前言
这里基于作为折线的字符轮廓属性提出了一些不错的想法。虽然这种方法存在一些固有缺陷，因为它们是分辨率和比例的函数，但我想提供一个进一步的中断。 我的算法仍然容易受到影响，但它可能会提供不同的视角。
理论
我建议的方法是通过轮廓中的拐点数来比较常见字符。在这种情况下，我所说的拐点是指连续折线段作为向量的叉积之间的符号变化。例如;考虑一个圆的折线轮廓，从中间 y 坐标和 x+ 最坐标开始。如果我们要围绕周界跟踪折线轮廓 CW（顺时针），则每个线段将递增地是先验的 CW 变换。如果在任何时候一个线段“远离”或“向外”，这个变换将是 CCW（逆时针）并且叉积将反转。因此，“粗糙”的圆会有拐点，“完美”或“平滑”的圆没有拐点。
算法
该算法使用Emgu.CV 遵循以下步骤。 C# 下面的代码：
- 图像被加载并通过阈值转换为二进制
- 然后对二值图像进行轮廓检测，这些轮廓按其边界框从左到右排序，因此它们的索引与它们轮廓的字符的出现顺序相匹配。
- 然后将每个轮廓重新指向相同数量的片段，以便标准化图像/字符之间的比例和分辨率差异。
- 每个轮廓都被“遍历”并计算其拐点数。
// [Some basic extensions are omitted for clarity]
// Load the images
Image<Rgb, byte> baseLineImage = new Image<Rgb, byte>("BaseLine.png");
Image<Rgb, byte> testCaseImage = new Image<Rgb, byte>("TestCase.png");
// Convert them to Gray Scale
Image<Gray, byte> baseLineGray = baseLineImage.Convert<Gray, byte>();
Image<Gray, byte> testCaseGray = testCaseImage.Convert<Gray, byte>();
// Threshold the images to binary
Image<Gray, byte> baseLineBinary = baseLineGray.ThresholdBinaryInv(new Gray(100), new Gray(255));
Image<Gray, byte> testCaseBinary = testCaseGray.ThresholdBinaryInv(new Gray(100), new Gray(255));
// Some dilation required on the test image so that the characters are continuous
testCaseBinary = testCaseBinary.Dilate(3);
// Extract the the contours from the images to isolate the character profiles
// and sort them left to right so as the indicies match the character order
VectorOfVectorOfPoint baseLineContours = new VectorOfVectorOfPoint();
Mat baseHierarchy = new Mat();
CvInvoke.FindContours(
baseLineBinary,
baseLineContours,
baseHierarchy,
RetrType.External,
ChainApproxMethod.ChainApproxSimple);
var baseLineContoursList = baseLineContours.ToList();
baseLineContoursList.Sort(new ContourComparer());
VectorOfVectorOfPoint testCaseContours = new VectorOfVectorOfPoint();
Mat testHierarchy = new Mat();
CvInvoke.FindContours(
testCaseBinary,
testCaseContours,
testHierarchy,
RetrType.External,
ChainApproxMethod.ChainApproxSimple);
var testCaseContoursList = testCaseContours.ToList();
testCaseContoursList.Sort(new ContourComparer());
var baseLineRepointedContours = RepointContours(baseLineContoursList, 50);
var testCaseRepointedContours = RepointContours(testCaseContoursList, 50);
var baseLineInflectionCounts = GetContourInflections(baseLineRepointedContours);
var testCaseInflectionCounts = GetContourInflections(testCaseRepointedContours);
拐点检测/计数
static List<List<Point>> GetContourInflections(List<VectorOfPoint> contours)
{
// A resultant list to return the inflection points
List<List<Point>> result = new List<List<Point>>();
// Calculate the forward to reverse cross product at each vertex
List<double> crossProducts;
// Points used to store 2D Vectors as X,Y (I,J)
Point priorVector, forwardVector;
foreach (VectorOfPoint contour in contours)
{
crossProducts = new List<double>();
for (int p = 0; p < contour.Size; p++)
{
// Determine the vector to the prior to this vertex
priorVector = p == 0 ?
priorVector = new Point()
{
X = contour[p].X - contour[contour.Size - 1].X,
Y = contour[p].Y - contour[contour.Size - 1].Y
} :
priorVector = new Point()
{
X = contour[p].X - contour[p - 1].X,
Y = contour[p].Y - contour[p - 1].Y
};
// Determine the vector to the next vector
// If this is the lst vertex, loop back to vertex 0
forwardVector = p == contour.Size - 1 ?
new Point()
{
X = contour[0].X - contour[p].X,
Y = contour[0].Y - contour[p].Y,
} :
new Point()
{
X = contour[p + 1].X - contour[p].X,
Y = contour[p + 1].Y - contour[p].Y,
};
// Calculate the cross product of the prior and forward vectors
crossProducts.Add(forwardVector.X * priorVector.Y - forwardVector.Y * priorVector.X);
}
// Given the calculated cross products, detect the inflection points
List<Point> inflectionPoints = new List<Point>();
for (int p = 1; p < contour.Size; p++)
{
// If there is a sign change between this and the prior cross product, an inflection,
// or change from CW to CCW bearing increments has occurred. To and from zero products
// are ignored
if ((crossProducts[p] > 0 && crossProducts[p-1] < 0) ||
(crossProducts[p] < 0 && crossProducts[p-1] > 0))
{
inflectionPoints.Add(contour[p]);
}
}
result.Add(inflectionPoints);
}
return result;
}
输出
L: Baseline Inflections:0 Testcase Inflections:22
E: Baseline Inflections:1 Testcase Inflections:16
M: Baseline Inflections:4 Testcase Inflections:15
P: Baseline Inflections:11 Testcase Inflections:17
E: Baseline Inflections:1 Testcase Inflections:10
R: Baseline Inflections:9 Testcase Inflections:16
轮廓（蓝色）和变形（红色）