我正在使用OpenCV来获取捕获帧的鸟瞰图。这是通过在飞机上提供棋盘格图案来完成的,该图案将形成鸟瞰图。
尽管看起来相机已经很漂亮了,但是我需要它完美以确定像素和厘米之间的关系。
在下一阶段,捕获帧将被扭曲。它给出了预期的结果:
但是,通过执行此转换,棋盘图案外部的数据将丢失。我需要的是旋转图像,而不是扭曲已知的四边形。
问题:如何以摄像机 Angular 旋转图像,使其自上而下?
一些代码来说明我当前正在做什么:
Size chessboardSize = new Size(12, 8); // Size of the chessboard
Size captureSize = new Size(1920, 1080); // Size of the captured frames
Size viewSize = new Size((chessboardSize.width / chessboardSize.height) * captureSize.height, captureSize.height); // Size of the view
MatOfPoint2f imageCorners; // Contains the imageCorners obtained in a earlier stage
Mat H; // Homography
找到角落的代码:
Mat grayImage = new Mat();
//Imgproc.resize(source, temp, new Size(source.width(), source.height()));
Imgproc.cvtColor(source, grayImage, Imgproc.COLOR_BGR2GRAY);
Imgproc.threshold(grayImage, grayImage, 0.0, 255.0, Imgproc.THRESH_OTSU);
imageCorners = new MatOfPoint2f();
Imgproc.GaussianBlur(grayImage, grayImage, new Size(5, 5), 5);
boolean found = Calib3d.findChessboardCorners(grayImage, chessboardSize, imageCorners, Calib3d.CALIB_CB_NORMALIZE_IMAGE + Calib3d.CALIB_CB_ADAPTIVE_THRESH + Calib3d.CALIB_CB_FILTER_QUADS);
if (found) {
determineHomography();
}
确定单应性的代码:
Point[] data = imageCorners.toArray();
if (data.length < chessboardSize.area()) {
return;
}
Point[] roi = new Point[] {
data[0 * (int)chessboardSize.width - 0], // Top left
data[1 * (int)chessboardSize.width - 1], // Top right
data[((int)chessboardSize.height - 1) * (int)chessboardSize.width - 0], // Bottom left
data[((int)chessboardSize.height - 0) * (int)chessboardSize.width - 1], // Bottom right
};
Point[] roo = new Point[] {
new Point(0, 0),
new Point(viewSize.width, 0),
new Point(0, viewSize.height),
new Point(viewSize.width, viewSize.height)
};
MatOfPoint2f objectPoints = new MatOfPoint2f(), imagePoints = new MatOfPoint2f();
objectPoints.fromArray(roo);
imagePoints.fromArray(roi);
Mat H = Imgproc.getPerspectiveTransform(imagePoints, objectPoints);
最后,捕获的帧将被扭曲:
Imgproc.warpPerspective(capture, view, H, viewSize);
最佳答案
[Edit2]更新了进度
可能还有更多的变化,所以我可以尝试以下方法:
您可以应用许多滤镜来消除图像中的噪点或使照明条件归一化(看起来您发布的图像不需要它)。然后,只需对图像进行二值化处理即可简化后续步骤。看到相关:
并将其坐标与拓扑存储在某个数组中
double pnt[col][row][2];
其中
(col,row)
是国际象棋棋盘索引,而[2]
存储(x,y)。您可以使用int
,但double/float
将避免在拟合过程中不必要的转换和舍入...通过扫描对角相邻像素,可以检测到拐角(除非倾斜/旋转接近45度),如下所示:
一个对角线应该使用一种颜色,而另一种应该使用不同的颜色。该模式将检测交叉点周围的点簇,因此找到这些点附近并计算其平均值。
如果您扫描整个图像,则
for
循环的上轴也会对点列表进行排序,因此无需进一步排序。在对网格拓扑的点进行平均排序/排序后(例如,按2个最近点之间的方向)为了使其坚固,我使用了旋转和倾斜的图像,因此拓扑检测有些棘手。经过一段时间的阐述,我得出以下结论:
p0
那应该确保在这一点上有邻居。
p
但忽略对角点(
|x/y| -> 1
+/-正方形比例)。从这一点开始,计算第一基 vector ,现在将其称为u
。 p
以与#2 相同的方式,但是这次也忽略了+/- u方向上的点(
