在二进制阈值图像中去除边缘连接的小破折号

问题描述

我有一个二进制阈值图像，我想从图像中删除小破折号.连接到圈子.我的主要重点是提取圆弧.

I have a binary threshold image and I want to remove small dashes from an image. which is connected to the circle. My main focus is to extract an arc of the circle.

• 原始图片:

• 将图像输出为:

推荐答案

由于我们具有空心形状，因此我采用了惰性方法.这是一种相对简单的问题的平滑方法(检查从内部到外部的距离)，但是由于它在简单的封闭形状的假设下运行，因此丢失了细节.如果这还不够好，请告诉我；有更复杂的方法可以完成您想要的操作，从而获得更清晰的结果.

I took the lazy approach to this since we have a hollow shape. This is a smoothing approach to the problem which is relatively simple (check the distance from inner to outer), but it loses details since it runs under the assumption of a simple closed shape. If this isn't good enough, let me know; there are more complicated methods for doing what you want that can give cleaner results.

因此，基本步骤如下:首先，使用findContours来获取形状的内层和外层(扩张直到得到两个，我们在这种情况下不必这样做，因为它已经做到了).

So the basic steps are this: First, use findContours to get the inner and outer layer of the shape (dilate until you get two, we didn't have to for this case since it already does that).

然后，计算从每个点到另一个轮廓上最接近点的距离.从图中可以很好地了解我们要在这里做什么.破折号是相对均匀图形的明显异常值.在这里，我将截止值手动设置为10，但是我们可以使用平均值和标准偏差来自动设置截止值.

Then, calculate the distance from the each point to the closest point on the other contour. From the graph you can get a pretty good idea of what we're going for here. The dashes are clear outliers to a relatively uniform graph. Here I manually set the cutoff to 10, but we could use averages and standard deviation to automatically set a cutoff.

一旦我们删除了离群点，我们就可以使用轮廓重新绘制形状.

Once we've removed the outlier points, we can just redraw the shape using the contours.

import cv2
import numpy as np

# returns a smoothed contour
def smoothed(contour, dists, cutoff):
    smooth_con = [];
    for a in range(len(dists)):
        if dists[a] < cutoff:
            smooth_con.append(contour[a]);
    return np.asarray(smooth_con);

# get the distance list for an array of points
def distList(src, other):
    dists = [];
    for point in src:
        point = point[0]; # drop extra brackets
        _, dist = closestPoint(point, other);
        dists.append(dist);
    return dists;

# returns squared distance of two points
def squaredDist(one, two):
    dx = one[0] - two[0];
    dy = one[1] - two[1];
    return dx*dx + dy*dy;

# find closest point (just do a linear search)
def closestPoint(point, arr):
    # init tracker vars
    closest = None;
    best_dist = 999999999;

    # linear search
    for other in arr:
        other = other[0]; # remove extra brackets
        dist = squaredDist(point, other);
        if dist < best_dist:
            closest = other;
            best_dist = dist;
    return closest, best_dist;

# load image
img = cv2.imread("circle_dashed.png");

# make a mask
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY);
mask = cv2.inRange(gray, 0, 100);

# get contours # OpenCV 3.4, if you're using OpenCV 2 or 4, it returns (contours, _)
_, contours, _ = cv2.findContours(mask, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE);
print(len(contours)); # we have two, inner and outer, no need to dilate first

# split
one = contours[0];
two = contours[1];

# get distances
one_dists = distList(one, two);
two_dists = distList(two, one);

# dump values greater than 10
smooth_one = smoothed(one, one_dists, 10);
smooth_two = smoothed(two, two_dists, 10);

# draw new contour
blank = np.zeros_like(mask);
cv2.drawContours(blank, [smooth_one], -1, (255), -1);
cv2.drawContours(blank, [smooth_two], -1, (0), -1);

# show
cv2.imshow("Image", img);
cv2.imshow("Smooth", blank);
cv2.waitKey(0);

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