Davide_sd's profile - activity

is it possible to mark two answers as both correct? truth is @kbarni answer nicely integrates with my current workflow.

This is the Python code for @pklab answer, in case someone is in a hurry...

import numpy as np
import cv2

def getJunctions(src):
    # the hit-and-miss kernels to locate 3-points junctions to be used in each directions
    # NOTE: float type is needed due to limitation/bug in warpAffine with signed char
    k1 = np.asarray([
        0,  1,  0,
        0,  1,  0,
        1,  0,  1], dtype=float).reshape((3, 3))
    k2 = np.asarray([
        1,  0,  0,
        0,  1,  0,
        1,  0,  1], dtype=float).reshape((3, 3))
    k3 = np.asarray([
        0, -1,  1,
        1,  1, -1,
        0,  1, 0], dtype=float).reshape((3, 3))

    # Some useful declarations
    tmp = np.zeros_like(src)
    ksize = k1.shape
    center = (ksize[1] / 2, ksize[0] / 2) # INVERTIRE 0 E 1??????
    # 90 degrees rotation matrix
    rotMat = cv2.getRotationMatrix2D(center, 90, 1)
    # dst accumulates all matches
    dst = np.zeros(src.shape, dtype=np.uint8)

    # Do hit & miss for all possible directions (0,90,180,270)
    for i in range(4):
        tmp = cv2.morphologyEx(src, cv2.MORPH_HITMISS, k1.astype(np.int8), tmp, (-1, -1), 1, cv2.BORDER_CONSTANT, 0)     
        dst = cv2.add(dst, tmp)
        tmp = cv2.morphologyEx(src, cv2.MORPH_HITMISS, k2.astype(np.int8), tmp, (-1, -1), 1, cv2.BORDER_CONSTANT, 0)
        dst = cv2.add(dst, tmp)
        tmp = cv2.morphologyEx(src, cv2.MORPH_HITMISS, k3.astype(np.int8), tmp, (-1, -1), 1, cv2.BORDER_CONSTANT, 0)
        dst = cv2.add(dst, tmp)
        # Rotate the kernels (90deg)
        k1 = cv2.warpAffine(k1, rotMat, ksize)
        k2 = cv2.warpAffine(k2, rotMat, ksize)
        k3 = cv2.warpAffine(k3, rotMat, ksize)

    return dst






# This is your sample image (objects are black)
src = np.asarray([0, 1, 1, 1, 1, 1, 0, 0,
        1, 0, 1, 1, 1, 0, 1, 1,
        1, 1, 0, 0, 0, 1, 1, 1,
        1, 1, 1, 0, 0, 0, 1, 1,
        1, 0, 0, 1, 1, 1, 0, 1,
        0, 1, 1, 1, 1, 1, 0, 0,
        0, 1, 1, 1, 1, 1, 1, 1], dtype=np.uint8).reshape((7,8))

src *= 255;
# Morphology logic is: white objects on black foreground
src = 255 - src;

# Get junctions
junctionsScore = getJunctions(src)

# Draw markers where junction score is non zero
dst = cv2.cvtColor(src, cv2.COLOR_GRAY2RGB)
# find the list of location of non-zero pixels
junctionsPoint = cv2.findNonZero(junctionsScore)

for pt in junctionsPoint:
    pt = pt[0]
    dst[pt[1], pt[0], :] = [0, 0, junctionsScore[pt[1], pt[0]]]

# show the result
winDst = "Dst"
winSrc = "Src"
winJunc = "Junctions"
cv2.namedWindow(winSrc, cv2.WINDOW_NORMAL | cv2.WINDOW_KEEPRATIO);
cv2.namedWindow(winJunc, cv2.WINDOW_NORMAL | cv2.WINDOW_KEEPRATIO);
cv2.namedWindow(winDst, cv2.WINDOW_NORMAL | cv2.WINDOW_KEEPRATIO);
scale = 24
cv2.resizeWindow(winSrc, scale*src.shape[1], scale*src.shape[0])
cv2.resizeWindow(winJunc, scale*src.shape[1], scale*src.shape[0])
cv2.resizeWindow(winDst, scale*src.shape[1], scale*src.shape[0])
cv2.imshow(winSrc, src)
cv2.imshow(winJunc, junctionsScore)
cv2.imshow(winDst, dst)
cv2.waitKey()

wow, hit and miss is really powerful! Thanks @pklab to show me this solution!!!

top notch! thank you for your help :D

unfortunately the contours are not straitgh lines, they come from random shapes, so i can't use the line equations. :(

I have two contour images and I would like to find the contours intersection. So far I've succesfully used the logical and operation, unfortunately I've fallen into this case:

As you can see, the logical and operation is going to fail because there isn't any intersection. If the red contours is in image A, and blue contour is in image B, I would like a way to return the two red pixel indicated by the arrow. Does openCV provide some function to do that? In case there isn't, how would you solve this problem?

PS: the contours come from random shapes, they are NOT straight lines!

Thank you.