将四边形图像提取到矩形 [英] extracting a quadrilateral image to a rectangle
问题描述
BOUNTY UPDATE
/目的地rect是我们的'in'quad
int dw = 300，dh = 250;
double [4] [4] = {{0,0}，{dw，0}，{dw，dh}，{0，dh}};
//源图像中的四元组是我们的'out'
double out [4] [5] = {{171,72}，{331,93}，{333,188}，{177,210} };
double homo [3] [6];
const int ret = mapQuadToQuad（in，out，homo）;
// homo可用于计算源中任何目标点
//的x，y，例如
for（int i = 0; i <4; i ++）{
double p1 [3] = {out [i] [0]，out [i] [7]，1};
double p2 [3];
transformMatrix（p1，p2，homo）;
p2 [0] / = p2 [2]; // x
p2 [1] / = p2 [2]; // y
printf（\t％2.2f\t％2.2f\\\
，p2 [0]，p2 [1]）;
}
这提供了一个转换，用于将目标中的点转换为源 - 您可以当然也可以用另一种方式来做，但是能够为混合做到这一点是很整洁的：
for（int y = 0; y
//目标像素的四个角，并混合
对于混音，我使用与随机点;即使在来源和目的地区域存在较大差异的情况下，它也能很好地发挥作用。 背景问题 在顶部的图像中，厢式货车侧面的标志并非正面对着相机。我想计算一下，尽可能使用我有的像素，看起来像是在面对什么。 我知道图像中四边形的角坐标，和目标矩形的大小。 我想象这是通过x轴和y轴的某种循环，同时在两个维度上同时执行Bresenham的行像源图像和目标图像中的像素重叠的混合类型 - 某种子像素混合类型？ 有什么方法，以及如何混合像素？ 有没有一个标准的方法呢？ Qt有 quadToQuad 新增11Jun：@ Will，这里是翻译。 （你知道一点？ 由此，您可以逐一映射像素，mapQuadToQuad（目标矩形，原始四边形）， BOUNTY UPDATE Following Denis's link, this is how to use the threeblindmiceandamonkey code: This provides a transform for converting points in destination to the source - you can of course do it the other way around, but it's tidy to be able to do this for the mixing: For the mixing, I'm using super-sampling with random points; it works very well, even when there is a big disparity in the source and destination area BACKGROUND QUESTION In the image at the top, the sign on the side of the van is not face-on to the camera. I want to calculate, as best I can with the pixels I have, what it'd look like face on. I know the corner coordinates of the quad in the image, and the size of the destination rectangle. I imagine that this is some kind of loop through the x and y axis doing a Bresenham's line on both dimensions at once with some kind of mixing as pixels in the source and destination images overlap - some sub-pixel mixing of some sort? What approaches are there, and how do you mix the pixels? Is there a standard approach for this? Look up Qt has quadToQuad
and can transform pixmaps with it, but I guess you don't have Qt ? Added 11Jun: @Will, here's translate.h in Python (which you know a bit ?
""" ...""" are multiline comments.) Bytheway, you could map the pixels one by one, mapQuadToQuad( target rect, orig quad ),
but without pixel interpolation it'll look terrible; OpenCV does it all.
这篇关于将四边形图像提取到矩形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！
quad to quadtransform ，例如
threeblindmiceandamonkey 。
在2d齐次坐标上进行3x3变换可以将任何4点（四元）
转换为任何其他四元组;
相反，任何fromquad和toquad，比如你的卡车的角落和一个目标矩形，
给出一个3×3的变换。
，并且可以用它转换pixmaps，但我猜您没有Qt？
从添加了10Jun：
labs.trolltech.com/page/Graphics/Examples
有一个很好的演示，当你移动角落时，它会四角形地显示一个像素图：
< h2>
...是多行注释。）
perstrans（）是关键;希望这是有道理的，如果不问的话。
但不包含像素插值它看起来很糟糕; OpenCV完成了这一切。
＃！/ usr / bin / env python
square< - > ; quad maps
from http://threeblindmiceandamonkey.com/?p=16 matrix.h
$ b $ from __future__ import division
import numpy as np
__date__ =2010-06-11 jun denis
def det2（a，b，c，d）：
返回a * d - b * c
def mapSquareToQuad（quad）：＃[4] [2]
SQ = np.zeros（（3,3））
px = quad [0,0] - quad [1,0] + quad [2,0] - quad [3,0]
py = quad [0,1] - quad [1,1] + quad [2,1] - quad [ 3,1]
如果abs（px）< 1e-10和abs（py） 1e-10：
SQ [0,0] = quad [1,0] - quad [0,0]
SQ [1,0] = quad [2,0] - quad [1， 0]
SQ [2,0] = quad [0,0]
SQ [0,1] = quad [1,1] - quad [0,1]
SQ [1] ，1] = quad [2,1] - quad [1,1]
SQ [2,1] = quad [0,1]
SQ [0,2] = 0.
