Fix typos in py_houghlines tutorial. Extract the python code in a specific folder.

pull/8024/head
catree 8 years ago
parent a8e5d1d9fd
commit e357e2c1b7
  1. 100
      doc/py_tutorials/py_imgproc/py_houghlines/py_houghlines.markdown
  2. 22
      samples/python/tutorial_code/imgProc/hough_line_transform/hough_line_transform.py
  3. 12
      samples/python/tutorial_code/imgProc/hough_line_transform/probabilistic_hough_line_transform.py

@ -5,36 +5,36 @@ Goal
----
In this chapter,
- We will understand the concept of Hough Tranform.
- We will see how to use it detect lines in an image.
- We will see following functions: **cv2.HoughLines()**, **cv2.HoughLinesP()**
- We will understand the concept of the Hough Transform.
- We will see how to use it to detect lines in an image.
- We will see the following functions: **cv2.HoughLines()**, **cv2.HoughLinesP()**
Theory
------
Hough Transform is a popular technique to detect any shape, if you can represent that shape in
The Hough Transform is a popular technique to detect any shape, if you can represent that shape in a
mathematical form. It can detect the shape even if it is broken or distorted a little bit. We will
see how it works for a line.
A line can be represented as \f$y = mx+c\f$ or in parametric form, as
\f$\rho = x \cos \theta + y \sin \theta\f$ where \f$\rho\f$ is the perpendicular distance from origin to the
line, and \f$\theta\f$ is the angle formed by this perpendicular line and horizontal axis measured in
counter-clockwise ( That direction varies on how you represent the coordinate system. This
representation is used in OpenCV). Check below image:
A line can be represented as \f$y = mx+c\f$ or in a parametric form, as
\f$\rho = x \cos \theta + y \sin \theta\f$ where \f$\rho\f$ is the perpendicular distance from the origin to the
line, and \f$\theta\f$ is the angle formed by this perpendicular line and the horizontal axis measured in
counter-clockwise (That direction varies on how you represent the coordinate system. This
representation is used in OpenCV). Check the image below:
![image](images/houghlines1.svg)
So if line is passing below the origin, it will have a positive rho and angle less than 180. If it
is going above the origin, instead of taking angle greater than 180, angle is taken less than 180,
So if the line is passing below the origin, it will have a positive rho and an angle less than 180. If it
is going above the origin, instead of taking an angle greater than 180, the angle is taken less than 180,
and rho is taken negative. Any vertical line will have 0 degree and horizontal lines will have 90
degree.
Now let's see how Hough Transform works for lines. Any line can be represented in these two terms,
\f$(\rho, \theta)\f$. So first it creates a 2D array or accumulator (to hold values of two parameters)
Now let's see how the Hough Transform works for lines. Any line can be represented in these two terms,
\f$(\rho, \theta)\f$. So first it creates a 2D array or accumulator (to hold the values of the two parameters)
and it is set to 0 initially. Let rows denote the \f$\rho\f$ and columns denote the \f$\theta\f$. Size of
array depends on the accuracy you need. Suppose you want the accuracy of angles to be 1 degree, you
array depends on the accuracy you need. Suppose you want the accuracy of angles to be 1 degree, you will
need 180 columns. For \f$\rho\f$, the maximum distance possible is the diagonal length of the image. So
taking one pixel accuracy, number of rows can be diagonal length of the image.
taking one pixel accuracy, the number of rows can be the diagonal length of the image.
Consider a 100x100 image with a horizontal line at the middle. Take the first point of the line. You
know its (x,y) values. Now in the line equation, put the values \f$\theta = 0,1,2,....,180\f$ and check
@ -42,57 +42,34 @@ the \f$\rho\f$ you get. For every \f$(\rho, \theta)\f$ pair, you increment value
in its corresponding \f$(\rho, \theta)\f$ cells. So now in accumulator, the cell (50,90) = 1 along with
some other cells.
Now take the second point on the line. Do the same as above. Increment the the values in the cells
Now take the second point on the line. Do the same as above. Increment the values in the cells
corresponding to `(rho, theta)` you got. This time, the cell (50,90) = 2. What you actually
do is voting the \f$(\rho, \theta)\f$ values. You continue this process for every point on the line. At
each point, the cell (50,90) will be incremented or voted up, while other cells may or may not be
voted up. This way, at the end, the cell (50,90) will have maximum votes. So if you search the
accumulator for maximum votes, you get the value (50,90) which says, there is a line in this image
at distance 50 from origin and at angle 90 degrees. It is well shown in below animation (Image
at a distance 50 from the origin and at angle 90 degrees. It is well shown in the below animation (Image
Courtesy: [Amos Storkey](http://homepages.inf.ed.ac.uk/amos/hough.html) )
![](images/houghlinesdemo.gif)
This is how hough transform for lines works. It is simple, and may be you can implement it using
This is how hough transform works for lines. It is simple, and may be you can implement it using
