]; i=255.0*correction(i/255.0,green control points,5); pic1.p[y][x].db[1]=i;
i=pic1.p[y][x].db[2]; i=255.0*correction(i/255.0,blue control points,5); pic1.p[y][x].db[2]=i;
}
- up are control points per R,G,B
- down left is original image
- down right is corrected image
这篇关于立方和cat色样条对图像的影响的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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