How to Convert Python program to Elixir code

How to convert following code to elixir code.
I have made this program to detect digit and its position in a grid.

I know how to make simple elixir programs but I dont know how to make such complex programs in elixir because I dont know the syntax and about the libraries used in elixir.

I also searched about evision but very less reference I got, but not sufficient…

Here is my Python code:-

img = cv2.imread('digit_1.jpeg')
imgGrey = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
_, thrash = cv2.threshold(imgGrey, 240, 255, cv2.THRESH_BINARY)
contours, _ = cv2.findContours(thrash, cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)
squares=0
for contour in contours:
    approx = cv2.approxPolyDP(contour, 0.01* cv2.arcLength(contour, True), True)
    if len(approx)==4:
    
        cv2.drawContours(img, [approx],0, (255, 0, 0),2)
    x = approx.ravel()[0]
    y = approx.ravel()[1] - 5
    if len(approx) == 4:
        x1 ,y1, w, h = cv2.boundingRect(approx)
        
        aspectRatio = float(w)/h
        if aspectRatio >= 0.95 and aspectRatio <= 1.05 and (y1+h)*(x1+w) <985000:
            squares +=1
#cv2.imshow("img", img)
for contour in contours:
    approx = cv2.approxPolyDP(contour, 0.01* cv2.arcLength(contour, True), True)
    if len(approx)==4:
        cv2.drawContours(img, [approx],0, (255, 0, 0),2)
    x = approx.ravel()[0]
    y = approx.ravel()[1] - 5
    if len(approx) == 4:
        x1 ,y1, w, h = cv2.boundingRect(approx)

Please help me to convert this code into elixir code.

That’s not just converting Python to Elixir code, you also need an image processing library.

This thread has a few references: Image processing library

Though I’d probably prefer the one described here: Image - an image processing library based upon Vix

…and by the same author, it even mentions eVision: GitHub - kipcole9/image: Image processing for Elixir

i want to do it using evision i have installed it in elixir but dont know which functions are available in evision similar to opencv so that i can use it accordingly

and what are there syntex, arguments and parameters.

The docs should provide that:
https://hexdocs.pm/evision/Evision.html#functions

2 Likes

thank u for this i think this will help me

def drawContours(path) do

       {:ok, imgGrey} =OpenCV.imread(path, flags: OpenCV.cv_IMREAD_GRAYSCALE())
       {:ok, {_, thrash}} =OpenCV.threshold(imgGrey, 240, 255, OpenCV.cv_THRESH_BINARY())
       {:ok, {contours, _}} =OpenCV.findContours(thrash, OpenCV.cv_RETR_TREE(), OpenCV.cv_CHAIN_APPROX_NONE())
       {:ok, d} = OpenCV.drawContours(thrash, contours, -1, [0, 255, 0])
       show(d)
       #IN PYHTON ->cv2.drawContours(image=image_copy, contours=contours, contourIdx=-1, color=(0, 255, 0), thickness=2, lineType=cv2.LINE_AA)

end

When i run this it shows threshold image only in output but not draw contours on image ??

As shown in your code, the python code is drawing on a copy of the original image (3-channel BGR) while you were drawing contours on the thresholded image (1-channel image).

So you can either draw contours on the original BGR image or change the color from {0, 255, 0} to maybe {128}, but it could be very hard to see the color {128} in a gray scale image.

1 Like
  1. After drawing contours in image now i want to crop the sudoku grid but i am confuse in how to send the coordinates of grid in Nx.slice to crop image because draw contours give reference only, i also change it in tensor but it gives whole image pixels.

  2. Also, i use CHAIN_APPROX_SIMPLE() in find contours to get 4 boundary points of grid but it is not working, and gives all contours. ??

For question 1, Nx.slice can only slice a rectangular area for you. To get the puzzle, you can use Evision.warpPerspective. There is an example of this function, evision/warp_perspective.livemd at main · cocoa-xu/evision · GitHub.

For question 2, there could be 2 possible reasons,

  • perhaps you forgot to change the second parameter of Evision.findContours/3 to Evision.cv_RETR_EXTERNAL()
  • you will need to find the contour that has shape {4, 1, 2}. And based on the sample image, I guess you were following a blog post and if so, I think you missed some intermediate steps.