SQ [1,2] = 0.
SQ [2,2] = 1.
返回SQ
else：
dx1 = quad [1,0] - quad [ 2,0]
dx2 = quad [3,0] - quad [2,0]
dy1 = quad [1,1] - quad [2,1]
dy2 = quad [ 3,1] - quad [2,1]
det = det2（dx1，dx2，dy1，dy2）
if det == 0 .:
return None
SQ [ 0] = det2（px，dx2，py，dy2）/ det
SQ [1,2] = det2（dx1，px，dy1，py）/ det
SQ [2,2] = 1.
SQ [0,0] = quad [1,0] - quad [0,0] + SQ [0,2] * quad [1,0]
SQ [1,0 ] = quad [3,0] - quad [0,0] + SQ [1,2] * quad [3,0]
SQ [2,0] = quad [0,0]
SQ [0,1] = quad [1,1] - quad [0,1] + SQ [0,2] * quad [1,1]
SQ [1,1] = quad [3,1] - quad [0,1] + SQ [1,2] * quad [3,1]
SQ [2,1] = quad [0,1]
返回SQ
＃................................... ............................................
def mapQuadToSquare （quad）：
return np.linalg.inv（mapSquareToQuad（quad））
def mapQuadToQuad（a，b）：
return np.dot（mapQuadToSquare（a）， mapSquareToQuad（b））
def perstrans（X，t）：
透视变换X Nx2，t 3x3：
[x0 y0 1] t = [a0 b0 w0] - > [a0 / w0 b0 / w0]
[x1 y1 1] t = [a1 b1 w1] - > [a1 / w1 b1 / w1]
...
x1 = np.vstack（（XT，np.ones（len（X））））
y = np.dot（tT，x1）
return（y [： - 1] / y [-1]）.T
＃........... .................................................. ..................
if __name__ ==__main__：
np.set_printoptions（2，threshold = 100，suppress = True）＃。 2f
sq = np.array（[[0,0]，[1,0]，[1,1]，[0,1]]）
quad = np.array （[[171,72]，[331,93]，[333,188]，[177,210]]）
printquad：，quad
printsquare to quad：， perstrans（sq，mapSquareToQuad（quad））
printquad to square：，perstrans（quad，mapQuadToSquare（quad））
dw，dh = 300，250
rect = np.array（[[0,0]，[dw，0]，[dw，dh]，[0，dh]]）
quadquad = mapQuadToQuad（quad，rect）
printquad四元变换：，quadquad
printquad to rect：，perstrans（quad，quadquad）
quad：[[171 72]
[331 93 ]
[333 188]
[177 210]]
square to quad：[[171. 72.]
[331. 93.]
[333. 188.]
[177. 210.]]
四方到四方：[[-0。 0]
[1. 0]
[1.1]
[0.1]]
quad到四元变换：[[1.29 -0.23 -0 。 ]
[-0.06 1.79 -0。 ]
[-217.24 -88.54 1.34]] $ b $四个矩形到矩形：[[0. 0.]
[300. 0]
[300. 250.]
[0. 250.]]
// the destination rect is our 'in' quad
int dw = 300, dh = 250;
double in[4][4] = {{0,0},{dw,0},{dw,dh},{0,dh}};
// the quad in the source image is our 'out'
double out[4][5] = {{171,72},{331,93},{333,188},{177,210}};
double homo[3][6];
const int ret = mapQuadToQuad(in,out,homo);
// homo can be used for calculating the x,y of any destination point
// in the source, e.g.
for(int i=0; i<4; i++) {
double p1[3] = {out[i][0],out[i][7],1};
double p2[3];
transformMatrix(p1,p2,homo);
p2[0] /= p2[2]; // x
p2[1] /= p2[2]; // y
printf("\t%2.2f\t%2.2f\n",p2[0],p2[1]);
}
for(int y=0; y<dh; y++) {
for(int x=0; x<dw; x++) {
// calc the four corners in source for this
// destination pixel, and mix
"quad to quad" transform, e.g.
threeblindmiceandamonkey.
