如何确定线条的宽度? [英] How to determine the width of the lines?

问题描述

我需要检测这些线的宽度:

I need to detect the width of these lines:

这些线是平行的，并且在它们上有一些噪声.

These lines are parallel and have some noise on them.

当前，我要做的是:

1.使用细化(ZhangSuen)查找中心

1.Find the center using thinning (ZhangSuen)

ZhanSuenThinning(binImage, thin);

2.计算距离变换

cv::distanceTransform(binImage, distImg, CV_DIST_L2, CV_DIST_MASK_5);

3.累计绕中心的一半距离

3.Accumulate the half distance around the center

double halfWidth = 0.0;
int count = 0;
for(int a = 0; a < thinImg.cols; a++)
    for(int b = 0; b < thinImg.rows; b++)
        if(thinImg.ptr<uchar>(b, a)[0] > 0)
        {
            halfWidth += distImg.ptr<float>(b, a)[0];
            count ++;
        }

4.最后，获取实际宽度

4.Finally, get the actual width

width = halfWidth / count * 2;

结果不是很好，在1-2像素左右是错误的.在更大的Image上，结果甚至更糟，有什么建议吗?

The result, isn't quite good, where it's wrong around 1-2 pixels. On bigger Image, the result is even worse, Any suggestion?

推荐答案

您可以采用条形码阅读器算法，这是实现此目的的更快方法.

You can adapt barcode reader algorithms which is the faster way to do it.

扫描水平和垂直线. 令X为与黑线的水平相交的长度，而Y为与垂直相交的长度的Y(如果有一些噪声，则可以计算多个X和Y的中值).

Scan horizontal and vertical lines. Lets X the length of the horizontal intersection with black line an Y the length of the vertical intersection (you can have it be calculating the median value of several X and Y if there are some noise).

X * Y / 2 = area
X²+Y² = hypotenuse²
hypotenuse * width / 2 = area

所以:宽度= 2 *面积/斜边

So : width = 2 * area / hypotenuse

您也可以使用PCA轻松找到角度.

EDIT : You can also easily find the angle by using PCA.

这篇关于如何确定线条的宽度?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！