|(u.v)|/(|u|.|v|) -> 1
+/-倾斜/旋转)。从这一点开始,计算第二基 vector ,现在将其称为v
。 我选择
u
vector 指向+x
,并且v
指向+y
方向。因此,具有更大|x|
值的基 vector 应为u
,而具有更大|y|
的值应为v
。因此,如有必要,请测试并交换。然后,如果符号错误,则取反。现在我们有了屏幕中间的基本 vector (更远的地方它们可能会改变)。 将
p0
点设置为(u=0,v=0)
作为起点。现在遍历所有尚未匹配的点p
。对于每个计算邻居的预测位置,方法是从其位置添加/减去基本 vector 。然后找到最靠近该位置的点,如果发现它应该是邻居,则将其(u,v)
坐标设置为原始点+/-1
的p
。现在更新这些点的基 vector ,并循环遍历整个过程,直到找不到新的匹配项为止。结果应该是大多数点都应该已经计算出了我们需要的(u,v)
坐标。 之后,您可以找到
min(u),min(v)
并将其移至(0,0)
,以便在需要时索引不为负。 例如:
pnt[i][j][0]=fx(i,j)
pnt[i][j][1]=fy(i,j)
其中
fx,fy
是多项式函数。您可以尝试任何拟合过程。我尝试使用approximation search进行三次多项式拟合,但结果不如本机的双三次插值(可能是由于测试图像的不均匀失真),所以我改用双三次插值而不是拟合。这比较简单,但是使逆运算变得非常困难,但是可以避免这样做,但会降低速度。如果仍然需要计算逆,请参见我正在使用像这样的简单插值三次方:
d1=0.5*(pp[2]-pp[0]);
d2=0.5*(pp[3]-pp[1]);
a0=pp[1];
a1=d1;
a2=(3.0*(pp[2]-pp[1]))-(2.0*d1)-d2;
a3=d1+d2+(2.0*(-pp[2]+pp[1])); }
coordinate = a0+(a1*t)+(a2*t*t)+(a3*t*t*t);
其中
pp[0..3]
是4个相应的已知控制点(我们的网格交叉),a0..a3
是计算的多项式系数,coordinate
是参数t
的曲线上的点。可以扩展到任意数量的尺寸。这条曲线的属性很简单,它是连续的,从
pp[1]
开始,到pp[2]
结束于t=<0.0,1.0>
。通过所有三次曲线共有的序列确保与相邻线段的连续性。 只需将
i,j
用作 float 值,步距约为像素大小的75%,以避免出现间隙。然后只需循环遍历所有位置(i,j)
计算(x,y)
并将(x,y)
中的源图像中的像素从ozt_code复制到(i*sz,j*sz)+/-offset
,其中sz
需要以像素为单位的网格大小。 这是 C++ :
//---------------------------------------------------------------------------
picture pic0,pic1; // pic0 - original input image,pic1 output
//---------------------------------------------------------------------------
struct _pnt
{
int x,y,n;
int ux,uy,vx,vy;
_pnt(){};
_pnt(_pnt& a){ *this=a; };
~_pnt(){};
_pnt* operator = (const _pnt *a) { x=a->x; y=a->y; return this; };
//_pnt* operator = (const _pnt &a) { ...copy... return this; };
};
//---------------------------------------------------------------------------
void vision()
{
pic1=pic0; // copy input image pic0 to pic1
pic1.enhance_range(); // maximize dynamic range of all channels
pic1.treshold_AND(0,127,255,0); // binarize (remove gray shades)
pic1&=0x00FFFFFF; // clear alpha channel for exact color matching
pic1.save("out_binarised.png");
int i0,i,j,k,l,x,y,u,v,ux,uy,ul,vx,vy,vl;
int qi[4],ql[4],e,us,vs,**uv;
_pnt *p,*q,p0;
List<_pnt> pnt;
// detect square crossings point clouds into pnt[]
pnt.allocate(512); pnt.num=0;
p0.ux=0; p0.uy=0; p0.vx=0; p0.vy=0;
for (p0.n=1,p0.y=2;p0.y<pic1.ys-2;p0.y++) // sorted by y axis, each point has usage n=1
for ( p0.x=2;p0.x<pic1.xs-2;p0.x++)
if (pic1.p[p0.y-2][p0.x+2].dd==pic1.p[p0.y+2][p0.x-2].dd)
if (pic1.p[p0.y-1][p0.x+1].dd==pic1.p[p0.y+1][p0.x-1].dd)
if (pic1.p[p0.y-1][p0.x+1].dd!=pic1.p[p0.y+1][p0.x+1].dd)
if (pic1.p[p0.y-1][p0.x-1].dd==pic1.p[p0.y+1][p0.x+1].dd)
if (pic1.p[p0.y-2][p0.x-2].dd==pic1.p[p0.y+2][p0.x+2].dd)