Numpy on your own. Below is an image which shows the accumulator. Bright spots at some locations
denotes they are the parameters of possible lines in the image. (Image courtesy: [Wikipedia](http://en.wikipedia.org/wiki/Hough_transform))
denote they are the parameters of possible lines in the image. (Image courtesy: [Wikipedia](http://en.wikipedia.org/wiki/Hough_transform) )
![](images/houghlines2.jpg)
Hough Transform in OpenCV
=========================
Everything explained above is encapsulated in the OpenCV function, \*\*cv2.HoughLines()\*\*. It simply returns an array of :math:(rho,
Everything explained above is encapsulated in the OpenCV function, **cv2.HoughLines()**. It simply returns an array of :math:(rho,
theta)\` values. \f$\rho\f$ is measured in pixels and \f$\theta\f$ is measured in radians. First parameter,
Input image should be a binary image, so apply threshold or use canny edge detection before finding
Input image should be a binary image, so apply threshold or use canny edge detection before
applying hough transform. Second and third parameters are \f$\rho\f$ and \f$\theta\f$ accuracies
respectively. Fourth argument is the threshold, which means minimum vote it should get for it to be
considered as a line. Remember, number of votes depend upon number of points on the line. So it
respectively. Fourth argument is the threshold, which means the minimum vote it should get to be
considered as a line. Remember, number of votes depends upon the number of points on the line. So it
represents the minimum length of line that should be detected.
@code{.py}
import cv2
import numpy as np
img = cv2.imread('sudoku.png')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
edges = cv2.Canny(gray,50,150,apertureSize = 3)
lines = cv2.HoughLines(edges,1,np.pi/180,200)
for line in lines:
rho,theta = line[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(img,(x1,y1),(x2,y2),(0,0,255),2)
cv2.imwrite('houghlines3.jpg',img)
@endcode
@include hough_line_transform.py
Check the results below:
![image](images/houghlines3.jpg)
@ -101,36 +78,23 @@ Probabilistic Hough Transform
-----------------------------
In the hough transform, you can see that even for a line with two arguments, it takes a lot of
computation. Probabilistic Hough Transform is an optimization of Hough Transform we saw. It doesn't
take all the points into consideration, instead take only a random subset of points and that is
sufficient for line detection. Just we have to decrease the threshold. See below image which compare
Hough Transform and Probabilistic Hough Transform in hough space. (Image Courtesy : [Franck
Bettinger's home page](http://phdfb1.free.fr/robot/mscthesis/node14.html)
computation. Probabilistic Hough Transform is an optimization of the Hough Transform we saw. It doesn't
take all the points into consideration. Instead, it takes only a random subset of points which is
sufficient for line detection. Just we have to decrease the threshold. See image below which compares
Hough Transform and Probabilistic Hough Transform in Hough space. (Image Courtesy :
[Franck Bettinger's home page](http://phdfb1.free.fr/robot/mscthesis/node14.html) )
![image](images/houghlines4.png)
OpenCV implementation is based on Robust Detection of Lines Using the Progressive Probabilistic
Hough Transform by Matas, J. and Galambos, C. and Kittler, J.V.. The function used is
Hough Transform by Matas, J. and Galambos, C. and Kittler, J.V. @cite Matas00. The function used is
**cv2.HoughLinesP()**. It has two new arguments.
- **minLineLength** - Minimum length of line. Line segments shorter than this are rejected.
- **maxLineGap** - Maximum allowed gap between line segments to treat them as single line.
- **maxLineGap** - Maximum allowed gap between line segments to treat them as a single line.
Best thing is that, it directly returns the two endpoints of lines. In previous case, you got only
the parameters of lines, and you had to find all the points. Here, everything is direct and simple.
@code{.py}
import cv2
import numpy as np
img = cv2.imread('sudoku.png')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
edges = cv2.Canny(gray,50,150,apertureSize = 3)
lines = cv2.HoughLinesP(edges,1,np.pi/180,100,minLineLength=100,maxLineGap=10)
for line in lines:
x1,y1,x2,y2 = line[0]
cv2.line(img,(x1,y1),(x2,y2),(0,255,0),2)
cv2.imwrite('houghlines5.jpg',img)
@endcode
@include probabilistic_hough_line_transform.py
See the results below:
![image](images/houghlines5.jpg)

@ -0,0 +1,22 @@
import cv2
import numpy as np
img = cv2.imread('../data/sudoku.png')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
edges = cv2.Canny(gray,50,150,apertureSize = 3)
lines = cv2.HoughLines(edges,1,np.pi/180,200)
for line in lines:
rho,theta = line[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(img,(x1,y1),(x2,y2),(0,0,255),2)
cv2.imwrite('houghlines3.jpg',img)

@ -0,0 +1,12 @@
import cv2
import numpy as np
img = cv2.imread('../data/sudoku.png')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
edges = cv2.Canny(gray,50,150,apertureSize = 3)
lines = cv2.HoughLinesP(edges,1,np.pi/180,100,minLineLength=100,maxLineGap=10)
for line in lines:
x1,y1,x2,y2 = line[0]
cv2.line(img,(x1,y1),(x2,y2),(0,255,0),2)
cv2.imwrite('houghlines5.jpg',img)
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