A 3x3 transform on 2d homogeneous coordinates can transform any 4 points (a quad)
to any other quad;
conversely, any fromquad and toquad, such as the corners of your truck and a target rectangle,
give a 3 x 3 transform.
Added 10Jun:
from labs.trolltech.com/page/Graphics/Examples
there's a nice demo which quad-to-quads a pixmap as you move the corners:
perstrans() is the key; hope that makes sense, if not ask. #!/usr/bin/env python
""" square <-> quad maps
from http://threeblindmiceandamonkey.com/?p=16 matrix.h
"""
from __future__ import division
import numpy as np
__date__ = "2010-06-11 jun denis"
def det2(a, b, c, d):
return a*d - b*c
def mapSquareToQuad( quad ): # [4][2]
SQ = np.zeros((3,3))
px = quad[0,0] - quad[1,0] + quad[2,0] - quad[3,0]
py = quad[0,1] - quad[1,1] + quad[2,1] - quad[3,1]
if abs(px) < 1e-10 and abs(py) < 1e-10:
SQ[0,0] = quad[1,0] - quad[0,0]
SQ[1,0] = quad[2,0] - quad[1,0]
SQ[2,0] = quad[0,0]
SQ[0,1] = quad[1,1] - quad[0,1]
SQ[1,1] = quad[2,1] - quad[1,1]
SQ[2,1] = quad[0,1]
SQ[0,2] = 0.
SQ[1,2] = 0.
SQ[2,2] = 1.
return SQ
else:
dx1 = quad[1,0] - quad[2,0]
dx2 = quad[3,0] - quad[2,0]
dy1 = quad[1,1] - quad[2,1]
dy2 = quad[3,1] - quad[2,1]
det = det2(dx1,dx2, dy1,dy2)
if det == 0.:
return None
SQ[0,2] = det2(px,dx2, py,dy2) / det
SQ[1,2] = det2(dx1,px, dy1,py) / det
SQ[2,2] = 1.
SQ[0,0] = quad[1,0] - quad[0,0] + SQ[0,2]*quad[1,0]
SQ[1,0] = quad[3,0] - quad[0,0] + SQ[1,2]*quad[3,0]
SQ[2,0] = quad[0,0]
SQ[0,1] = quad[1,1] - quad[0,1] + SQ[0,2]*quad[1,1]
SQ[1,1] = quad[3,1] - quad[0,1] + SQ[1,2]*quad[3,1]
SQ[2,1] = quad[0,1]
return SQ
#...............................................................................
def mapQuadToSquare( quad ):
return np.linalg.inv( mapSquareToQuad( quad ))
def mapQuadToQuad( a, b ):
return np.dot( mapQuadToSquare(a), mapSquareToQuad(b) )
def perstrans( X, t ):
""" perspective transform X Nx2, t 3x3:
[x0 y0 1] t = [a0 b0 w0] -> [a0/w0 b0/w0]
[x1 y1 1] t = [a1 b1 w1] -> [a1/w1 b1/w1]
...
"""
x1 = np.vstack(( X.T, np.ones(len(X)) ))
y = np.dot( t.T, x1 )
return (y[:-1] / y[-1]) .T
#...............................................................................
if __name__ == "__main__":
np.set_printoptions( 2, threshold=100, suppress=True ) # .2f
sq = np.array([[0,0], [1,0], [1,1], [0,1]])
quad = np.array([[171, 72], [331, 93], [333, 188], [177, 210]])
print "quad:", quad
print "square to quad:", perstrans( sq, mapSquareToQuad(quad) )
print "quad to square:", perstrans( quad, mapQuadToSquare(quad) )
dw, dh = 300, 250
rect = np.array([[0, 0], [dw, 0], [dw, dh], [0, dh]])
quadquad = mapQuadToQuad( quad, rect )
print "quad to quad transform:", quadquad
print "quad to rect:", perstrans( quad, quadquad )
"""
quad: [[171 72]
[331 93]
[333 188]
[177 210]]
square to quad: [[ 171. 72.]
[ 331. 93.]
[ 333. 188.]
[ 177. 210.]]
quad to square: [[-0. 0.]
[ 1. 0.]
[ 1. 1.]
[ 0. 1.]]
quad to quad transform: [[ 1.29 -0.23 -0. ]
[ -0.06 1.79 -0. ]
[-217.24 -88.54 1.34]]
quad to rect: [[ 0. 0.]
[ 300. 0.]
[ 300. 250.]
[ 0. 250.]]
"""