pnt.add(p0);
// merge close points (deleted point has n=0)
for (p=pnt.dat,i=0;i<pnt.num;i++,p++)
if (p->n) // skip deleted points
for (p0=*p,j=i+1,q=p+1;j<pnt.num;j++,q++) // scan all remaining points
if (q->n) // skip deleted points
{
if (q->y>p0.y+4) continue; // scan only up do y distance <=4 (clods are not bigger then that)
x=p0.x-q->x; x*=x; // compute distance^2
y=p0.y-q->y; y*=y; x+=y;
if (x>25) continue; // skip too distant points
p->x+=q->x; // add coordinates (average)
p->y+=q->y;
p->n++; // increase ussage
q->n=0; // mark current point as deleted
}
// divide the average coordinates and delete marked points
for (p=pnt.dat,i=0,j=0;i<pnt.num;i++,p++)
if (p->n) // skip deleted points
{
p->x/=p->n;
p->y/=p->n;
p->n=1;
pnt.dat[j]=*p; j++;
} pnt.num=j;
// n is now encoded (u,v) so set it as unmatched (u,v) first
#define uv2n(u,v) ((((v+32768)&65535)<<16)|((u+32768)&65535))
#define n2uv(n) { u=n&65535; u-=32768; v=(n>>16)&65535; v-=32768; }
for (p=pnt.dat,i=0;i<pnt.num;i++,p++) p->n=0;
// p0,i0 find point near middle of image
x=pic1.xs>>2;
y=pic1.ys>>2;
for (p=pnt.dat,i=0;i<pnt.num;i++,p++)
if ((p->x>=x)&&(p->x<=x+x+x)
&&(p->y>=y)&&(p->y<=y+y+y)) break;
p0=*p; i0=i;
// q,j find closest point to p0
vl=pic1.xs+pic1.ys; k=0;
for (p=pnt.dat,i=0;i<pnt.num;i++,p++)
if (i!=i0)
{
x=p->x-p0.x;
y=p->y-p0.y;
l=sqrt((x*x)+(y*y));
if (abs(abs(x)-abs(y))*5<l) continue; // ignore diagonals
if (l<=vl) { k=i; vl=l; } // remember smallest distance
}
q=pnt.dat+k; j=k;
ux=q->x-p0.x;
uy=q->y-p0.y;
ul=sqrt((ux*ux)+(uy*uy));
// q,k find closest point to p0 not in u direction
vl=pic1.xs+pic1.ys; k=0;
for (p=pnt.dat,i=0;i<pnt.num;i++,p++)
if (i!=i0)
{
x=p->x-p0.x;
y=p->y-p0.y;
l=sqrt((x*x)+(y*y));
if (abs(abs(x)-abs(y))*5<l) continue; // ignore diagonals
if (abs((100*ux*y)/((x*uy)+1))>75) continue;// ignore paralel to u directions
if (l<=vl) { k=i; vl=l; } // remember smallest distance
}
q=pnt.dat+k;
vx=q->x-p0.x;
vy=q->y-p0.y;
vl=sqrt((vx*vx)+(vy*vy));
// normalize directions u -> +x, v -> +y
if (abs(ux)<abs(vx))
{
x=j ; j =k ; k =x;
x=ux; ux=vx; vx=x;
x=uy; uy=vy; vy=x;
x=ul; ul=vl; vl=x;
}
if (abs(vy)<abs(uy))
{
x=ux; ux=vx; vx=x;
x=uy; uy=vy; vy=x;
x=ul; ul=vl; vl=x;
}
x=1; y=1;
if (ux<0) { ux=-ux; uy=-uy; x=-x; }
if (vy<0) { vx=-vx; vy=-vy; y=-y; }
// set (u,v) encoded in n for already found points
p0.n=uv2n(0,0); // middle point
p0.ux=ux; p0.uy=uy;
p0.vx=vx; p0.vy=vy;
pnt.dat[i0]=p0;
p=pnt.dat+j; // p0 +/- u basis vector
p->n=uv2n(x,0);
p->ux=ux; p->uy=uy;
p->vx=vx; p->vy=vy;
p=pnt.dat+k; // p0 +/- v basis vector
p->n=uv2n(0,y);
p->ux=ux; p->uy=uy;
p->vx=vx; p->vy=vy;
// qi[k],ql[k] find closest point to p0
#define find_neighbor \
for (ql[k]=0x7FFFFFFF,qi[k]=-1,q=pnt.dat,j=0;j<pnt.num;j++,q++) \
{ \
x=q->x-p0.x; \
y=q->y-p0.y; \
l=(x*x)+(y*y); \
if (ql[k]>=l) { ql[k]=l; qi[k]=j; } \
}
// process all matched points
for (e=1;e;)
for (e=0,p=pnt.dat,i=0;i<pnt.num;i++,p++)
if (p->n)
{
// prepare variables
ul=(p->ux*p->ux)+(p->uy*p->uy);
vl=(p->vx*p->vx)+(p->vy*p->vy);
// find neighbors near predicted position p0
k=0; p0.x=p->x-p->ux; p0.y=p->y-p->uy; find_neighbor; if (ql[k]<<1>ul) qi[k]=-1; // u-1,v
k++; p0.x=p->x+p->ux; p0.y=p->y+p->uy; find_neighbor; if (ql[k]<<1>ul) qi[k]=-1; // u+1,v
k++; p0.x=p->x-p->vx; p0.y=p->y-p->vy; find_neighbor; if (ql[k]<<1>vl) qi[k]=-1; // u,v-1
k++; p0.x=p->x+p->vx; p0.y=p->y+p->vy; find_neighbor; if (ql[k]<<1>vl) qi[k]=-1; // u,v+1
// update local u,v basis vectors for found points (and remember them)
n2uv(p->n); ux=p->ux; uy=p->uy; vx=p->vx; vy=p->vy;
k=0; if (qi[k]>=0) { q=pnt.dat+qi[k]; if (!q->n) { e=1; q->n=uv2n(u-1,v); q->ux=-(q->x-p->x); q->uy=-(q->y-p->y); } ux=q->ux; uy=q->uy; }
k++; if (qi[k]>=0) { q=pnt.dat+qi[k]; if (!q->n) { e=1; q->n=uv2n(u+1,v); q->ux=+(q->x-p->x); q->uy=+(q->y-p->y); } ux=q->ux; uy=q->uy; }
k++; if (qi[k]>=0) { q=pnt.dat+qi[k]; if (!q->n) { e=1; q->n=uv2n(u,v-1); q->vx=-(q->x-p->x); q->vy=-(q->y-p->y); } vx=q->vx; vy=q->vy; }
k++; if (qi[k]>=0) { q=pnt.dat+qi[k]; if (!q->n) { e=1; q->n=uv2n(u,v+1); q->vx=+(q->x-p->x); q->vy=+(q->y-p->y); } vx=q->vx; vy=q->vy; }
// copy remembered local u,v basis vectors to points where are those missing
k=0; if (qi[k]>=0) { q=pnt.dat+qi[k]; if (!q->vy) { q->vx=vx; q->vy=vy; }}
k++; if (qi[k]>=0) { q=pnt.dat+qi[k]; if (!q->vy) { q->vx=vx; q->vy=vy; }}
k++; if (qi[k]>=0) { q=pnt.dat+qi[k]; if (!q->ux) { q->ux=ux; q->uy=uy; }}
k++; if (qi[k]>=0) { q=pnt.dat+qi[k]; if (!q->ux) { q->ux=ux; q->uy=uy; }}
}
// find min,max (u,v)
ux=0; uy=0; vx=0; vy=0;
for (p=pnt.dat,i=0;i<pnt.num;i++,p++)
if (p->n)
{
n2uv(p->n);
if (ux>u) ux=u;
if (vx>v) vx=v;
if (uy<u) uy=u;
if (vy<v) vy=v;
}
// normalize (u,v)+enlarge and create topology table
us=uy-ux+1;
vs=vy-vx+1;
uv=new int*[us];
for (u=0;u<us;u++) uv[u]=new int[vs];
for (u=0;u<us;u++)
for (v=0;v<vs;v++)
uv[u][v]=-1;
for (p=pnt.dat,i=0;i<pnt.num;i++,p++)
if (p->n)
{
n2uv(p->n);
u-=ux; v-=vx;
p->n=uv2n(u,v);
uv[u][v]=i;
}
// bi-cubic interpolation
double a0,a1,a2,a3,d1,d2,pp[4],qx[4],qy[4],t,fu,fv,fx,fy;
// compute cubic curve coefficients a0..a3 from 1D points pp[0..3]
