Python OpenCV 线检测以检测图像中的“X”符号
Python OpenCV line detection to detect `X` symbol in image
我有一张图片，我需要在其中检测到行内的 X 符号。
图片：
如上图所示，一行中有一个 X 符号。我想知道符号的 X 和 Y 坐标。有没有办法在这张图片中找到这个符号或者它太小了？
import cv2
import numpy as np
def calculateCenterSpot(results):
startX, endX = results[0][0], results[0][2]
startY, endY = results[0][1], results[0][3]
centerSpotX = (endX - startX) / 2 + startX
centerSpotY = (endY - startY) / 2 + startY
return [centerSpotX, centerSpotY]
img = cv2.imread('crop_1.png')
res2 = img.copy()
cords = [[1278, 704, 1760, 1090]]
center = calculateCenterSpot(cords)
cv2.circle(img, (int(center[0]), int(center[1])), 1, (0,0,255), 30)
cv2.line(img, (int(center[0]), 0), (int(center[0]), img.shape[0]), (0,255,0), 10)
cv2.line(img, (0, int(center[1])), (img.shape[1], int(center[1])), (255,0,0), 10)
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
# You can either use threshold or Canny edge for HoughLines().
_, thresh = cv2.threshold(gray,0,255,cv2.THRESH_BINARY_INV+cv2.THRESH_OTSU)
#edges = cv2.Canny(gray, 50, 150, apertureSize=3)
# Perform HoughLines tranform.
lines = cv2.HoughLines(thresh,0.5,np.pi/180,1000)
for line in lines:
for rho,theta in line:
a = np.cos(theta)
b = np.sin(theta)
x0 = a*rho
y0 = b*rho
x1 = int(x0 + 5000*(-b))
y1 = int(y0 + 5000*(a))
x2 = int(x0 - 5000*(-b))
y2 = int(y0 - 5000*(a))
if x2 == int(center[0]):
cv2.circle(img, (x2,y1), 100, (0,0,255), 30)
if y2 == int(center[1]):
print('hell2o')
# cv2.line(res2,(x1,y1),(x2,y2),(0,0,255),2)
#Display the result.
cv2.imwrite('h_res1.png', img)
cv2.imwrite('h_res3.png', res2)
cv2.imwrite('image.png', img)
我已经用 HoughLines 尝试过，但没有成功。
如果您有多个需要检测此 X 符号的图像，并且如果此 X 符号始终相同且具有相同的尺寸，则您可以 运行您正在卷积的二维 convolution over each image, where the kernel 是您试图检测的孤立的 X 符号。然后，您可以检查此二维卷积的输出以获得最大强度的像素，其归一化坐标 (x/w,y/h) 很可能对应于输入图像中 X 符号的归一化坐标。这是二维卷积的数学表达式：
在 opencv 中，您可以定义 your own kernel（确保只保留十字架，背景中没有其他任何东西），然后将其应用于您的图像。
不使用 cv2.HoughLines()，另一种方法是使用 template matching。这个想法是在更大的图像中搜索和找到模板图像的位置。为了执行此方法，模板在输入图像上滑动（类似于 2D 卷积），在输入图像上执行比较方法以确定像素相似性。这是模板匹配背后的基本思想。不幸的是，这种基本方法存在缺陷，因为它 仅在模板图像大小与要在输入图像中找到的所需项目 相同时才有效。因此，如果您的模板图像小于要在输入图像中找到的所需区域，则此方法将不起作用。
为了绕过这个限制，我们可以使用 np.linspace() 动态重新缩放图像以获得更好的模板匹配。在每次迭代中，我们调整输入图像的大小并跟踪比率。我们继续调整大小，直到模板图像大小大于调整后的图像，同时跟踪最高相关值。较高的相关值意味着更好的匹配。一旦我们遍历了各种尺度，我们就会找到匹配度最大的比率，然后计算边界框的坐标以确定 ROI。
使用这个截图模板图像
这是结果
import cv2
import numpy as np
# Resizes a image and maintains aspect ratio
def maintain_aspect_ratio_resize(image, width=None, height=None, inter=cv2.INTER_AREA):
# Grab the image size and initialize dimensions
dim = None
(h, w) = image.shape[:2]
# Return original image if no need to resize
if width is None and height is None:
return image
# We are resizing height if width is none
if width is None:
# Calculate the ratio of the height and construct the dimensions
r = height / float(h)
dim = (int(w * r), height)
# We are resizing width if height is none
else:
# Calculate the ratio of the 0idth and construct the dimensions
r = width / float(w)
dim = (width, int(h * r))
# Return the resized image
return cv2.resize(image, dim, interpolation=inter)
# Load template, convert to grayscale, perform canny edge detection
template = cv2.imread('template.png')
template = cv2.cvtColor(template, cv2.COLOR_BGR2GRAY)
template = cv2.Canny(template, 50, 200)
(tH, tW) = template.shape[:2]
cv2.imshow("template", template)
