Filling an outlined circle

Question

I have a collection of images with a circle drawn as a white outline. However, I want to fill the complete circle with white color. What is a fast way to do it? The following is the sample of the image:

I have tried using nested loops to achieve this, but it takes a lot of time, and I have around 1.5 million images. The following is my code:

roundRobinIndex = 0
new_image = np.zeros((img_w, img_h))
for row in range(540):
    for column in range(800):
        if image[row,column] == 255:
            roundRobinIndex = (roundRobinIndex + 1) % 2
        if roundRobinIndex == 1:
            new_image[row, column] = 255

Answer 1

Use cv2.fillPoly() to fill the circle contour

import cv2

image = cv2.imread('1.png', 0)
thresh = cv2.threshold(image, 0, 255, cv2.THRESH_BINARY + cv2.THRESH_OTSU)[1]
cnts = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]
cv2.fillPoly(image, cnts, [255,255,255])

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

Note: The Otsu's threshold could be removed for slightly faster performance since the input image is already a binary image, you could directly find contours on the grayscale image

Answer 2

I tried finding the bounding box of the white outline, and getting its centre, then floodfilling with white from there outwards.

#!/usr/bin/env python3

import cv2

def findfill(image):
    thresh = cv2.threshold(image, 0, 255, cv2.THRESH_BINARY + cv2.THRESH_OTSU)[1]
    cnts = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
    cnts = cnts[0] if len(cnts) == 2 else cnts[1]
    cv2.fillPoly(image, cnts, [255,255,255])

def me(image):
    x,y,w,h = cv2.boundingRect(image)
    cv2.floodFill(image,None,(int(x+w/2),int(y+h/2)),255)
    return image

image = cv2.imread('BLYmz.png', 0)

%timeit findfill(image)
%timeit me(image)

This seems to give the same results and run 2.5x faster:

findfill
810 µs ± 2.94 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

me
343 µs ± 1.06 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Of course, if you have 1.5 million to do, I would recommend some parallel processing too :-)