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How to sort contours of a grid using OpenCV python?

Time:05-11

I'm trying to sort the following squares inside the checkers board grid.

enter image description here

I have the valid contours, inside a NumPy array.

Here's a snippet of the code, on how I get the valid squares contours.

  # Find contours and find squares with contour area filtering   shape approximation
            cnts = cv2.findContours(invert, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
            r = 0
            cnts = cnts[0] if len(cnts) == 2 else cnts[1]
            sort_contours(cnts, "bottom-to-top")
            sort_contours(cnts, "left-to-right")
            valid_cnts = []
            v = []
            areas = []
            for c in cnts:
                area = cv2.contourArea(c)
                peri = cv2.arcLength(c, True)
                approx = cv2.approxPolyDP(c, 0.02 * peri, True)
                if len(approx) == 4 and area > 150 and area < 15000:
                    areas.append(area)
                    x, y, w, h = cv2.boundingRect(c)
                    s = img[y:y   h, x:x   w]
                    imgStr = "squares/square"   str(r)   ".png"
                    v.insert(r, [x, y, w, h])
                    cv2.imwrite(imgStr, s)
                    cv2.drawContours(original, [c], -1, (36, 255, 12), 2)
                    cv2.drawContours(mask, [c], -1, (255, 255, 255), -1)
                    valid_cnts.insert(r, c)
                    r = r   1

My aim also is to sort them from left to right then bottom to up. So I can then recognize each piece on them. This is my current sorting function:

def sort_contours(cnts, method="left-to-right"):
    # initialize the reverse flag and sort index
    reverse = False
    i = 0
    # handle if we need to sort in reverse
    if method == "right-to-left" or method == "bottom-to-top":
        reverse = True
    # handle if we are sorting against the y-coordinate rather than
    # the x-coordinate of the bounding box
    if method == "top-to-bottom" or method == "bottom-to-top":
        i = 1
    # construct the list of bounding boxes and sort them from top to
    # bottom
    boundingBoxes = [cv2.boundingRect(c) for c in cnts]
    (cnts, boundingBoxes) = zip(*sorted(zip(cnts, boundingBoxes),
                                        key=lambda b: b[1][i], reverse=reverse))
    # return the list of sorted contours and bounding boxes
    return (cnts, boundingBoxes)

Unfortunately, it does not work, I think it has to do with the camera angle. Because when I crop the photo to 64 squares they do not appear in the order I desire. If anyone can guide me on how to sort them correctly and precisely it would be great!

CodePudding user response:

enter image description here

The idea is after finding contours on the thresholded image, we utilize enter image description here

import cv2
from imutils import contours

# Load image, grayscale, gaussian blur, Otsu's threshold
image = cv2.imread("1.jpg")
original = image.copy()
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
blur = cv2.GaussianBlur(gray, (5,5), 0)
thresh = cv2.threshold(blur, 0, 255, cv2.THRESH_BINARY   cv2.THRESH_OTSU)[1]

# Find all contour and sort from top-to-bottom or bottom-to-top
cnts, _ = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)[-2:]
(cnts, _) = contours.sort_contours(cnts, method="bottom-to-top")

# Take each row of 8 and sort from left-to-right
checkerboard_row = []
row = []
for (i, c) in enumerate(cnts, 1):
    row.append(c)
    if i % 8 == 0:  
        (cnts, _) = contours.sort_contours(row, method="left-to-right")
        checkerboard_row.append(cnts)
        row = []

# Draw text
number = 0
for row in checkerboard_row:
    for c in row:
        M = cv2.moments(c)
        x = int(M['m10']/M['m00'])
        y = int(M['m01']/M['m00'])
        cv2.putText(original, "{}".format(number   1), (x - 20,y), cv2.FONT_HERSHEY_SIMPLEX, 0.7, (255,50,10), 2)
        number  = 1

cv2.imshow('original', original)
cv2.waitKey()

Note: You could also change the sort direction such as right-to-left or top-to-bottom and so on

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