Python 检查两个轮廓是否相交?
我有两个轮廓(Python 检查两个轮廓是否相交?,python,opencv,opencv-contour,Python,Opencv,Opencv Contour,我有两个轮廓(cont1和cont2)是从cv2.findContours()收到的。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 我尝试过不同的方法,并且已经尝试过与 if ((cont1 & cont2).area() > 0): 。。。 但是得到的错误是数组没有方法“Area()” 。。。 cont1array=cv2.findContentours(二进制1,cv2.RETR\u列表,cv2.CHAIN\u近似值\u简单)[0] cont2
cont1
和cont2
)是从cv2.findContours()
收到的。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
我尝试过不同的方法,并且已经尝试过与
if ((cont1 & cont2).area() > 0):
。。。
但是得到的错误是数组没有方法“Area()”
。。。
cont1array=cv2.findContentours(二进制1,cv2.RETR\u列表,cv2.CHAIN\u近似值\u简单)[0]
cont2array=cv2.findContentours(二进制2,cv2.RETR\u列表,cv2.CHAIN\u近似值\u简单)[0]
...
对于cont1阵列中的cont1:
对于cont2阵列中的cont2:
打印(“续1”)
印刷品(续1)
打印(类型(续1))
打印(“续二”)
印刷品(续二)
打印(类型(续2))
>如果cont1和cont2相交:#我不知道如何检查相交
打印(“是的,它们相交”)
其他:
打印(“不,它们不相交”)
#cont1
# [[172 302]
# [261 301]
# [262 390]
# [173 391]]
#
#cont2
# [[ 0 0]
# [ 0 699]
# [499 699]
# [499 0]]
#
一旦从cv2.findContours()
获得两条等高线,就可以使用按位和
操作来检测交点。具体来说,我们可以使用。其思想是为每个轮廓创建两个单独的图像,然后对其使用逻辑和操作。任何具有正值(1
或True
)的点都将是交点。因此,由于您只希望获得是否存在交点的布尔值,我们可以检查相交图像以查看是否存在单个正值。基本上,如果整个阵列为False
,则轮廓之间没有相交。但如果只有一个为真
,则轮廓会接触,从而相交
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()
def ccw(A,B,C):
return (C[1]-A[1]) * (B[0]-A[0]) > (B[1]-A[1]) * (C[0]-A[0])
def contour_intersect(cnt_ref,cnt_query):
## Contour is a list of points
## Connect each point to the following point to get a line
## If any of the lines intersect, then break
for ref_idx in range(len(cnt_ref)-1):
## Create reference line_ref with point AB
A = cnt_ref[ref_idx][0]
B = cnt_ref[ref_idx+1][0]
for query_idx in range(len(cnt_query)-1):
## Create query line_query with point CD
C = cnt_query[query_idx][0]
D = cnt_query[query_idx+1][0]
## Check if line intersect
if ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D):
## If true, break loop earlier
return True
return False
示例
原始图像
检测轮廓
现在,我们将两个检测到的轮廓传递给函数,并获得该相交数组:
[[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]]
我们检查交叉点
数组以查看True
是否存在。我们将在轮廓相交处获得True
或1
,在轮廓不相交处获得False
或0
return intersection.any()
因此,我们获得
假的
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)
nathancy的答案是可行的,但在性能方面会受到影响,如示例中所示,创建3个图像副本以绘制轮廓,因此在执行时间方面比较缓慢
我的备选答案如下:
def contour_intersect(cnt_ref,cnt_query, edges_only = True):
intersecting_pts = []
## Loop through all points in the contour
for pt in cnt_query:
x,y = pt[0]
## find point that intersect the ref contour
## edges_only flag check if the intersection to detect is only at the edges of the contour
if edges_only and (cv2.pointPolygonTest(cnt_ref,(x,y),True) == 0):
intersecting_pts.append(pt[0])
elif not(edges_only) and (cv2.pointPolygonTest(cnt_ref,(x,y),True) >= 0):
intersecting_pts.append(pt[0])
if len(intersecting_pts) > 0:
return True
else:
return False
编辑强>
测试此代码后,意识到当轮廓没有两个相似点时,此检查失败。因此,我重写了检查两条等高线相交的算法
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()
def ccw(A,B,C):
return (C[1]-A[1]) * (B[0]-A[0]) > (B[1]-A[1]) * (C[0]-A[0])
def contour_intersect(cnt_ref,cnt_query):
## Contour is a list of points
## Connect each point to the following point to get a line
## If any of the lines intersect, then break
for ref_idx in range(len(cnt_ref)-1):
## Create reference line_ref with point AB
A = cnt_ref[ref_idx][0]
B = cnt_ref[ref_idx+1][0]
for query_idx in range(len(cnt_query)-1):
## Create query line_query with point CD
C = cnt_query[query_idx][0]
D = cnt_query[query_idx+1][0]
## Check if line intersect
if ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D):
## If true, break loop earlier
return True
return False
您可以使用边界矩形作为第一个粗糙而廉价的步骤,然后再进行细化。我会在单个遮罩图像上绘制每个填充轮廓并计算遮罩交点,但这可能太慢了。在现实世界中,两个不同的轮廓永远不会相交(请参见,例如)。在计算机的近似计算中,一些轮廓可能相交,但其有用性取决于具体情况。你能多说一点你正在试图解决的潜在问题吗?这可能会让人们提供更多有用的答案。这里有一张儿童游戏的图片(如乐透网格)。在游戏中,他们把筷子放在图片上(像冰激凌棒一样)。我需要确定哪些细胞在棍子下面。cont1是这个游戏网格中所有单元格的轮廓(在它们被放在棍子上之前)。cont2是图片中所有木棒的轮廓。在代码中,我做了一个检查:如果单元的轮廓与棒的轮廓交叉,则单元被棒闭合。因此,我需要知道如何确定两个等高线相交的事实。因为在本文中,每个等高线都只是一个点向量,所以在将它们转换为集合(例如,如中所述)后,您能否找到两个点向量的交点?您可以在交叉点处返回True
或者更简单地返回交叉点。any()
谢谢,它似乎可以工作,但是如果一个电路位于另一个电路内,然后我得到了一个错误的答案,我会推荐这个答案,而不是选择的答案,因为这会带来巨大的性能提升。这只适用于cnt\u查询
角位于cnt\u ref
区域内的情况。必须在cnt_查询中使用cv2.CHAIN_APPROX_NONE
参数才能使其正常工作。@justas建议使用cv2.CHAIN_APPROX_,以便减少处理轮廓时的内存占用,如本文所强调的