問題描述
這個問題與這個問題有關: 函數.我寫了一個快速腳本來展示如何做到這一點.如您所見，這在 Python 中非常簡單.這是測試圖片:
這是變形后的結果:
這里是代碼:
導入 cv2從 scipy.interpolate 導入網格數據將 numpy 導入為 npgrid_x, grid_y = np.mgrid[0:149:150j, 0:149:150j]目的地 = 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]])源 = 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(目標，源，(grid_x，grid_y)，方法='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")扭曲 = cv2.remap(原始，map_x_32，map_y_32，cv2.INTER_CUBIC)cv2.imwrite("warped.png", 扭曲)我想你可以用谷歌搜索一下 griddata 是做什么的.簡而言之，它進行插值，在這里我們使用它來將稀疏映射轉換為密集映射，因為 cv2.remap 需要密集映射.我們只需要將值轉換為 float32，因為 OpenCV 抱怨 float64 類型.請告訴我進展如何.
更新:如果你不想依賴Scipy，一種方法是在你的代碼中實現2d插值功能，例如看Scipy中griddata的源代碼或更簡單的像這樣 http://inasafe.readthedocs.org/en/latest/_modules/engine/interpolation2d.html 僅依賴于 numpy.不過，我建議為此使用 Scipy 或其他庫，盡管我明白為什么只需要 CV2 和 numpy 對于這樣的情況可能會更好.我想聽聽您的最終代碼如何解決數獨問題.
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:
Now, i want to transform this into a perfect square of size (450,450) as given below:
(Never mind the brightness difference of two images).
Question:
How can i do this in OpenCV-Python? I am using cv2 version.
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.
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