如何可靠地检测条形码的4个角？

Question

我尝试使用Python + zbar模块检测此Code128条形码：

（图片下载链接here）。

这有效：

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')    

它更好，我们有3分：（596,210），（482,211），（596,212）。但它增加了一个难度（找到最佳阈值 - 这里100 - 自动为每个新图像）。

尽管如此，我们还没有条形码的四个角落。

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

注意：

• 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

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

  

• 有趣的解决方案：Real-time barcode detection in video with Python and OpenCV但是该方法的局限性（参见文章：条形码应该关闭等等）限制了潜在的使用。此外，我还在寻找一个可立即使用的库。

• 有趣的解决方案2：Detecting Barcodes in Images with Python and OpenCV但同样，它似乎不是一个生产就绪的解决方案，而是一个正在进行中的研究。实际上，我在这张图片上尝试了他们的代码，但检测并没有产生成功的结果。必须注意的是，它没有考虑到条形码的任何规格考虑到检测（事实是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。使图像失败的关键因素是阈值处理。如果您将参数225向下移至55，您将获得更好的结果。

我已经重新编写了代码，在这里和那里做了一些调整。如果您愿意，原始代码很好。 documentation for OpenCV is quite good，非常好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)

edge.jpg

thresh.jpg

closed.jpg

found.jpg

从控制台输出：

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