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Replace irregular shapes with one white pixel

I have some binary images (only 0 or 255 values) with many irregular shapes like this:
I would like to replace each shape with one white pixel in the center of the shape. The rest of the pixels should be black.
What is the best way to do that?

You can do it like this:
load image and ensure it is thresholded
find contours and iterate over them finding centroid of each
If you use OpenCV that might look something like this:
#!/usr/bin/env python3
import cv2
import numpy as np
# Load image as greyscale
im = cv2.imread('irregular.png', cv2.IMREAD_GRAYSCALE)
# Make empty black output image, same size
result = np.zeros_like(im)
# Ensure binary
_, im = cv2.threshold(im,127,255,0)
# Find contours and iterate over them
_, contours, hierarchy = cv2.findContours(im,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
for i, c in enumerate(contours):
# Calculate moments for this contour to get centroid
M = cv2.moments(c)
# Calculate x,y coordinates of centre
cX = int(M["m10"] / M["m00"])
cY = int(M["m01"] / M["m00"])
# Mark centroid in white, and tell user
result[cY,cX] = 255
print(f'Centroid {i+1} at location: {cX},{cY}')
# Save output file
cv2.imwrite('result.png', result)
Sample Output
Centroid 1 at location: 191,354
Centroid 2 at location: 572,185

I am trying to measure land plot area Using OpenCV in Python

SO far I have been able to perform medianBlur and Edge Detection, Now I want to o further remove noise from the image, Matlabs region **property functions ** was used to remove all white regions that had a total pixel area of less than the mean pixel area value. How can I implement this on python
import matplotlib.image as mpimg
import numpy as np
import cv2
import os
import math
from collections import defaultdict
from matplotlib import pyplot as plt
import imutils
#import generalized_hough
gray = cv2.cvtColor(image, cv2.COLOR_RGB2GRAY)
print(gray.shape)
blur = cv2.bilateralFilter(gray,9,75,75)
median = cv2.medianBlur(gray,5)
# display input and output image
titles = ["bilateral Smoothing","median bulr"]
images = [ blur, median]
plt.figure(figsize=(20, 20))
for i in range(2):
plt.subplot(1,2,i+1)
plt.imshow(images[i])
plt.title(titles[i])
plt.xticks([]), plt.yticks([])
plt.show()
sobelX = cv2.Sobel(median,cv2.cv2.CV_64F, 1, 0)
sobelY = cv2.Sobel(median,cv2.cv2.CV_64F, 0,1)
sobelX = np.uint8(np.absolute(sobelX))
sobelY = np.uint8(np.absolute(sobelY))
SobelCombined = cv2.bitwise_or(sobelX,sobelY)
cv2.imshow('img', SobelCombined)
cv2.waitKey(0)
cv2.destroyAllWindows()
Here is a Matlab code that works for the same task.
close all
%upload image of farm
figure,
farm = imread('small_farms.JPG');%change this to the file path of image
imshow(farm);%this shows the original image
%convert the image to grayscale for 2D manipulation
gfarm = rgb2gray(farm);
figure,
imshow(gfarm);%show grayscaled image
%median filters take a m*n area around a coordinate and
%find the median pixel value and set that coordinate to that
%pixel value. It's a method of removing noise or details in an
%image. may want to tune dimensions of filter.
A = medfilt2(gfarm,[4 4]);
figure,
imshow(A);
%perform a logarithmic edge detection filter,
%this picks out the edges of the image, log setting
%was found to wrok best, although 'Sobel' can also be tried
B = edge(A,'log');
%show results of the edge filter
figure,
imshow(B,[]);
%find the areas of the lines made
areas = regionprops(B,'Area');
%find the mean and one standard deviation
men = mean([areas.Area])+0*std([areas.Area]);
%find max pixel area
big = max([areas.Area]);
%remove regions that are too small
C = bwpropfilt(B,'Area',[men big]);
%perform a dilation on the remaining pixels, this
%helps fill in gaps. The size and shape of the dilation
%can be tuned below.
SE = strel('square',4);
C = imdilate(C,SE);
areas2 = regionprops(C,'Area');
%place white border around image to find areas of farms
%that go off the picture
[h,w] = size(C);
C(1,:) = 1;
C(:,1) = 1;
C(h,:) = 1;
C(:,w) = 1;
C = C<1;
%fill in holes
C = imfill(C,'holes');
%show final processed image
figure,imshow(C);
%the section below is for display purpose
%it creates the boundaries of the image and displays them
%in a rainbow fashion
figure,
[B,L,n,A] = bwboundaries(C,'noholes');
imshow(label2rgb(L, #jet, [.5 .5 .5]))
hold on
for k = 1:length(B)
boundary = B{k};
plot(boundary(:,2), boundary(:,1), 'w', 'LineWidth', 2)
end
%The section below prints out the areas of each found
%region by pixel values. These values need to be scaled
%by the real measurements of the images to get relevant
%metrics
centers = regionprops(C,'Centroid','Area');
for k=1:length(centers)
if(centers(k).Area > mean([centers.Area])-std([areas.Area]))
text(centers(k).Centroid(1),centers(k).Centroid(2),string(centers(k).Area));
end
end

