检查两个轮廓是否相交？

小编典典

检查两个轮廓是否相交？

python

我收到了2个轮廓（cont1和cont2）cv2.findContours()。我怎么知道它们是否相交？我不需要坐标，我只需要一个布尔值True或False。

我尝试了不同的方式，并且已经尝试与

if ((cont1 & cont2).area() > 0):

…但是得到的错误是数组没有方法“ Area（）”

...
cont1array = cv2.findContours(binary1, cv2.RETR_LIST, cv2.CHAIN_APPROX_SIMPLE)[0]
cont2array = cv2.findContours(binary2, cv2.RETR_LIST, cv2.CHAIN_APPROX_SIMPLE)[0]
...

for cont1 in cont1array:
  for cont2 in cont2array:
    print("cont1")
    print(cont1)
    print(type(cont1))
    print("cont2")
    print(cont2)
    print(type(cont2))
>   if cont1 and cont2 intersect: #i dont know how check intersect
      print("yes they intersect")
    else:
      print("no they do not intersect")

# cont1
# [[172 302]
#  [261 301]
#  [262 390]
#  [173 391]]
# <class 'numpy.ndarray'>
# cont2
# [[  0   0]
#  [  0 699]
#  [499 699]
#  [499   0]]
# <class 'numpy.ndarray'>

阅读 857

2021-01-20

共1个答案

小编典典

一旦有了的两个轮廓cv2.findContours()，就可以使用按位AND运算来检测相交。具体来说，我们可以使用np.logical_and()。想法是为每个轮廓创建两个单独的图像，然后AND对它们使用逻辑运算。具有正值（1或True）的任何点都将是交点。因此，由于您只想获取是否存在相交的布尔值，因此我们可以检查相交的图像以查看是否存在单个正值。本质上，如果整个数组都False存在，则轮廓之间就没有交集。但是，如果只有一个True，则轮廓会接触并因此相交。

def contourIntersect(original_image, contour1, contour2):
    # Two separate contours trying to check intersection on
    contours = [contour1, contour2]

    # Create image filled with zeros the same size of original image
    blank = np.zeros(original_image.shape[0:2])

    # Copy each contour into its own image and fill it with '1'
    image1 = cv2.drawContours(blank.copy(), contours, 0, 1)
    image2 = cv2.drawContours(blank.copy(), contours, 1, 1)

    # Use the logical AND operation on the two images
    # Since the two images had bitwise and applied to it,
    # there should be a '1' or 'True' where there was intersection
    # and a '0' or 'False' where it didnt intersect
    intersection = np.logical_and(image1, image2)

    # Check if there was a '1' in the intersection
    return intersection.any()

例

原始图片

检测到轮廓

现在，我们将两个检测到的轮廓传递给函数，并获得此交集数组：

[[False False False ... False False False]
 [False False False ... False False False]
 [False False False ... False False False]
 ...
 [False False False ... False False False]
 [False False False ... False False False]
 [False False False ... False False False]]

我们检查intersection数组以查看是否True存在。我们将获得一个True或1轮廓相交的地方，False或0它们不相交的地方。

return intersection.any()

这样我们得到

假

完整代码

import cv2
import numpy as np

def contourIntersect(original_image, contour1, contour2):
    # Two separate contours trying to check intersection on
    contours = [contour1, contour2]

    # Create image filled with zeros the same size of original image
    blank = np.zeros(original_image.shape[0:2])

    # Copy each contour into its own image and fill it with '1'
    image1 = cv2.drawContours(blank.copy(), contours, 0, 1)
    image2 = cv2.drawContours(blank.copy(), contours, 1, 1)

    # Use the logical AND operation on the two images
    # Since the two images had bitwise AND applied to it,
    # there should be a '1' or 'True' where there was intersection
    # and a '0' or 'False' where it didnt intersect
    intersection = np.logical_and(image1, image2)

    # Check if there was a '1' in the intersection array
    return intersection.any()

original_image = cv2.imread("base.png")
image = original_image.copy()

cv2.imshow("original", image)
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
cv2.imshow("gray", gray)
blurred = cv2.GaussianBlur(gray, (5,5), 0)
cv2.imshow("blur", blurred)
threshold = cv2.threshold(blurred, 60, 255, cv2.THRESH_BINARY)[1]
cv2.imshow("thresh", threshold)

contours = cv2.findContours(threshold.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
# Depending on OpenCV version, number of arguments return by cv.findContours 
# is either 2 or 3
contours = contours[1] if len(contours) == 3 else contours[0]

contour_list = []
for c in contours:
    contour_list.append(c)
    cv2.drawContours(image, [c], 0, (0,255,0), 2)

print(contourIntersect(original_image, contour_list[0], contour_list[1]))
cv2.imshow("contour", image)
cv2.waitKey(0)

2021-01-20