Question

我有一组反映图像上已知位置的坐标。我们称之为模板图像。它有一个独特的条形码和方向标记（也在坐标数组中）。

打印，扫描图像并将其反馈到我的应用程序中进行检测。在打印和扫描期间，图像可以以三种方式转换;翻译，轮换和扩展。

假设我可以在失真图像上找到方向标记，我如何使用矩阵变换来获得剩余坐标的相对位置？

我在SO before上发布了这个问题但却太复杂了，无法理解我想要的内容。

修改

namespace MatrixTest
{
using System;
using System.Drawing;
using System.Drawing.Drawing2D;
using System.Collections.Generic;

public static class Program
{
public static void Main ()
{
Template template = new Template(); // Original template image.
Document document = new Document(); // Printed and scanned distorted image.

template.CreateTemplateImage();

// The template image is printed and scanned. This method generates an example scan or this question.
document.CreateDistortedImageFromTemplateImage();
// Stuck here.
document.Transform();
// Draw transformed points on the image to verify that transformation is successful.
document.DrawPoints();

System.Diagnostics.Process.Start(new System.IO.FileInfo(System.Reflection.Assembly.GetExecutingAssembly().Location).Directory.FullName);
}
}

public class Page
{
public Bitmap Image { get; set; }
public Point[] Markers = new Point[3]; // Orientation markers: 1=TopLeft, 2=TopRight, 3=BottomRight.
public Point[] Points = new Point[100]; // Coordinates to transform in the TemplateScanned derived class!
}

// This class represents the originalk template image.
public class Template: Page
{
public Template ()
{
this.Image = new Bitmap(300, 400);

// Known dimentions for marker rectangles.
this.Markers[0] = new Point(10, 10);
this.Markers[1] = new Point(this.Image.Width - 20 - 10, 10);
this.Markers[2] = new Point(this.Image.Width - 20 - 10, this.Image.Height - 20 - 10);

// Known points of interest. Consider them hardcoded.
int index = 0;
for (int y = 0; y < 10; y++)
for (int x = 0; x < 10; x++)
this.Points[index++] = new Point((this.Image.Width / 10) + (x * 20), (this.Image.Height / 10) + (y * 20));
}

public void CreateTemplateImage ()
{
using (Graphics graphics = Graphics.FromImage(this.Image))
{
graphics.Clear(Color.White);

for (int i = 0; i < this.Markers.Length; i++)
graphics.FillRectangle(Brushes.Black, this.Markers[i].X, this.Markers[i].Y, 20, 20);

for (int i = 0; i < this.Points.Length; i++)
graphics.DrawRectangle(Pens.Red, this.Points[i].X, this.Points[i].Y, 5, 5);
}

this.Image.Save("Document Original.png");
}
}

// This class represents the scanned image.
public class Document: Page
{
public struct StructTransformation
{
public float AngleOfRotation;
public SizeF ScaleRatio;
public SizeF TranslationOffset;
}

private Template Template = new Template();
private StructTransformation Transformation = new StructTransformation();

public Document ()
{
this.Template = new Template();
this.Transformation = new StructTransformation { AngleOfRotation = 5f, ScaleRatio = new SizeF(.8f, .7f), TranslationOffset = new SizeF(100f, 30f) };

this.Template.CreateTemplateImage();

// Copy points from template.
for (int i = 0; i < this.Template.Markers.Length; i++)
this.Markers[i] = this.Template.Markers[i];

for (int i = 0; i < this.Points.Length; i++)
this.Points[i] = this.Template.Points[i];
}

// Just distorts the original template image as if it had been read from a scanner.
public void CreateDistortedImageFromTemplateImage ()
{
// Distort coordinates.
Matrix matrix = new Matrix();
matrix.Rotate(this.Transformation.AngleOfRotation);
matrix.Scale(this.Transformation.ScaleRatio.Width, this.Transformation.ScaleRatio.Height);
matrix.Translate(this.Transformation.TranslationOffset.Width, this.Transformation.TranslationOffset.Height);
matrix.TransformPoints(this.Markers);
matrix.TransformPoints(this.Points);

// Distort and save image for visual reference.
this.Image = new Bitmap(this.Template.Image.Width, this.Template.Image.Height);
using (Graphics graphics = Graphics.FromImage(this.Image))
{
graphics.Clear(Color.White);
graphics.RotateTransform(this.Transformation.AngleOfRotation);
graphics.ScaleTransform(this.Transformation.ScaleRatio.Width, this.Transformation.ScaleRatio.Height);
graphics.TranslateTransform(this.Transformation.TranslationOffset.Width, this.Transformation.TranslationOffset.Height);
graphics.DrawImage(this.Template.Image, 0, 0);
}
this.Image.Save("Document Scanned.png");
}

public void Transform ()
{
// The rectangles of the ScannedDcoument are not known at this time. They would obviously be relative to the three orientation markers.
//    I can't figure out how to use the following code properly i.e. using Matrix to apply all three transformations.
Matrix matrix = new Matrix();
matrix.Rotate(-this.Transformation.AngleOfRotation);
matrix.Scale(1f/this.Transformation.ScaleRatio.Width, 1f/this.Transformation.ScaleRatio.Height);
matrix.Translate(-this.Transformation.TranslationOffset.Width, -this.Transformation.TranslationOffset.Height);
matrix.TransformPoints(this.Markers);
matrix.TransformPoints(this.Points);
}

public void DrawPoints ()
{
using (Graphics graphics = Graphics.FromImage(this.Image))
{
graphics.Clear(Color.White);

for (int i = 0; i < this.Markers.Length; i++)
graphics.FillRectangle(Brushes.Blue, this.Markers[i].X, this.Markers[i].Y, 20, 20);

for (int i = 0; i < this.Points.Length; i++)
graphics.DrawRectangle(Pens.Purple, this.Points[i].X, this.Points[i].Y, 5, 5);
}
this.Image.Save("Document Fixed.png");
}
}
}

Answer 1

我假设您要将图像转换为单位平方（（0,0） - （1.0,1.0））你需要三个点，一个是原点，另一个是x轴（1.0,0），另一个是y轴（0,1.0）。

在原始坐标系中：

起源是（Ox，Oy）
X轴是（X1，Y2）
Y轴是（X2，Y2）
X轴相对于原点（X1-Ox，Y1-Oy）将缩短为（RX1，RY1）
相对于原点的Y轴（X2-ox，Y2-Oy）将缩短为（RX2，RY2）

首先，我们将在齐次坐标中将原点移动到（0,0），变换矩阵将是

(1   0   -Ox)
(0   1   -Oy)
(0   0    1)

从新空间到旧空间的转换由以下矩阵表示：

(RX1   RX2   0)
(RY1   RY2   0)
( 0    0     1)

因为你想要从旧空间到新空间的逆变换，我们需要反转这个矩阵：让我们将（RX1 * RY2-RX2 * RY1）缩短为D

(RY2/D   -RX2/D   0)
(-RY1/D   RX1/D   0)
(  0       0      1)

现在，您可以先将两个矩阵相乘，然后再使用第二个矩阵来转换基础。

C＃中的矩阵/坐标转换

1 个答案: