I'm using Hough Lines to do corner detection for this image. i plan to find the intersection of the lines as the corner.
This is the image.
Unfortunately, Hough return lots of lines for each line I expect
How do I tune the Hough Lines so there is only four lines each corresponds to actual line on the image?
OpenCVs hough transform really could use some better Non-Maximum Suppression. Without that, you get this phenomenon of duplicate lines. Unfortunately I know of no easy way to tune that, besides reimplementing your own hough transform. (Which is a valid option. Hough transform is fairly simple)
Fortunately it is easy to fix in post-processing:
For the non-probabilistic hough transform, OpenCv will return the lines in order of their confidence, with the strongest line first. So simply take the first four lines that differ strongly in either rho or theta.
so, add the first line found by HoughLines into a new List: strong_lines
for each line found by HoughLines:
test whether its rho and theta are close to any strong_line (e.g. rho is within 50 pixels and theta is within 10° of the other line)
if not, put it into the list of strong_lines
if you have found 4 strong_lines, break
I implemented the approach described by HugoRune and though I would share my code as an example of how I implemented this. I used a tolerance of 5 degrees and 10 pixels.
strong_lines = np.zeros([4,1,2])
minLineLength = 2
maxLineGap = 10
lines = cv2.HoughLines(edged,1,np.pi/180,10, minLineLength, maxLineGap)
n2 = 0
for n1 in range(0,len(lines)):
for rho,theta in lines[n1]:
if n1 == 0:
strong_lines[n2] = lines[n1]
n2 = n2 + 1
else:
if rho < 0:
rho*=-1
theta-=np.pi
closeness_rho = np.isclose(rho,strong_lines[0:n2,0,0],atol = 10)
closeness_theta = np.isclose(theta,strong_lines[0:n2,0,1],atol = np.pi/36)
closeness = np.all([closeness_rho,closeness_theta],axis=0)
if not any(closeness) and n2 < 4:
strong_lines[n2] = lines[n1]
n2 = n2 + 1
EDIT: The code was updated to reflect the comment regarding a negative rho value
Collect the intersection of all line
for (int i = 0; i < lines.size(); i++)
{
for (int j = i + 1; j < lines.size(); j++)
{
cv::Point2f pt = computeIntersectionOfTwoLine(lines[i], lines[j]);
if (pt.x >= 0 && pt.y >= 0 && pt.x < image.cols && pt.y < image.rows)
{
corners.push_back(pt);
}
}
}
You can google the algorithm to find the intersection of two lines.
Once you collect all the intersection points you can easily determine the min max which will give you top-left and bottom right points. From these two points you can easily get the rectangle.
Here Sorting 2d point array to find out four corners & http://opencv-code.com/tutorials/automatic-perspective-correction-for-quadrilateral-objects/ Refer these two links.
Here is a complete solution written in python 2.7.x using OpenCV 2.4.
It is based on ideas from this thread.
Method: Detect all lines. Assume that the Hough function returns highest ranked lines first. Filter the lines to keep those that are separated by some minimum distance and/or angle.
Image of all Hough lines:
https://i.ibb.co/t3JFncJ/all-lines.jpg
Filtered lines:
https://i.ibb.co/yQLNxXT/filtered-lines.jpg
Code:
http://codepad.org/J57oVIzs
"""
Detect the best 4 lines for a rounded rectangle.
"""
import numpy as np
import cv2
input_image = cv2.imread("image.jpg")
def drawLines(img, lines):
"""
Draw lines on an image
"""
for line in lines:
for rho,theta in line:
a = np.cos(theta)
b = np.sin(theta)
x0 = a*rho
y0 = b*rho
x1 = int(x0 + 1000*(-b))
y1 = int(y0 + 1000*(a))
x2 = int(x0 - 1000*(-b))
y2 = int(y0 - 1000*(a))
cv2.line(img, (x1,y1), (x2,y2), (0,0,255), 1)
input_image_grey = cv2.cvtColor(input_image, cv2.COLOR_BGR2GRAY)
edged = input_image_grey
rho = 1 # 1 pixel
theta = 1.0*0.017 # 1 degree
threshold = 100
lines = cv2.HoughLines(edged, rho, theta, threshold)
# Fix negative angles
num_lines = lines.shape[1]
for i in range(0, num_lines):
line = lines[0,i,:]
rho = line[0]
theta = line[1]
if rho < 0:
rho *= -1.0
theta -= np.pi
line[0] = rho
line[1] = theta
# Draw all Hough lines in red
img_with_all_lines = np.copy(input_image)
drawLines(img_with_all_lines, lines)
cv2.imshow("Hough lines", img_with_all_lines)
cv2.waitKey()
cv2.imwrite("all_lines.jpg", img_with_all_lines)
# Find 4 lines with unique rho & theta:
num_lines_to_find = 4
filtered_lines = np.zeros([1, num_lines_to_find, 2])
if lines.shape[1] < num_lines_to_find:
print("ERROR: Not enough lines detected!")
# Save the first line
filtered_lines[0,0,:] = lines[0,0,:]
print("Line 1: rho = %.1f theta = %.3f" % (filtered_lines[0,0,0], filtered_lines[0,0,1]))
idx = 1 # Index to store the next unique line
# Initialize all rows the same
for i in range(1,num_lines_to_find):
filtered_lines[0,i,:] = filtered_lines[0,0,:]
# Filter the lines
num_lines = lines.shape[1]
for i in range(0, num_lines):
line = lines[0,i,:]
rho = line[0]
theta = line[1]
# For this line, check which of the existing 4 it is similar to.
closeness_rho = np.isclose(rho, filtered_lines[0,:,0], atol = 10.0) # 10 pixels
closeness_theta = np.isclose(theta, filtered_lines[0,:,1], atol = np.pi/36.0) # 10 degrees
similar_rho = np.any(closeness_rho)
similar_theta = np.any(closeness_theta)
similar = (similar_rho and similar_theta)
if not similar:
print("Found a unique line: %d rho = %.1f theta = %.3f" % (i, rho, theta))
filtered_lines[0,idx,:] = lines[0,i,:]
idx += 1
if idx >= num_lines_to_find:
print("Found %d unique lines!" % (num_lines_to_find))
break
# Draw filtered lines
img_with_filtered_lines = np.copy(input_image)
drawLines(img_with_filtered_lines, filtered_lines)
cv2.imshow("Filtered lines", img_with_filtered_lines)
cv2.waitKey()
cv2.imwrite("filtered_lines.jpg", img_with_filtered_lines)
The above approach (proposed by #HugoRune's and implemented by #Onamission21) is correct but has a little bug. cv2.HoughLines may return negative rho and theta upto pi. Notice for example that the line (r0,0) is very close to the line (-r0,pi-epsilon) but they would not be found in the above closeness test.
I simply treated negative rhos by applying rho*=-1, theta-=pi before closeness calculations.
Related
I am currently trying to simulate an optical flow using the following equation:
Below is a basic example where I have a 7x7 image where the central pixel is illuminated. The velocity I am applying is a uniform x-velocity of 2.
using Interpolations
using PrettyTables
# Generate grid
nx = 7 # Image will be 7x7 pixels
x = zeros(nx, nx)
yy = repeat(1:nx, 1, nx) # grid of y-values
xx = yy' # grid of x-values
# In this example x is the image I in the above equation
x[(nx-1)÷2 + 1, (nx-1)÷2 + 1] = 1.0 # set central pixel equal to 1
# Initialize velocity
velocity = 2;
vx = velocity .* ones(nx, nx); # vx=2
vy = 0.0 .* ones(nx, nx); # vy=0
for t in 1:1
# create 2d grid interpolator of the image
itp = interpolate((collect(1:nx), collect(1:nx)), x, Gridded(Linear()));
# create 2d grid interpolator of vx and vy
itpvx = interpolate((collect(1:nx), collect(1:nx)), vx, Gridded(Linear()));
itpvy = interpolate((collect(1:nx), collect(1:nx)), vy, Gridded(Linear()));
∇I_x = Array{Float64}(undef, nx, nx); # Initialize array for ∇I_x
∇I_y = Array{Float64}(undef, nx, nx); # Initialize array for ∇I_y
∇vx_x = Array{Float64}(undef, nx, nx); # Initialize array for ∇vx_x
∇vy_y = Array{Float64}(undef, nx, nx); # Initialize array for ∇vy_y
for i=1:nx
for j=1:nx
# gradient of image in x and y directions
Gx = Interpolations.gradient(itp, i, j);
∇I_x[i, j] = Gx[2];
∇I_y[i, j] = Gx[1];
Gvx = Interpolations.gradient(itpvx, i, j) # gradient of vx in both directions
Gvy = Interpolations.gradient(itpvy, i, j) # gradient of vy in both directions
∇vx_x[i, j] = Gvx[2];
∇vy_y[i, j] = Gvy[1];
end
end
v∇I = (vx .* ∇I_x) .+ (vy .* ∇I_y) # v dot ∇I
I∇v = x .* (∇vx_x .+ ∇vy_y) # I dot ∇v
x = x .- (v∇I .+ I∇v) # I(x, y, t+dt)
pretty_table(x)
end
What I expect is that the illuminated pixel in x will shift two pixels to the right in x_predicted. What I am seeing is the following:
where the original illuminated pixel's value is moved to the neighboring pixel twice rather than being shifted two pixels to the right. I.e. the neighboring pixel goes from being 0 to 2 and the original pixel goes from a value of 1 to -1. I'm not sure if I'm messing up the equation or if I'm thinking of velocity in the wrong way here. Any ideas?
Without looking into it too deeply, I think there are a couple of potential issues here:
Violation of the Courant Condition
The code you originally posted (I've edited it now) simulates a single timestep. I would not expect a cell 2 units away from your source cell to be activated in a single timestep. Doing so would voilate the Courant condition. From wikipedia:
The principle behind the condition is that, for example, if a wave is moving across a discrete spatial grid and we want to compute its amplitude at discrete time steps of equal duration, then this duration must be less than the time for the wave to travel to adjacent grid points.
The Courant condition requires that uΔt/Δx <= 1 (for an explicit time-marching solver such as the one you've implemented). Plugging in u=2, Δt=1, Δx=1 gives 2, which is greater than 1, so you have a mathematical problem. The general way of fixing this problem is to make Δt smaller. You probably want something like:
x = x .- Δt*(v∇I .+ I∇v) # I(x, y, t+dt)
Missing gradients?
I'm a little concerned about what's going on here:
Gvx = Interpolations.gradient(itpvx, i, j) # gradient of vx in both directions
Gvy = Interpolations.gradient(itpvy, i, j) # gradient of vy in both directions
∇vx_x[i, j] = Gvx[2];
∇vy_y[i, j] = Gvy[1];
You're able to pull two gradients out of both Gvx and Gvy, but you're only using one from each of them. Does that mean you're throwing information away?
https://scicomp.stackexchange.com/ is likely to provide better help with this.
(There is a solid line at C and a faint line at T)
I want to detect the line at T. Currently I am using opencv to locate the qr code and rotate the image until the qr code is upright. Then I calculate the approximate location of the C and T mark by using the coordinates of the qr code. Then my code will scan along the y axis down and detect there are difference in the Green and Blue values.
My problem is, even if the T line is as faint as shown, it should be regarded as positive. How could I make a better detection?
I cropped out just the white strip since I assume you have a way of finding it already. Since we're looking for red, I changed to the LAB colorspace and looked on the "a" channel.
Note: all images of the strip have been transposed (np.transpose) for viewing convenience, it's not that way in the code.
the A channel
I did a linear reframe to improve the contrast
The image is super noisy. Again, I'm not sure if this is from the camera or the jpg compression. I averaged each row to smooth out some of the nonsense.
