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How to remove black part from the image?

I have stitched two images together using OpenCV functions and C++. Now I am facing a problem that the final image contains a large black part.

The final image should be a rectangle containing the effective part. My image is the following:

enter image description here

How can I remove the black section?

like image 900
deepak_k Avatar asked May 17 '12 09:05

deepak_k


2 Answers

mevatron's answer is one way where amount of black region is minimised while retaining full image.

Another option is removing complete black region where you also loose some part of image, but result will be a neat looking rectangular image. Below is the Python code.

Here, you find three main corners of the image as below:

enter image description here

I have marked those values. (1,x2), (x1,1), (x3,y3). It is based on the assumption that your image starts from (1,1).

Code :

First steps are same as mevatron's. Blur the image to remove noise, threshold the image, then find contours.

import cv2
import numpy as np

img = cv2.imread('office.jpg')
img = cv2.resize(img,(800,400))

gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
gray = cv2.medianBlur(gray,3)

ret,thresh = cv2.threshold(gray,1,255,0)
contours,hierarchy = cv2.findContours(thresh,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)

Now find the biggest contour which is your image. It is to avoid noise in case if any (Most probably there won't be any). Or you can use mevatron's method.

max_area = -1
best_cnt = None

for cnt in contours:
    
    area = cv2.contourArea(cnt)
    if area > max_area:
        max_area = area
        best_cnt = cnt

Now approximate the contour to remove unnecessary points in contour values found, but it preserve all corner values.

approx = cv2.approxPolyDP(best_cnt,0.01*cv2.arcLength(best_cnt,True),True)

Now we find the corners.

First, we find (x3,y3). It is farthest point. So x3*y3 will be very large. So we find products of all pair of points and select the pair with maximum product.

far = approx[np.product(approx,2).argmax()][0]

Next (1,x2). It is the point where first element is one,then second element is maximum.

ymax = approx[approx[:,:,0]==1].max()

Next (x1,1). It is the point where second element is 1, then first element is maximum.

xmax = approx[approx[:,:,1]==1].max()

Now we find the minimum values in (far.x,xmax) and (far.y, ymax)

x = min(far[0],xmax)
y = min(far[1],ymax)

If you draw a rectangle with (1,1) and (x,y), you get result as below:

enter image description here

So you crop the image to correct rectangular area.

img2 = img[:y,:x].copy()

Below is the result:

enter image description here

See, the problem is that you lose some parts of the stitched image.

like image 154
Abid Rahman K Avatar answered Sep 21 '22 18:09

Abid Rahman K


You can do this with threshold, findContours, and boundingRect.

So, here is a quick script doing this with the python interface.

stitched = cv2.imread('stitched.jpg', 0)
(_, mask) = cv2.threshold(stitched, 1.0, 255.0, cv2.THRESH_BINARY);

# findContours destroys input
temp = mask.copy()
(contours, _) = cv2.findContours(temp, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)

# sort contours by largest first (if there are more than one)
contours = sorted(contours, key=lambda contour:len(contour), reverse=True)
roi = cv2.boundingRect(contours[0])

# use the roi to select into the original 'stitched' image
stitched[roi[1]:roi[3], roi[0]:roi[2]]

Ends up looking like this: enter image description here

NOTE : Sorting may not be necessary with raw imagery, but using the compressed image caused some compression artifacts to show up when using a low threshold, so that is why I post-processed with sorting.

Hope that helps!

like image 29
mevatron Avatar answered Sep 21 '22 18:09

mevatron