#define cubic_init { d1=0.5*(pp[2]-pp[0]); d2=0.5*(pp[3]-pp[1]); a0=pp[1]; a1=d1; a2=(3.0*(pp[2]-pp[1]))-(2.0*d1)-d2; a3=d1+d2+(2.0*(-pp[2]+pp[1])); }
// compute cubic curve cordinates =f(t)
#define cubic_xy (a0+(a1*t)+(a2*t*t)+(a3*t*t*t));
// safe access to grid (u,v) point copies it to p0
// points utside grid are computed by mirroring
#define point_uv(u,v) \
{ \
if ((u>=0)&&(u<us)&&(v>=0)&&(v<vs)) p0=pnt.dat[uv[u][v]]; \
else{ \
int uu=u,vv=v; \
if (uu<0) uu=0; \
if (uu>=us) uu=us-1; \
if (vv<0) vv=0; \
if (vv>=vs) vv=vs-1; \
p0=pnt.dat[uv[uu][vv]]; \
uu=u-uu; vv=v-vv; \
p0.x+=(uu*p0.ux)+(vv*p0.vx); \
p0.y+=(uu*p0.uy)+(vv*p0.vy); \
} \
}
//----------------------------------------
//--- Debug draws: -----------------------
//----------------------------------------
// debug recolor white to gray to emphasize debug render
pic1.recolor(0x00FFFFFF,0x00404040);
// debug draw basis vectors
for (p=pnt.dat,i=0;i<pnt.num;i++,p++)
{
pic1.bmp->Canvas->Pen->Color=clRed;
pic1.bmp->Canvas->Pen->Width=1;
pic1.bmp->Canvas->MoveTo(p->x,p->y);
pic1.bmp->Canvas->LineTo(p->x+p->ux,p->y+p->uy);
pic1.bmp->Canvas->Pen->Color=clBlue;
pic1.bmp->Canvas->MoveTo(p->x,p->y);
pic1.bmp->Canvas->LineTo(p->x+p->vx,p->y+p->vy);
pic1.bmp->Canvas->Pen->Width=1;
}
// debug draw crossings
AnsiString s;
pic1.bmp->Canvas->Font->Height=12;
pic1.bmp->Canvas->Brush->Style=bsClear;
for (p=pnt.dat,i=0;i<pnt.num;i++,p++)
{
n2uv(p->n);
if (p->n)
{
pic1.bmp->Canvas->Font->Color=clWhite;
s=AnsiString().sprintf("%i,%i",u,v);
}
else{
pic1.bmp->Canvas->Font->Color=clGray;
s=i;
}
x=p->x-(pic1.bmp->Canvas->TextWidth(s)>>1);
y=p->y-(pic1.bmp->Canvas->TextHeight(s)>>1);
pic1.bmp->Canvas->TextOutA(x,y,s);
}
pic1.bmp->Canvas->Brush->Style=bsSolid;
pic1.save("out_topology.png");
// debug draw of bi-cubic interpolation fit/coveradge with half square step
pic1=pic0;
pic1.treshold_AND(0,200,0x40,0); // binarize (remove gray shades)
pic1.bmp->Canvas->Pen->Color=clAqua;
pic1.bmp->Canvas->Brush->Color=clBlue;
for (fu=-1;fu<double(us)+0.01;fu+=0.5)
for (fv=-1;fv<double(vs)+0.01;fv+=0.5)
{
u=floor(fu);
v=floor(fv);
// 4x cubic curve in v direction
t=fv-double(v);
for (i=0;i<4;i++)
{
point_uv(u-1+i,v-1); pp[0]=p0.x;
point_uv(u-1+i,v+0); pp[1]=p0.x;
point_uv(u-1+i,v+1); pp[2]=p0.x;
point_uv(u-1+i,v+2); pp[3]=p0.x;
cubic_init; qx[i]=cubic_xy;
point_uv(u-1+i,v-1); pp[0]=p0.y;
point_uv(u-1+i,v+0); pp[1]=p0.y;
point_uv(u-1+i,v+1); pp[2]=p0.y;
point_uv(u-1+i,v+2); pp[3]=p0.y;
cubic_init; qy[i]=cubic_xy;
}
// 1x cubic curve in u direction on the resulting 4 points
t=fu-double(u);
for (i=0;i<4;i++) pp[i]=qx[i]; cubic_init; fx=cubic_xy;