I graphed the intensities (x-vals were the row index)
Use a mean filter to smooth out the graph
I ran a mountain climber algorithm to look for peaks and valleys
And then I filtered for peaks with a climb greater than 10 (the second highest peak has a climb of 25.5, the third highest is 4.4).
Using these peaks we can determine that there are two lines and they are (about) here:
import cv2
import numpy as np
import matplotlib.pyplot as plt
# returns direction of gradient
# 1 if positive, -1 if negative, 0 if flat
def getDirection(one, two):
dx = two - one;
if dx == 0:
return 0;
if dx > 0:
return 1;
return -1;
# detects and returns peaks and valleys
def mountainClimber(vals, minClimb):
# init trackers
last_valley = vals[0];
last_peak = vals[0];
last_val = vals[0];
last_dir = getDirection(vals[0], vals[1]);
# get climbing
peak_valley = []; # index, height, climb (positive for peaks, negative for valleys)
for a in range(1, len(vals)):
# get current direction
sign = getDirection(last_val, vals[a]);
last_val = vals[a];
# if not equal, check gradient
if sign != 0:
if sign != last_dir:
# change in gradient, record peak or valley
# peak
if last_dir > 0:
last_peak = vals[a];
climb = last_peak - last_valley;
climb = round(climb, 2);
peak_valley.append([a, vals[a], climb]);
else:
# valley
last_valley = vals[a];
climb = last_valley - last_peak;
climb = round(climb, 2);
peak_valley.append([a, vals[a], climb]);
# change direction
last_dir = sign;
# filter out very small climbs
filtered_pv = [];
for dot in peak_valley:
if abs(dot[2]) > minClimb:
filtered_pv.append(dot);
return filtered_pv;
# run an mean filter over the graph values
def meanFilter(vals, size):
fil = [];
filtered_vals = [];
for val in vals:
fil.append(val);
# check if full
if len(fil) >= size:
# pop front
fil = fil[1:];
filtered_vals.append(sum(fil) / size);
return filtered_vals;
# averages each row (also gets graph values while we're here)
def smushRows(img):
vals = [];
h,w = img.shape[:2];
for y in range(h):
ave = np.average(img[y, :]);
img[y, :] = ave;
vals.append(ave);
return vals;
# linear reframe [min1, max1] -> [min2, max2]
def reframe(img, min1, max1, min2, max2):
copy = img.astype(np.float32);
copy -= min1;
copy /= (max1 - min1);
copy *= (max2 - min2);
copy += min2;
return copy.astype(np.uint8);
# load image
img = cv2.imread("strip.png");
# resize
scale = 2;
h,w = img.shape[:2];
h = int(h*scale);
w = int(w*scale);
img = cv2.resize(img, (w,h));
# lab colorspace
lab = cv2.cvtColor(img, cv2.COLOR_BGR2LAB);
l,a,b = cv2.split(lab);
# stretch contrast
low = np.min(a);
high = np.max(a);
a = reframe(a, low, high, 0, 255);
# smush and get graph values
vals = smushRows(a);
# filter and round values
mean_filter_size = 20;
filtered_vals = meanFilter(vals, mean_filter_size);
for ind in range(len(filtered_vals)):
filtered_vals[ind] = round(filtered_vals[ind], 2);
# get peaks and valleys
pv = mountainClimber(filtered_vals, 1);
# pull x and y values
pv_x = [ind[0] for ind in pv];
pv_y = [ind[1] for ind in pv];
# find big peaks
big_peaks = [];
for dot in pv:
if dot[2] > 10: # climb filter size
big_peaks.append(dot);
print(big_peaks);
# make plot points for the two best
tops_x = [dot[0] for dot in big_peaks];
tops_y = [dot[1] for dot in big_peaks];
# plot
x = [index for index in range(len(filtered_vals))];
fig, ax = plt.subplots()
ax.plot(x, filtered_vals);
ax.plot(pv_x, pv_y, 'og');
ax.plot(tops_x, tops_y, 'vr');
plt.show();
# draw on original image
h,w = img.shape[:2];
for dot in big_peaks:
y = int(dot[0] + mean_filter_size / 2.0); # adjust for mean filter cutting
cv2.line(img, (0, y), (w,y), (100,200,0), 2);
# show
cv2.imshow("a", a);
cv2.imshow("strip", img);
cv2.waitKey(0);
Edit:
I was wondering why the lines seemed so off, then I realized that I forgot to account for the fact that the meanFilter reduces the size of the list (it cuts from the front and back). I've updated to take that into account.