# Load original image, convert to grayscale
original_image = cv2.imread('1.png')
gray = cv2.cvtColor(original_image, cv2.COLOR_BGR2GRAY)
found = None
# Dynamically rescale image for better template matching
for scale in np.linspace(0.1, 3.0, 20)[::-1]:
# Resize image to scale and keep track of ratio
resized = maintain_aspect_ratio_resize(gray, width=int(gray.shape[1] * scale))
r = gray.shape[1] / float(resized.shape[1])
# Stop if template image size is larger than resized image
if resized.shape[0] < tH or resized.shape[1] < tW:
break
# Detect edges in resized image and apply template matching
canny = cv2.Canny(resized, 50, 200)
detected = cv2.matchTemplate(canny, template, cv2.TM_CCOEFF)
(_, max_val, _, max_loc) = cv2.minMaxLoc(detected)
# Uncomment this section for visualization
'''
clone = np.dstack([canny, canny, canny])
cv2.rectangle(clone, (max_loc[0], max_loc[1]), (max_loc[0] + tW, max_loc[1] + tH), (0,255,0), 2)
cv2.imshow('visualize', clone)
cv2.waitKey(0)
'''
# Keep track of correlation value
# Higher correlation means better match
if found is None or max_val > found[0]:
found = (max_val, max_loc, r)
# Compute coordinates of bounding box
(_, max_loc, r) = found
(start_x, start_y) = (int(max_loc[0] * r), int(max_loc[1] * r))
(end_x, end_y) = (int((max_loc[0] + tW) * r), int((max_loc[1] + tH) * r))
# Draw bounding box on ROI
cv2.rectangle(original_image, (start_x, start_y), (end_x, end_y), (0,255,0), 2)
cv2.imshow('detected', original_image)
cv2.imwrite('detected.png', original_image)
cv2.waitKey(0)
对于多个模板图像，您可以对您拥有的不同模板图像的数量使用 for 循环，然后使用阈值方法找到多个模板匹配项。
for i in range(templateAmount):
template = cv2.imread('template{}.png'.format(i),0)
w, h = template.shape[::-1]
res = cv2.matchTemplate(img_gray,template,cv2.TM_CCOEFF_NORMED)
threshold = 0.8
loc = np.where( res >= threshold)
for pt in zip(*loc[::-1]):
cv2.rectangle(img_rgb, pt, (pt[0] + w, pt[1] + h), (0,0,255), 2)
我有一张图片，我需要在其中检测到行内的 X 符号。
图片：
如上图所示，一行中有一个 X 符号。我想知道符号的 X 和 Y 坐标。有没有办法在这张图片中找到这个符号或者它太小了？
import cv2
import numpy as np
def calculateCenterSpot(results):
startX, endX = results[0][0], results[0][2]
startY, endY = results[0][1], results[0][3]
centerSpotX = (endX - startX) / 2 + startX
centerSpotY = (endY - startY) / 2 + startY
return [centerSpotX, centerSpotY]
img = cv2.imread('crop_1.png')
res2 = img.copy()
cords = [[1278, 704, 1760, 1090]]
center = calculateCenterSpot(cords)
cv2.circle(img, (int(center[0]), int(center[1])), 1, (0,0,255), 30)
cv2.line(img, (int(center[0]), 0), (int(center[0]), img.shape[0]), (0,255,0), 10)
cv2.line(img, (0, int(center[1])), (img.shape[1], int(center[1])), (255,0,0), 10)
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
# You can either use threshold or Canny edge for HoughLines().
_, thresh = cv2.threshold(gray,0,255,cv2.THRESH_BINARY_INV+cv2.THRESH_OTSU)
#edges = cv2.Canny(gray, 50, 150, apertureSize=3)
# Perform HoughLines tranform.
lines = cv2.HoughLines(thresh,0.5,np.pi/180,1000)
for line in lines:
for rho,theta in line:
a = np.cos(theta)
b = np.sin(theta)
x0 = a*rho
y0 = b*rho
x1 = int(x0 + 5000*(-b))
y1 = int(y0 + 5000*(a))
x2 = int(x0 - 5000*(-b))
y2 = int(y0 - 5000*(a))
if x2 == int(center[0]):
cv2.circle(img, (x2,y1), 100, (0,0,255), 30)
if y2 == int(center[1]):
print('hell2o')
# cv2.line(res2,(x1,y1),(x2,y2),(0,0,255),2)
#Display the result.