for (i=0;i<4;i++) pp[i]=qy[i]; cubic_init; fy=cubic_xy;
t=1.0;
pic1.bmp->Canvas->Ellipse(fx-t,fy-t,fx+t,fy+t);
}
pic1.save("out_fit.png");
// linearizing of original image
DWORD col;
double grid_size=32.0; // linear grid square size in pixels
double grid_step=0.01; // u,v step <= 1 pixel
pic1.resize((us+1)*grid_size,(vs+1)*grid_size); // resize target image
pic1.clear(0); // clear target image
for (fu=-1;fu<double(us)+0.01;fu+=grid_step) // copy/transform source image to target
for (fv=-1;fv<double(vs)+0.01;fv+=grid_step)
{
u=floor(fu);
v=floor(fv);
// 4x cubic curve in v direction
t=fv-double(v);
for (i=0;i<4;i++)
{
point_uv(u-1+i,v-1); pp[0]=p0.x;
point_uv(u-1+i,v+0); pp[1]=p0.x;
point_uv(u-1+i,v+1); pp[2]=p0.x;
point_uv(u-1+i,v+2); pp[3]=p0.x;
cubic_init; qx[i]=cubic_xy;
point_uv(u-1+i,v-1); pp[0]=p0.y;
point_uv(u-1+i,v+0); pp[1]=p0.y;
point_uv(u-1+i,v+1); pp[2]=p0.y;
point_uv(u-1+i,v+2); pp[3]=p0.y;
cubic_init; qy[i]=cubic_xy;
}
// 1x cubic curve in u direction on the resulting 4 points
t=fu-double(u);
for (i=0;i<4;i++) pp[i]=qx[i]; cubic_init; fx=cubic_xy; x=fx;
for (i=0;i<4;i++) pp[i]=qy[i]; cubic_init; fy=cubic_xy; y=fy;
// here (x,y) contains source image coordinates coresponding to grid (fu,fv) so copy it to col
col=0; if ((x>=0)&&(x<pic0.xs)&&(y>=0)&&(y<pic0.ys)) col=pic0.p[y][x].dd;
// compute liner image coordinates (x,y) by scaling (fu,fv)
fx=(fu+1.0)*grid_size; x=fx;
fy=(fv+1.0)*grid_size; y=fy;
// copy col to it
if ((x>=0)&&(x<pic1.xs)&&(y>=0)&&(y<pic1.ys)) pic1.p[y][x].dd=col;
}
pic1.save("out_linear.png");
// release memory and cleanup macros
for (u=0;u<us;u++) delete[] uv[u]; delete[] uv;
#undef uv2n
#undef n2uv
#undef find_neighbor
#undef cubic_init
#undef cubic_xy
#undef point_uv(u,v)
}
//---------------------------------------------------------------------------
抱歉,我知道很多代码,但至少我已经尽我所能评论了。为了简化和易于理解,未对代码进行优化,最终的图像线性化可以更快地编写。我也手动在那部分代码中选择了
grid_size
和grid_step
。应该从图像和已知的物理属性计算得出。我将自己的
picture
类用于图像,因此一些成员是:xs,ys
图像大小(以像素为单位)p[y][x].dd
是(x,y)
位置的像素,为32位整数类型clear(color)
-清除整个图像resize(xs,ys)
-将图像调整为新分辨率bmp
-具有 Canvas 访问权限的VCL封装的GDI位图我还使用了我的动态列表模板,因此:
List<double> xxx;
与double xxx[];
相同xxx.add(5);
将5
添加到列表的末尾xxx[7]
访问数组元素(安全)xxx.dat[7]
访问数组元素(不安全但快速直接访问)xxx.num
是数组xxx.reset()
清除数组并设置xxx.num = 0 xxx.allocate(100)
为100
项目预分配空间这是子结果输出图像。为了使内容更健壮,我将输入图像更改为更失真的图像:
为了使外观更悦目,我将白色重新着色为灰色。 红色行是本地
u
的基础,而蓝色是v
本地的 vector 。白色2D vector 数字是拓扑(u,v)
的坐标,而灰色标量数字是pnt[]
中的拓扑的交叉索引,但点不匹配。[说明]
这种方法不适用于45度左右的旋转。在这种情况下,您需要将交叉检测从交叉模式更改为正模式,并且拓扑条件和方程式也会有所变化。更不用说u,v方向选择。
关于java - OpenCV鸟瞰 View 而不会丢失数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39316126/