I am new to opencv-python. I have found the lines in image through houghtransformP. The lines drawn from hough transform are discontinued and are giving multiple lines. I need to draw only one line for the edges and find the 'distance' between lines which are found.
The output image is shown below
"""
Created on Fri Nov 8 11:41:16 2019
#author: romanth.chowan
"""
import cv2
import numpy as np
import math
def getSlopeOfLine(line):
xDis = line[0][2] - line[0][0]
if (xDis == 0):
return None
return (line[0][3] - line[0][1]) / xDis
if __name__ == '__main__':
inputFileName_ =r"C:\Users\romanth.chowan\Desktop\opencv\stent spec\2prox.jpeg"
img = cv2.imread(inputFileName_)
img1=cv2.GaussianBlur(img,(5,5),0)
gray = cv2.cvtColor(img1, cv2.COLOR_BGR2GRAY)
edges = cv2.Laplacian(gray,cv2.CV_8UC1) # Laplacian Edge Detection
lines = cv2.HoughLinesP(edges, 1, np.pi / 180, 300, 10, 10)
print(len(lines))
parallelLines = []
for a in lines:
for b in lines:
if a is not b:
slopeA = getSlopeOfLine(b)
slopeB = getSlopeOfLine(b)
if slopeA is not None and slopeB is not None:
if 0 <= abs(slopeA - slopeB) <= 10:
parallelLines.append({'lineA': a, 'lineB': b})
for pairs in parallelLines:
lineA = pairs['lineA']
lineB = pairs['lineB']
leftx, boty, rightx, topy = lineA[0]
cv2.line(img, (leftx, boty), (rightx, topy), (0, 0, 255), 2)
left_x, bot_y, right_x, top_y = lineB[0]
cv2.line(img, (left_x, bot_y), (right_x, top_y), (0, 0, 255), 2)
cv2.imwrite('linesImg.jpg', img)
output image after drawing lines:
It's mostly geometric task, not specific to OpenCV.
For each line you have two points (x1,y1) and (x2,y2) which are already used in your getSlopeOfLine(line) method.
You can denote each line in form:
ax + by + c = 0
To do that use two known line's points:
(y1 - y2)x + (x2 - x1)y + (x1y2 - x2y1) = 0
Note than parallel lines have same a and b while different c.
And than measure distance between any two of them (distance between non-parallel lines is equal to zero since they have a crossing point) :
d = abs(c2 - c1) / sqrt(a*a + b*b)
In Euclidean geometry line may be denoted in several ways and one may suit specific task better than another.
Currently you evaluate line's slope, from formula above we can get:
y = (-b / a)x - c / b
same to (b has another meaning now)
y = kx + b
Or using two line's points:
y = (x1 - x2) / (y1 - y2) * x + (x1y2 - x2y1)
Where k is line's slope (tan(alpha)) and b is shift.
Now you just match parallel lines (one with close k). You can take in account line's shift to merge several parallel lines into one.
