为什么cv2膨胀实际上不会影响我的形象? [英] Why doesn't cv2 dilate actually affect my image?

问题描述

因此，我正在使用python和opencv2生成二进制(很好，真正的灰度级，8位，用作二进制)图像，向该图像写入少量多边形，然后使用内核对图像进行扩展.但是，无论我使用什么内核，我的源映像和目标映像始终总是相同的.有什么想法吗?

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!!")

我得到的只是EPIC FAIL！

谢谢！

推荐答案

因此，事实证明问题出在内核和映像的创建上.我相信openCV希望'uint8'作为内核和映像的数据类型.在这种情况下，我使用dtype='int'创建了内核，默认为'int64'.另外，我将图像创建为'int8'，而不是'uint8'.不知何故，这并没有引发异常，但以令人惊讶的方式导致了扩张失败.

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.

将以上两行更改为

binary_image = np.zeros(image.shape,dtype='uint8')

kernel = np.ones((11,11),'uint8')

解决了问题，现在我得到了EUREKA！哇！

Fixed the problem, and now I get EUREKA! Hooray!

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