将四边形图像提取为矩形 [英] extracting a quadrilateral image to a rectangle
问题描述
赏金更新
按照丹尼斯的链接，这就是方法使用 threeblindmiceandamonkey 代码:
Following Denis's link, this is how to use the threeblindmiceandamonkey code:
// 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(" %2.2f %2.2f
",p2[0],p2[1]);
}
这提供了将目标中的点转换为源的转换 - 当然您可以反过来做，但能够为混合执行此操作很整洁:
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(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
对于混合，我使用 超级采样 和随机点；即使源和目标区域存在很大差异，它也能很好地工作
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
背景问题
在顶部的图像中，货车侧面的标志没有正对着相机.我想用我拥有的像素尽我所能计算出它的脸是什么样子.
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.
我知道图像中四边形的角坐标，以及目标矩形的大小.
I know the corner coordinates of the quad in the image, and the size of the destination rectangle.
我想象这是一种通过 x 和 y 轴的循环，同时在两个维度上执行 Bresenham 线，并在源图像和目标图像中的像素重叠时进行某种混合 - 某种子像素混合?
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?
有哪些方法，如何混合像素?
What approaches are there, and how do you mix the pixels?
对此有标准方法吗?
推荐答案
查找quad to quad"转换，例如threeblindmiceandamonkey.
二维齐次坐标上的 3x3 变换可以变换任何 4 个点(四边形)到任何其他四边形；相反，任何 fromquad 和 toquad，例如卡车的角和目标矩形，进行 3 x 3 变换.
Look up "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.
Qt 有 quadToQuad并且可以用它转换像素图，但我猜你没有 Qt 吗?
新增 10Jun:来自 labs.trolltech.com/page/Graphics/Examples有一个很好的演示，当你移动角落时，它可以将像素图从四边形变成四边形:
Qt has quadToQuad
and can transform pixmaps with it, but I guess you don't have Qt ?
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:
添加了 11Jun:@Will，这是 Python 中的 translate.h(你知道一点吗?""" ...""" 是多行注释.)perstrans() 是关键；希望这是有道理的，如果不是问.
Added 11Jun: @Will, here's translate.h in Python (which you know a bit ?
""" ...""" are multiline comments.)
perstrans() is the key; hope that makes sense, if not ask.
顺便说一下，你可以将像素一一映射，mapQuadToQuad(target rect, orig quad)，但如果没有像素插值，它看起来会很糟糕；OpenCV 做到了这一切.
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.
#!/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.]]
"""
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