python - 如何可靠地检测条码的 4 个角？

Question

我正在尝试检测这个 Code128使用 Python + zbar 模块的条形码:

(图片下载链接here)

这个有效:

import cv2, numpy
import zbar
from PIL import Image 
import matplotlib.pyplot as plt

scanner = zbar.ImageScanner()
pil = Image.open("000.jpg").convert('L')
width, height = pil.size    
plt.imshow(pil); plt.show()
image = zbar.Image(width, height, 'Y800', pil.tobytes())
result = scanner.scan(image)

for symbol in image:
    print symbol.data, symbol.type, symbol.quality, symbol.location, symbol.count, symbol.orientation

但只检测到一个点:(596, 210)。

如果我应用黑白阈值:

pil = Image.open("000.jpg").convert('L')
pil = pil .point(lambda x: 0 if x<100 else 255, '1').convert('L')    

比较好，我们有3个点:(596, 210), (482, 211), (596, 212)。但这又增加了一个困难(自动为每个新图像找到最佳阈值 - 这里是 100)。

仍然，我们没有条形码的 4 个角。

问题:如何使用 Python 可靠地找到图像上条形码的 4 个角？(也许还有 OpenCV 或其他库？)

注意事项:

• 它是可能的，这是一个很好的例子(但遗憾的是不是评论中提到的开源):

  Object detection, very fast and robust blurry 1D barcode detection for real-time applications

  角点检测似乎非常好而且非常快，即使条形码只是整个图像的一小部分(这对我来说很重要)。

  

• 有趣的解决方案:Real-time barcode detection in video with Python and OpenCV但该方法存在局限性(参见文章中:条形码应关闭等)限制了潜在用途。此外，我更想为此寻找一个现成的库。

• 有趣的解决方案 2:Detecting Barcodes in Images with Python and OpenCV但同样，它看起来不像是一个生产就绪的解决方案，而更像是一个正在进行的研究。事实上，我在这张图片上尝试了他们的代码，但检测没有产生成功的结果。必须注意的是，它在检测时不考虑条形码的任何规范(事实上有一个 start/stop symbol 等)

  import numpy as np
  import cv2
  image = cv2.imread("000.jpg")
  gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
  gradX = cv2.Sobel(gray, ddepth = cv2.CV_32F, dx = 1, dy = 0, ksize = -1)
  gradY = cv2.Sobel(gray, ddepth = cv2.CV_32F, dx = 0, dy = 1, ksize = -1)
  gradient = cv2.subtract(gradX, gradY)
  gradient = cv2.convertScaleAbs(gradient)
  blurred = cv2.blur(gradient, (9, 9))
  (_, thresh) = cv2.threshold(blurred, 225, 255, cv2.THRESH_BINARY)
  kernel = cv2.getStructuringElement(cv2.MORPH_RECT, (21, 7))
  closed = cv2.morphologyEx(thresh, cv2.MORPH_CLOSE, kernel)
  closed = cv2.erode(closed, None, iterations = 4)
  closed = cv2.dilate(closed, None, iterations = 4)
  (_, cnts, _) = cv2.findContours(closed.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
  c = sorted(cnts, key = cv2.contourArea, reverse = True)[0]
  rect = cv2.minAreaRect(c)
  box = np.int0(cv2.boxPoints(rect))
  cv2.drawContours(image, [box], -1, (0, 255, 0), 3)
  cv2.imshow("Image", image)
  cv2.waitKey(0)
  

Answer 1

解决方案 2 非常好。使它在图像上失败的关键因素是阈值。如果将参数 225 降低到 55，您将获得更好的结果。

我重新编写了代码，在各处进行了一些调整。如果您愿意，原始代码很好。 documentation for OpenCV is quite good ，还有很好的Python tutorials .

import numpy as np
import cv2

image = cv2.imread("barcode.jpg")
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)

# equalize lighting
clahe = cv2.createCLAHE(clipLimit=2.0, tileGridSize=(8,8))
gray = clahe.apply(gray)

# edge enhancement
edge_enh = cv2.Laplacian(gray, ddepth = cv2.CV_8U, 
                         ksize = 3, scale = 1, delta = 0)
cv2.imshow("Edges", edge_enh)
cv2.waitKey(0)
retval = cv2.imwrite("edge_enh.jpg", edge_enh)

# bilateral blur, which keeps edges
blurred = cv2.bilateralFilter(edge_enh, 13, 50, 50)

# use simple thresholding. adaptive thresholding might be more robust
(_, thresh) = cv2.threshold(blurred, 55, 255, cv2.THRESH_BINARY)
cv2.imshow("Thresholded", thresh)
cv2.waitKey(0)
retval = cv2.imwrite("thresh.jpg", thresh)

# do some morphology to isolate just the barcode blob
kernel = cv2.getStructuringElement(cv2.MORPH_RECT, (9, 9))
closed = cv2.morphologyEx(thresh, cv2.MORPH_CLOSE, kernel)
closed = cv2.erode(closed, None, iterations = 4)
closed = cv2.dilate(closed, None, iterations = 4)
cv2.imshow("After morphology", closed)
cv2.waitKey(0)
retval = cv2.imwrite("closed.jpg", closed)

# find contours left in the image
(_, cnts, _) = cv2.findContours(closed.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
c = sorted(cnts, key = cv2.contourArea, reverse = True)[0]
rect = cv2.minAreaRect(c)
box = np.int0(cv2.boxPoints(rect))
cv2.drawContours(image, [box], -1, (0, 255, 0), 3)
print(box)
cv2.imshow("found barcode", image)
cv2.waitKey(0)
retval = cv2.imwrite("found.jpg", image)

边缘.jpg

脱粒.jpg

关闭.jpg

找到.jpg

控制台输出:

[[596 249]
 [470 213]
 [482 172]
 [608 209]]