python - Image transformation in OpenCV

Question

This question is related to this question: How to remove convexity defects in sudoku square

I was trying to implement nikie's answer in Mathematica to OpenCV-Python. But i am stuck at the final step of procedure.

ie I got the all intersection points in square like below:

enter image description here

Now, i want to transform this into a perfect square of size (450,450) as given below:

enter image description here

(Never mind the brightness difference of two images).

Question: How can i do this in OpenCV-Python? I am using cv2 version.

As he suggested in his answer, just use standard transforms (the one you tried to use in your first question) for each individual cell ... — etarion
I used warp perspective which used 4 corners to transform. Are you saying to use the same taking 4 points of each cell? isn't it a time consuming process? — Abid Rahman K
Are you talking embedded programming? I guess not, since you're using python ... on a desktop machine, no, it should not be noticeable, since it's only 81 cells ... — etarion
@AbidRahmanK: I haven't tested that, but I'd expect that the running time of the perspective warp transformation is proportional to the number of output pixels. So if it was fast enough to transform the whole grid, it should be fast enough transform 81 grid cells that are only 1/81 of the grid size. — Niki

fireant fireant · Accepted Answer · 2012-04-29T18:34:39

Apart from etarion's suggestion, you could also use the remap function. I wrote a quick script to show how you can do this. As you see coding this is really easy in Python. This is the test image:

distorted image

and this is the result after warping:

warped image

And here is the code:

import cv2
from scipy.interpolate import griddata
import numpy as np

grid_x, grid_y = np.mgrid[0:149:150j, 0:149:150j]
destination = np.array([[0,0], [0,49], [0,99], [0,149],
                  [49,0],[49,49],[49,99],[49,149],
                  [99,0],[99,49],[99,99],[99,149],
                  [149,0],[149,49],[149,99],[149,149]])
source = np.array([[22,22], [24,68], [26,116], [25,162],
                  [64,19],[65,64],[65,114],[64,159],
                  [107,16],[108,62],[108,111],[107,157],
                  [151,11],[151,58],[151,107],[151,156]])
grid_z = griddata(destination, source, (grid_x, grid_y), method='cubic')
map_x = np.append([], [ar[:,1] for ar in grid_z]).reshape(150,150)
map_y = np.append([], [ar[:,0] for ar in grid_z]).reshape(150,150)
map_x_32 = map_x.astype('float32')
map_y_32 = map_y.astype('float32')

orig = cv2.imread("tmp.png")
warped = cv2.remap(orig, map_x_32, map_y_32, cv2.INTER_CUBIC)
cv2.imwrite("warped.png", warped)

I suppose you can google and find what griddata does. In short, it does interpolation and here we use it to convert sparse mappings to dense mappings as cv2.remap requires dense mappings. We just need to convert to the values to float32 as OpenCV complains about the float64 type. Please let me know how it goes.

Update: If you don't want to rely on Scipy, one way is to implement the 2d interpolation function in your code, for example, see the source code of griddata in Scipy or a simpler one like this http://inasafe.readthedocs.org/en/latest/_modules/engine/interpolation2d.html which depends only on numpy. Though, I'd suggest to use Scipy or another library for this, though I see why requiring only CV2 and numpy may be better for a case like this. I'd like to hear how your final code solves Sudokus.

python - Image transformation in OpenCV

2 Answers