How to clear numbers from the image using openCV? - python-3.x

Question

I'm trying to remove numbers which are laying inside the circular part of image, numbers are in black in color and background varies between red,yellow, blue and green.
I am using opencv to remove those numbers. I used a mask which extracts numbers from image, with help of cv2.inpaint tried to remove those numbers from images.
For my further analysis I required to have clear image. But my current approach gives distorted image and numbers are not completely removed.
I tried changing the threshold values, lowering will neglect numbers from dark shaded area such as from green and red.
import cv2
img = cv2.imread('scan_1.jpg')
mask = cv2.threshold(img,50,255,cv2.THRESH_BINARY_INV)[1][:,:,0]
cv2.imshow('mask', mask)
cv2.waitKey(0)
cv2.destroyAllWindows()
dst = cv2.inpaint(img, mask, 5, cv2.INPAINT_TELEA)
cv2.imshow('dst',dst)
cv2.waitKey(0)
cv2.destroyAllWindows()
cv2.imwrite('ost_1.jpg',dst)
Input images: a) scan_1.jpg
b) scan_2.jpg
Output images: a) ost_1.jpg
b) ost_2.jpg
Expected Image: Circles can ignored, but something similar to it is required.

Answer 1

Here is my attempt, a better/easier solution might be acquired if you do not care about preserving texts outside of your circle.
import cv2
import numpy as np
# connectivity method used for finding connected components, 4 vs 8
CONNECTIVITY = 4
# HSV threshold for finding black pixels
H_THRESHOLD = 179
S_THRESHOLD = 255
V_THRESHOLD = 150
# read image
img = cv2.imread("a1.jpg")
img_height = img.shape[0]
img_width = img.shape[1]
# save a copy for creating resulting image
result = img.copy()
# convert image to grayscale
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
# found the circle in the image
circles = cv2.HoughCircles(gray, cv2.HOUGH_GRADIENT, 1.7, minDist= 100, param1 = 48, param2 = 100, minRadius=70, maxRadius=100)
# draw found circle, for visual only
circle_output = img.copy()
# check if we found exactly 1 circle
num_circles = len(circles)
print("Number of found circles:{}".format(num_circles))
if (num_circles != 1):
print("invalid number of circles found ({}), should be 1".format(num_circles))
exit(0)
# save center position and radius of found circle
circle_x = 0
circle_y = 0
circle_radius = 0
if circles is not None:
# convert the (x, y) coordinates and radius of the circles to integers
circles = np.round(circles[0, :]).astype("int")
for (x, y, radius) in circles:
circle_x, circle_y, circle_radius = (x, y, radius)
cv2.circle(circle_output, (circle_x, circle_y), circle_radius, (255, 0, 0), 4)
print("circle center:({},{}), radius:{}".format(x,y,radius))
# keep a median filtered version of image, will be used later
median_filtered = cv2.medianBlur(img, 21)
# Convert BGR to HSV
hsv = cv2.cvtColor(img, cv2.COLOR_BGR2HSV)
# define range of black color in HSV
lower_val = np.array([0,0,0])
upper_val = np.array([H_THRESHOLD,S_THRESHOLD,V_THRESHOLD])
# Threshold the HSV image to get only black colors
mask = cv2.inRange(hsv, lower_val, upper_val)
# find connected components
components = cv2.connectedComponentsWithStats(mask, CONNECTIVITY, cv2.CV_32S)
# apply median filtering to found components
#centers = components[3]
num_components = components[0]
print("Number of found connected components:{}".format(num_components))
labels = components[1]
stats = components[2]
for i in range(1, num_components):
left = stats[i, cv2.CC_STAT_LEFT] - 10
top = stats[i, cv2.CC_STAT_TOP] - 10
width = stats[i, cv2.CC_STAT_WIDTH] + 10
height = stats[i, cv2.CC_STAT_HEIGHT] + 10
# iterate each pixel and replace them if
#they are inside circle
for row in range(top, top+height+1):
for col in range(left, left+width+1):
dx = col - circle_x
dy = row - circle_y
if (dx*dx + dy*dy <= circle_radius * circle_radius):
result[row, col] = median_filtered[row, col]
# smooth the image, may be necessary?
#result = cv2.blur(result, (3,3))
# display image(s)
cv2.imshow("img", img)
cv2.imshow("gray", gray)
cv2.imshow("found circle:", circle_output)
cv2.imshow("mask", mask)
cv2.imshow("result", result)
cv2.waitKey(0)
cv2.destroyAllWindows()
Result for a1: