如何摆脱 bw 图像中的点并留下连接的像素？

Question

我有如下图片（图片全黑，不要介意边框）

我只想留下（1 像素宽）线，并去掉所有点（其中一些实际上是几个像素）。

我知道这很容易，但我现在没有想法:(

我尝试过的：

用于打开的形态学操作（侵蚀 + 扩张）- 让我一无所有或保持原样

blob 检测 - 做 this 可能会在膨胀后工作，但它崩溃没有错误（windows 异常），我没有时间调试 opencv。

喜欢任何可行的想法。

Answer 1

我会使用自定义侵蚀函数来计算 3x3 window 中的非黑色像素，如果有，像素值不变，如果没有，像素将变为黑色。

在 C++ 中它看起来像（编辑：我添加了一个基于模糊的方法，它更快）：

#include <opencv2/opencv.hpp>
#include <chrono>

int main(void)
{
  cv::Mat input = cv::imread("salt.png", cv::IMREAD_GRAYSCALE);
  imshow("input", input);

  cv::Mat thresh;
  // threshold the input just to be sure to get only white or black pixels.
  cv::threshold(input, thresh, 128, 255, cv::THRESH_BINARY);

  cv::Mat mask = cv::Mat::ones(3,3,CV_8UC1);
  mask.at<unsigned char>(1,1)=0;
  cv::imshow("mask", mask);
  cv::Mat output1 = thresh;
  cv::Mat output2 = thresh;

  { // first method : iterate over all pixels (very slow but didactic)
    auto start = std::chrono::high_resolution_clock::now();
    for(int r=1; r<thresh.rows-1; r++)
    {
      for(int c=1; c<thresh.cols-1; c++)
      {
        cv::Rect roi(c-1,r-1,3,3);
        cv::Mat window = thresh(roi);
        cv::Scalar val = cv::sum(window.mul(mask));
        if(val[0]==0)
          output1.at<unsigned char>(r,c) = 0;
      }
    }
    auto end = std::chrono::high_resolution_clock::now();
    auto duration = std::chrono::duration_cast<std::chrono::microseconds>(end-start);
    std::cout << "iterating method: " << duration.count() << std::endl; // 116965µs

    cv::imshow("output", output1);
  }

  { // optimized method : use blur and threshold
    cv::Mat blur, output2=thresh, thresh2;
    auto start = std::chrono::high_resolution_clock::now();

    cv::blur(thresh, blur, cv::Size(3,3));

    cv::threshold(blur, thresh2, 60, 255, cv::THRESH_BINARY);
    output2.mul(thresh2);
    auto end = std::chrono::high_resolution_clock::now();
    auto duration = std::chrono::duration_cast<std::chrono::microseconds>(end-start);
    std::cout << "blur method: " << duration.count() << std::endl; // 378µs

    cv::imshow("blur", output2);
  }

  auto diff = (output1 - output2) * 255;
  std::cout << "diff norm: " << cv::norm(diff) << std::endl; // 0

  while(1)
  {
    if(27 ==cv::waitKey(0))
      break;
  }
  return 0;
}

输出：

Answer 2

我最后做了以下，只是为了完成它。我相信可以做一些更聪明的事情：

    thread_defect_mask_clean = thread_defect_mask_noisy.copy()
    ret, connected_components_labels = cv2.connectedComponents(thread_defect_mask_noisy.astype('uint8'), connectivity=8)
    for label in range(1, ret):
        label_count = np.count_nonzero(label == connected_components_labels)
        if label_count < self._min_thread_defect_size:
            thread_defect_mask_clean[label == connected_components_labels] = 0

这会计算每个连接组件的大小，并强制执行阈值（此处为 5）

Answer 3

[更新] - 没注意到它需要 1x1 像素线

就像你说的 - 我忽略了边框

import cv2
import numpy as np
import matplotlib.pyplot as plt


img = cv2.imread("dumpster/eLIRj.png")
cv2.imwrite("dumpster/masking/1.png", img)

# convert to grayscale (single channel)
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
cv2.imwrite("dumpster/masking/2.png", gray)

# otsu thresholding
otsu = cv2.threshold(
    gray, 0, 255, cv2.THRESH_BINARY + cv2.THRESH_OTSU)[1]
cv2.imwrite("dumpster/masking/3.png", otsu)

# erode with horizontal and vertical lines (2x1 pixels)
kernel = np.ones((2,1), dtype=np.uint8) 
erosion = cv2.erode(otsu, kernel, iterations = 1)
kernel = np.ones((1,2), dtype=np.uint8) 
erosion = erosion + cv2.erode(otsu, kernel, iterations = 1)
cv2.imwrite("dumpster/masking/4.png", erosion)

# dilate whats left with 3x3 kernel
kernel = np.ones((3,3), dtype=np.uint8)
dilated = cv2.dilate(erosion, kernel, iterations=1)
cv2.imwrite("dumpster/masking/5.png", dilated)

# create binary mask and apply it on thresholded, grayscale image
mask = dilated / 255
final = otsu * mask
cv2.imwrite("dumpster/masking/6.png", final)

原图：

膨胀：

决赛：

如何摆脱 bw 图像中的点并留下连接的像素？

How to get rid of points in bw image and be left with connected pixels?

python

opencv

image-processing

image-morphology

cv2