Mathematica to OpenCV-Python
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:
and this is the result after warping:
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.