I am implementing the Hough circle transform and trying my code on a binary image that contains only one circle circumference, however for any radius I try, I get the same number of accumulated points, here is the code:
y0array, x0array= np.nonzero(image1)
r=8
acc_cells = np.zeros((100,100), dtype=np.uint64)
for i in range( len(x0array)):
y0= y0array[i]
x0= x0array[i]
for angle in range(0,360):
b = int(y0 - (r * s[angle]) ) //s is an array of sine of angles from 0 to 360
a = int(x0 - (r * c[angle]) ) //c is an array of cosine of angles from 0 to 360
if a >= 0 and a < 100 and b >= 0 and b < 100:
acc_cells[a, b] += 1
acc_cell_max = np.amax(acc_cells)
print(r, acc_cell_max)
Why is this behaviour happening?
You have to find out the center of the circles as you did. you have to find each edge coordinates
You can check python implementation of hough circles in detectCircles function
https://github.com/PavanGJ/Circle-Hough-Transform/blob/master/main.py
Also, take a look at hough circle implementation of Matlab functions
http://www.mathworks.com/matlabcentral/fileexchange/4985-circle-detection-via-standard-hough-transform
function [y0detect,x0detect,Accumulator] = houghcircle(Imbinary,r,thresh)
%HOUGHCIRCLE - detects circles with specific radius in a binary image. This
%is just a standard implementaion of Hough transform for circles in order
%to show how this method works.
%
%Comments:
% Function uses Standard Hough Transform to detect circles in a binary image.
% According to the Hough Transform for circles, each pixel in image space
% corresponds to a circle in Hough space and vise versa.
% upper left corner of image is the origin of coordinate system.
%
%Usage: [y0detect,x0detect,Accumulator] = houghcircle(Imbinary,r,thresh)
%
%Arguments:
% Imbinary - A binary image. Image pixels with value equal to 1 are
% candidate pixels for HOUGHCIRCLE function.
% r - Radius of the circles.
% thresh - A threshold value that determines the minimum number of
% pixels that belong to a circle in image space. Threshold must be
% bigger than or equal to 4(default).
%
%Returns:
% y0detect - Row coordinates of detected circles.
% x0detect - Column coordinates of detected circles.
% Accumulator - The accumulator array in Hough space.
%
%Written by :
% Amin Sarafraz
% Computer Vision Online
% http://www.computervisiononline.com
% amin#computervisiononline.com
%
% Acknowledgement: Thanks to CJ Taylor and Peter Bone for their constructive comments
%
%May 5,2004 - Original version
%November 24,2004 - Modified version,faster and better performance (suggested by CJ Taylor)
%Aug 31,2012 - Implemented suggestion by Peter Bone/ Better documentation
if nargin == 2
thresh = 4; % set threshold to default value
end
if thresh < 4
error('HOUGHCIRCLE:: Treshold value must be bigger or equal to 4');
end
%Voting
Accumulator = zeros(size(Imbinary)); % initialize the accumulator
[yIndex xIndex] = find(Imbinary); % find x,y of edge pixels
numRow = size(Imbinary,1); % number of rows in the binary image
numCol = size(Imbinary,2); % number of columns in the binary image
r2 = r^2; % square of radius, to prevent its calculation in the loop
for cnt = 1:numel(xIndex)
low=xIndex(cnt)-r;
high=xIndex(cnt)+r;
if (low<1)
low=1;
end
if (high>numCol)
high=numCol;
end
for x0 = low:high
yOffset = sqrt(r2-(xIndex(cnt)-x0)^2);
y01 = round(yIndex(cnt)-yOffset);
y02 = round(yIndex(cnt)+yOffset);
if y01 < numRow && y01 >= 1
Accumulator(y01,x0) = Accumulator(y01,x0)+1;
end
if y02 < numRow && y02 >= 1
Accumulator(y02,x0) = Accumulator(y02,x0)+1;
end
end
end
% Finding local maxima in Accumulator
y0detect = []; x0detect = [];
AccumulatorbinaryMax = imregionalmax(Accumulator);
[Potential_y0 Potential_x0] = find(AccumulatorbinaryMax == 1);
Accumulatortemp = Accumulator - thresh;