Automatic Skew correction using opencv [duplicate]

This question already has answers here:
Python OpenCV skew correction for OCR
(3 answers)
Closed 10 months ago.
I want a way to automatically detect and correct skew of a image of a receipt,
I tried to find variance between the rows for various angles of rotation and choose the angle which has the the maximum variance.
To calculate variance I did the following:
1.For each row I calculated the sum of the pixels values and stored it in a list.
2.Found the the variance of the list using np.var(list)
src = cv.imread(f_name, cv.IMREAD_GRAYSCALE)
blurred=median = cv.medianBlur(src,9)
ret,thresh2 = cv.threshold(src,127,255,cv.THRESH_BINARY_INV)
height, width = thresh2.shape[:2]
print(height,width)
res=[-1,0]
for angle in range(0,100,10):
rotated_temp=deskew(thresh2,angle)
cv.imshow('rotated_temp',rotated_temp)
cv.waitKey(0)
height,width=rotated_temp.shape[:2]
li=[]
for i in range(height):
sum=0
for j in range(width):
sum+=rotated_temp[i][j]
li.append(sum)
curr_variance=np.var(li)
print(curr_variance,angle)
if(curr_variance>res[0]):
res[0]=curr_variance
res[1]=angle
print(res)
final_rot=deskew(src,res[1])
cv.imshow('final_rot',final_rot)
cv.waitKey(0)
However the variance for a skewed image is coming to be more than the properly aligned image,is there any way to correct this
variance for the horizontal text aligned image(required):122449908.009789
variance for the vertical text aligned image :1840071444.404522
I have tried using HoughLines However since the spacing between the text is too less vertical lines are detected,hence this also fails
Any modifications or other approaches are appreciated

Working code for skew correction
import matplotlib.pyplot as plt
import numpy as np
from PIL import Image as im
from scipy.ndimage import interpolation as inter
input_file = r'E:\flaskV8\test1.jpg'
img = im.open(input_file)
convert to binary
wd, ht = img.size
pix = np.array(img.convert('1').getdata(), np.uint8)
bin_img = 1 - (pix.reshape((ht, wd)) / 255.0)
plt.imshow(bin_img, cmap='gray')
plt.savefig(r'E:\flaskV8\binary.png')
def find_score(arr, angle):
data = inter.rotate(arr, angle, reshape=False, order=0)
hist = np.sum(data, axis=1)
score = np.sum((hist[1:] - hist[:-1]) ** 2)
return hist, score
delta = 1
limit = 5
angles = np.arange(-limit, limit+delta, delta)
scores = []
for angle in angles:
hist, score = find_score(bin_img, angle)
scores.append(score)
best_score = max(scores)
best_angle = angles[scores.index(best_score)]
print('Best angle: {}'.format(best_angle))
data = inter.rotate(bin_img, best_angle, reshape=False, order=0)
img = im.fromarray((255 * data).astype("uint8")).convert("RGB")
img.save(r'E:\flaskV8\skew_corrected.png')