cv2.imwrite('h_res1.png', img)
cv2.imwrite('h_res3.png', res2)
cv2.imwrite('image.png', img)
我已经用 HoughLines 尝试过，但没有成功。
如果您有多个需要检测此 X 符号的图像，并且如果此 X 符号始终相同且具有相同的尺寸，则您可以 运行您正在卷积的二维 convolution over each image, where the kernel 是您试图检测的孤立的 X 符号。然后，您可以检查此二维卷积的输出以获得最大强度的像素，其归一化坐标 (x/w,y/h) 很可能对应于输入图像中 X 符号的归一化坐标。这是二维卷积的数学表达式：
不使用 cv2.HoughLines()，另一种方法是使用 template matching。这个想法是在更大的图像中搜索和找到模板图像的位置。为了执行此方法，模板在输入图像上滑动（类似于 2D 卷积），在输入图像上执行比较方法以确定像素相似性。这是模板匹配背后的基本思想。不幸的是，这种基本方法存在缺陷，因为它 仅在模板图像大小与要在输入图像中找到的所需项目 相同时才有效。因此，如果您的模板图像小于要在输入图像中找到的所需区域，则此方法将不起作用。
为了绕过这个限制，我们可以使用 np.linspace() 动态重新缩放图像以获得更好的模板匹配。在每次迭代中，我们调整输入图像的大小并跟踪比率。我们继续调整大小，直到模板图像大小大于调整后的图像，同时跟踪最高相关值。较高的相关值意味着更好的匹配。一旦我们遍历了各种尺度，我们就会找到匹配度最大的比率，然后计算边界框的坐标以确定 ROI。
使用这个截图模板图像
这是结果
import cv2
import numpy as np
# Resizes a image and maintains aspect ratio
def maintain_aspect_ratio_resize(image, width=None, height=None, inter=cv2.INTER_AREA):
# Grab the image size and initialize dimensions
dim = None
(h, w) = image.shape[:2]
# Return original image if no need to resize
if width is None and height is None:
return image
# We are resizing height if width is none
if width is None:
# Calculate the ratio of the height and construct the dimensions
r = height / float(h)
dim = (int(w * r), height)
# We are resizing width if height is none
else:
# Calculate the ratio of the 0idth and construct the dimensions
r = width / float(w)
dim = (width, int(h * r))
# Return the resized image
return cv2.resize(image, dim, interpolation=inter)
# Load template, convert to grayscale, perform canny edge detection
template = cv2.imread('template.png')
template = cv2.cvtColor(template, cv2.COLOR_BGR2GRAY)
template = cv2.Canny(template, 50, 200)
(tH, tW) = template.shape[:2]
cv2.imshow("template", template)
# Load original image, convert to grayscale
original_image = cv2.imread('1.png')
gray = cv2.cvtColor(original_image, cv2.COLOR_BGR2GRAY)
found = None
# Dynamically rescale image for better template matching
for scale in np.linspace(0.1, 3.0, 20)[::-1]:
# Resize image to scale and keep track of ratio
resized = maintain_aspect_ratio_resize(gray, width=int(gray.shape[1] * scale))
r = gray.shape[1] / float(resized.shape[1])
# Stop if template image size is larger than resized image
if resized.shape[0] < tH or resized.shape[1] < tW:
break
# Detect edges in resized image and apply template matching
canny = cv2.Canny(resized, 50, 200)
detected = cv2.matchTemplate(canny, template, cv2.TM_CCOEFF)
(_, max_val, _, max_loc) = cv2.minMaxLoc(detected)
# Uncomment this section for visualization
'''
clone = np.dstack([canny, canny, canny])
cv2.rectangle(clone, (max_loc[0], max_loc[1]), (max_loc[0] + tW, max_loc[1] + tH), (0,255,0), 2)
cv2.imshow('visualize', clone)
cv2.waitKey(0)
'''
# Keep track of correlation value
# Higher correlation means better match
if found is None or max_val > found[0]:
found = (max_val, max_loc, r)
# Compute coordinates of bounding box
(_, max_loc, r) = found
(start_x, start_y) = (int(max_loc[0] * r), int(max_loc[1] * r))
(end_x, end_y) = (int((max_loc[0] + tW) * r), int((max_loc[1] + tH) * r))
# Draw bounding box on ROI
cv2.rectangle(original_image, (start_x, start_y), (end_x, end_y), (0,255,0), 2)
cv2.imshow('detected', original_image)
cv2.imwrite('detected.png', original_image)
cv2.waitKey(0)
对于多个模板图像，您可以对您拥有的不同模板图像的数量使用 for 循环，然后使用阈值方法找到多个模板匹配项。
for i in range(templateAmount):
template = cv2.imread('template{}.png'.format(i),0)
w, h = template.shape[::-1]
res = cv2.matchTemplate(img_gray,template,cv2.TM_CCOEFF_NORMED)
threshold = 0.8
loc = np.where( res >= threshold)
for pt in zip(*loc[::-1]):
cv2.rectangle(img_rgb, pt, (pt[0] + w, pt[1] + h), (0,0,255), 2)