Why doesn't cv2 dilate actually affect my image?
opencv, python
Solution
So, it turns out the problem was in the creation of both the kernel and the image. I believe that openCV expects `'uint8'` as a data type for both the kernel and the image. In this particular case, I created the kernel with `dtype='int'`, which defaults to `'int64'`. Additionally, I created the image as `'int8'`, not `'uint8'`. Somehow this did not trigger an exception, but caused the dilation to fail in a surprising fashion.
Changing the above two lines to
binary_image = np.zeros(image.shape,dtype='uint8')
kernel = np.ones((11,11),'uint8')
Fixed the problem, and now I get `EUREKA`! Hooray!
Problem
So, I'm generating a binary (well, really gray scale, 8bit, used as binary) image with python and opencv2, writing a small number of polygons to the image, and then dilating the image using a kernel. However, my source and destination image always end up the same, no matter what kernel I use. Any thoughts? ``` from matplotlib import pyplot import numpy as np import cv2 binary_image = np.zeros(image.shape,dtype='int8') for rect in list_of_rectangles: cv2.fillConvexPoly(binary_image, np.array(rect), 255) kernel = np.ones((11,11),'int') dilated = cv2.dilate(binary_image,kernel) if np.array_equal(dilated, binary_image): print("EPIC FAIL!!") else: print("eureka!!") ``` All I get is `EPIC FAIL`! Thanks!