finding edge in tilted image with Canny

I'm trying to find the tilt angle in a series of images which look like the created example data below. There should be a clear edge which is visible by eye. However I'm struggling in extracting the edges so far. Is Canny the right way of finding the edge here or is there a better way of finding the edge?
import cv2 as cv
import numpy as np
import matplotlib.pyplot as plt
from scipy.ndimage.filters import gaussian_filter
# create data
xvals = np.arange(0,2000)
yvals = 10000 * np.exp((xvals - 1600)/200) + 100
yvals[1600:] = 100
blurred = gaussian_filter(yvals, sigma=20)
# create image
img = np.tile(blurred,(2000,1))
img = np.swapaxes(img,0,1)
# rotate image
rows,cols = img.shape
M = cv.getRotationMatrix2D((cols/2,rows/2),3.7,1)
img = cv.warpAffine(img,M,(cols,rows))
# convert to uint8 for Canny
img_8 = cv.convertScaleAbs(img,alpha=(255.0/65535.0))
fig,ax = plt.subplots(3)
ax[0].plot(xvals,blurred)
ax[1].imshow(img)
# find edge
ax[2].imshow(cv.Canny(img_8, 20, 100, apertureSize=5))

You can find the angle by transforming your image to binary (cv2.threshold(cv2.THRESH_BINARY)) then search for contours.
When you locate your contour (line) then you can fit a line on your contour cv2.fitLine() and get two points of your line. My math is not very good but I think that in linear equation the formula goes f(x) = k*x + n and you can get k out of those two points (k = (y2-y1)/(x2-x1)) and finally the angle phi = arctan(k). (If I'm wrong please correct it)
You can also use the rotated bounding rectangle - cv2.minAreaRect() - which already returns the angle of the rectangle (rect = cv2.minAreaRect() --> rect[2]). Hope it helps. Cheers!
Here is an example code:
import cv2
import numpy as np
import math
img = cv2.imread('angle.png')
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
ret, threshold = cv2.threshold(gray,170,255,cv2.THRESH_BINARY)
im, contours, hierarchy = cv2.findContours(threshold,cv2.RETR_TREE,cv2.CHAIN_APPROX_NONE)
for c in contours:
area = cv2.contourArea(c)
perimeter = cv2.arcLength(c, False)
if area < 10001 and 100 < perimeter < 1000:
# first approach - fitting line and calculate with y=kx+n --> angle=tan^(-1)k
rows,cols = img.shape[:2]
[vx,vy,x,y] = cv2.fitLine(c, cv2.DIST_L2,0,0.01,0.01)
lefty = int((-x*vy/vx) + y)
righty = int(((cols-x)*vy/vx)+y)
cv2.line(img,(cols-1,righty),(0,lefty),(0,255,0),2)
(x1, y1) = (cols-1, righty)
(x2, y2) = (0, lefty)
k = (y2-y1)/(x2-x1)
angle = math.atan(k)*180/math.pi
print(angle)
#second approch - cv2.minAreaRect --> returns center (x,y), (width, height), angle of rotation )
rect = cv2.minAreaRect(c)
box = cv2.boxPoints(rect)
box = np.int0(box)
cv2.drawContours(img,[box],0,(0,0,255),2)
print(rect[2])
cv2.imshow('img2', img)
Original image:
Output:
-3.8493663478518627
-3.7022125720977783

tribol,
it seems like you can take the gradient image G = |Gx| + |Gy| (normalize it to some known range), calc its Histogram and take the top bins of it. it will give you approx mask of the line. Then you can do line fitting. It'll give you a good initial guess.

A very simple way of doing it is as follows... adjust my numbers to suit your knowledge of the data.
Normalise your image to a scale of 0-255.
Choose two points A and B, where A is 10% of the image width in from the left side and B is 10% in from the right side. The distance AB is now 0.8 x 2000, or 1600 px.
Go North from point A sampling your image till you exceed some sensible threshold that means you have met the tilted line. Note the Y value at this point, as YA.
Do the same, going North from point B till you meet the tilted line. Note the Y value at this point, as YB.
The angle you seek is:
tan-1((YB-YA)/1600)

