I use OpenCV and Python and I want to remove the "+" sign that is repeated on my image. The following is an example of an image in question.  The goal is to produce the same image, but with the "+" signs removed. How can I achieve this? I've tried using the below code to achieve this. img = cv2.imread(image_path) gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY) img_inverted = cv2.bitwise_not(gray) thresh = cv2.adaptiveThreshold(img_inverted, 255, cv2.ADAPTIVE_THRESH_MEAN_C, cv2.THRESH_BINARY, 15, -2) cross_structure = cv2.getStructuringElement(cv2.MORPH_CROSS, (3, 3)) detected_cross = cv2.morphologyEx(thresh, cv2.MORPH_OPEN, cross_structure) mask = detected_cross kernel = cv2.getStructuringElement(cv2.MORPH_RECT, (3, 3)) dilated_mask = cv2.dilate(mask, kernel, iterations=1) output_telea = cv2.inpaint(img, dilated_mask, 3, cv2.INPAINT_TELEA)  But it seems like the detected_cross isn't really detecting the expected objects, and as a result, it returns the same threshold image. This is the result I end up with. 
This is basically implementing the comment of fmw42; that is, create the mask by thresholding for white regions (also I increased your dilation to 2 iterations for a visually slightly better result): import cv2 import numpy as np image_path = ... # TODO: adjust as necessary img = cv2.imread(image_path) mask = (cv2.cvtColor(img, cv2.COLOR_BGR2GRAY) > 250).astype(np.uint8) kernel = cv2.getStructuringElement(cv2.MORPH_RECT, (3, 3)) dilated_mask = cv2.dilate(mask, kernel, iterations=2) output_telea = cv2.inpaint(img, dilated_mask, 3, cv2.INPAINT_TELEA) cv2.imshow("result", output_telea) cv2.waitKey(0) cv2.destroyAllWindows()  Result:  In general, I agree with Cris Luengo's comments though:  If you have multiple such images where the crosses are always in the same place, create a mask once (possibly manually) and reuse it. Even better, try to remove the crosses at image acquisition time (i.e. don't acquire them in the first place) rather than as a post-processing step. That is to say, if you have control over the acquisition process, of course. 