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pythonopencvcaptchacontour

OpenCV findContours() just returning one external contour


I'm trying to isolate letters in a captcha, I managed to filter a captcha and that result in this black and white image:

enter image description here

But when I tried to separate the letters with findContours method of OpenCV it just found a external contour that wraps my entire image, resulting in this image (black contour outside image).

enter image description here

I'm using this code with Python 3 and OpenCV 3.4.2.17:

img = threshold_image(img)
cv2.imwrite("images/threshold.png", img)

image, contours, _ = cv2.findContours(img, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_NONE)

for i, contour in enumerate(contours):
    area = cv2.contourArea(contour)
    cv2.drawContours(img, contours, i, (0, 0, 0), 3)

cv2.imwrite('images/output3.png', img)

I just want my final result is 5 contours outside each character.


Solution

  • The contour to be extracted should be in white, background being black. I have modified your code a bit, eliminated the lines which were not adding any value.

    import cv2
    img = cv2.imread('image_to_be_read',0)
    backup = img.copy()   #taking backup of the input image
    backup = 255-backup    #colour inversion
    

    I am using RETR_TREE as the contour retrieval mode, which retrieves all the contours and creates full family hierarchy list. Please find the documentation for the same here

    _, contours, _ = cv2.findContours(backup, cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)
    

    In opencv4, the finContours method has been changed. Please use:

    contours, _ = cv2.findContours(backup, cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)
    

    Then iterate through contours and draw the rectangle around the contours

    for i, contour in enumerate(contours):
         x, y, w, h = cv2.boundingRect(contour)
         cv2.rectangle(img, (x, y), (x + w, y + h), (0, 255, 0), 1)
    

    Save the image

    cv2.imwrite('output3.png', img)
    

    I got result which looks like this -

    enter image description here