Thresholding as suggested by kavko didn't work that well, as the intensity varied from image to image (I could of course consider the histogram for each image to imrove this approach). I ended up with taking the maximum of the gradient in the y-direction:
def rotate_image(image):
blur = ndimage.gaussian_filter(image, sigma=10) # blur image first
grad = np.gradient(blur, axis= 0) # take gradient along y-axis
grad[grad>10000]=0 # filter unreasonable high values
idx_maxline = np.argmax(grad, axis=0) # get y-indices of max slope = indices of edge
mean = np.mean(idx_maxline)
std = np.std(idx_maxline)
idx = np.arange(idx_maxline.shape[0])
idx_filtered = idx[(idx_maxline < mean+std) & (idx_maxline > mean - std)] # filter positions where highest slope is at different position(blobs)
slope, intercept, r_value, p_value, std_err = stats.linregress(idx_filtered, idx_maxline[idx_filtered])
out = ndimage.rotate(image,slope*180/np.pi, reshape = False)
return out
out = rotate_image(img)
plt.imshow(out)

Python - Perspective transform for OpenCV from a rotation angle

I'm working on depth map with OpenCV. I can obtain it but it is reconstructed from the left camera origin and there is a little tilt of this latter and as you can see on the figure, the depth is "shifted" (the depth should be close and no horizontal gradient):
I would like to express it as with a zero angle, i try with the warp perspective function as you can see below but i obtain a null field...
P = np.dot(cam,np.dot(Transl,np.dot(Rot,A1)))
dst = cv2.warpPerspective(depth, P, (2048, 2048))
with :
#Projection 2D -> 3D matrix
A1 = np.zeros((4,3))
A1[0,0] = 1
A1[0,2] = -1024
A1[1,1] = 1
A1[1,2] = -1024
A1[3,2] = 1
#Rotation matrice around the Y axis
theta = np.deg2rad(5)
Rot = np.zeros((4,4))
Rot[0,0] = np.cos(theta)
Rot[0,2] = -np.sin(theta)
Rot[1,1] = 1
Rot[2,0] = np.sin(theta)
Rot[2,2] = np.cos(theta)
Rot[3,3] = 1
#Translation matrix on the X axis
dist = 0
Transl = np.zeros((4,4))
Transl[0,0] = 1
Transl[0,2] = dist
Transl[1,1] = 1
Transl[2,2] = 1
Transl[3,3] = 1
#Camera Intrisecs matrix 3D -> 2D
cam = np.concatenate((C1,np.zeros((3,1))),axis=1)
cam[2,2] = 1
P = np.dot(cam,np.dot(Transl,np.dot(Rot,A1)))
dst = cv2.warpPerspective(Z0_0, P, (2048*3, 2048*3))
EDIT LATER :
You can download the 32MB field dataset here: https://filex.ec-lille.fr/get?k=cCBoyoV4tbmkzSV5bi6. Then, load and view the image with:
from matplotlib import pyplot as plt
import numpy as np
img = np.load('testZ0.npy')
plt.imshow(img)
plt.show()

I have got a rough solution in place. You can modify it later.
I used the mouse handling operations available in OpenCV to crop the region of interest in the given heatmap.
(Did I just say I used a mouse to crop the region?) Yes, I did. To learn more about mouse functions in OpenCV SEE THIS. Besides, there are many other SO questions that can help you in this regard.:)
Using those functions I was able to obtain the following:
Now to your question of removing the tilt. I used the homography principal by taking the corner points of the image above and using it on a 'white' image of a definite size. I used the cv2.findHomography() function for this.
Now using the cv2.warpPerspective() function in OpenCV, I was able to obtain the following:
Now you can the required scale to this image as you wanted.
CODE:
I have also attached some snippets of code for your perusal:
#First I created an image of white color of a definite size
back = np.ones((435, 379, 3)) # size
back[:] = (255, 255, 255) # white color
Next I obtained the corner points pts_src on the tilted image below :
pts_src = np.array([[25.0, 2.0],[403.0,22.0],[375.0,436.0],[6.0,433.0]])
I wanted the points above to be mapped to the points 'pts_dst' given below :
pts_dst = np.array([[2.0, 2.0], [379.0, 2.0], [379.0, 435.0],[2.0, 435.0]])
Now I used the principal of homography:
h, status = cv2.findHomography(pts_src, pts_dst)
Finally I mapped the original image to the white image using perspective transform.
fin = cv2.warpPerspective(img, h, (back.shape[1],back.shape[0]))
# img -> original tilted image.
# back -> image of white color.
Hope this helps! I also got to learn a great deal from this question.
Note: The points fed to the 'cv2.findHomography()' must be in float.
For more info on Homography , visit THIS PAGE