for cnt = 1:numel(Potential_y0)
if Accumulatortemp(Potential_y0(cnt),Potential_x0(cnt)) >= 0
y0detect = [y0detect;Potential_y0(cnt)];
x0detect = [x0detect;Potential_x0(cnt)];
end
end
currently I'm working on a project where I try to find the corners of the rectangle's surface in a photo using OpenCV (Python, Java or C++)
I've selected desired surface by filtering color, then I've got mask and passed it to the cv2.findContours:
cnts, _ = cv2.findContours(mask, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
cnt = sorted(cnts, key = cv2.contourArea, reverse = True)[0]
peri = cv2.arcLength(cnt, True)
approx = cv2.approxPolyDP(cnt, 0.02*peri, True)
if len(approx) == 4:
cv2.drawContours(mask, [approx], -1, (255, 0, 0), 2)
This gives me an inaccurate result:
Using cv2.HoughLines I've managed to get 4 straight lines that accurately describe the surface. Their intersections are exactly what I need:
edged = cv2.Canny(mask, 10, 200)
hLines = cv2.HoughLines(edged, 2, np.pi/180, 200)
lines = []
for rho,theta in hLines[0]:
a = np.cos(theta)
b = np.sin(theta)
x0 = a*rho
y0 = b*rho
x1 = int(x0 + 1000*(-b))
y1 = int(y0 + 1000*(a))
x2 = int(x0 - 1000*(-b))
y2 = int(y0 - 1000*(a))
cv2.line(mask, (x1,y1), (x2,y2), (255, 0, 0), 2)
lines.append([[x1,y1],[x2,y2]])
The question is: is it possible to somehow tweak findContours?
Another solution would be to find coordinates of intersections. Any clues for this approach are welcome :)
Can anybody give me a hint how to solve this problem?
Finding intersection is not so trivial problem as it seems to be, but before the intersection points will be found, following problems should be considered:
The most important thing is to choose the right parameters for the HoughLines function, since it can return from 0 to an infinite numbers of lines (we need 4 parallel)
Since we do not know in what order these lines go, they need to be compared with each other
Because of the perspective, parallel lines are no longer parallel, so each line will have a point of intersection with the others. A simple solution would be to filter the coordinates located outside the photo. But it may happen that an undesirable intersection will be within the photo.
The coordinates should be sorted. Depending on the task, it could be done in different ways.
cv2.HoughLines will return an array with the values of rho and theta for each line.
Now the problem becomes a system of equations for all lines in pairs:
def intersections(edged):
# Height and width of a photo with a contour obtained by Canny
h, w = edged.shape
hl = cv2.HoughLines(edged,2,np.pi/180,190)[0]
# Number of lines. If n!=4, the parameters should be tuned
n = hl.shape[0]
# Matrix with the values of cos(theta) and sin(theta) for each line
T = np.zeros((n,2),dtype=np.float32)
# Vector with values of rho
R = np.zeros((n),dtype=np.float32)
T[:,0] = np.cos(hl[:,1])
T[:,1] = np.sin(hl[:,1])
R = hl[:,0]
# Number of combinations of all lines
c = n*(n-1)/2
# Matrix with the obtained intersections (x, y)
XY = np.zeros((c,2))
# Finding intersections between all lines
for i in range(n):
for j in range(i+1, n):
XY[i+j-1, :] = np.linalg.inv(T[[i,j],:]).dot(R[[i,j]])
# filtering out the coordinates outside the photo
XY = XY[(XY[:,0] > 0) & (XY[:,0] <= w) & (XY[:,1] > 0) & (XY[:,1] <= h)]
# XY = order_points(XY) # obtained points should be sorted
return XY
here is the result:
It is possible to:
select the longest contour
break it into segments and group them by gradient
Fit lines to the largest four groups
Find intersection points
But then, Hough transform does nearly the same thing. Is there any particular reason for not using it?
Intersection points of lines are very easy to calculate. A high-school coordinate geometry lesson can provide you with the algorithm.