opencv之膨胀与腐蚀

腐蚀和膨胀 Erosion/Dilation

erosion/dilation,用白话说,就是让图像亮的区域收缩和扩张.

原理

  • 我们定义一个卷积核矩阵.这个矩阵可以是任何形状的,但通常而言,是矩形或者圆形的.同时要定义一个锚点位置.
  • 用这个卷积核矩阵挨个地划过原始图像矩阵,同时更改锚点位置的像素值.
  • 锚点位置的像素值更改为卷积核矩阵覆盖的有效像素值中的最大值/最小值(分别对应膨胀/腐蚀).
    什么叫"有效"像素值呢?就是卷积核中不为0的那些位置.用公式表达的话,即:

膨胀和腐蚀,说白了就是个求"卷积核所表示的局部"的最大值最小值的过程.

我们来看一个例子:

import cv2
import numpy as np
def test1():
    img = np.zeros((10,10,1),np.uint8)
    img[3:7,3:7,:] = 255
    img[4:6,4:6,:] = 200
    
    kernel1 = cv2.getStructuringElement(cv2.MORPH_RECT,(3,3))
    erosion_dst = cv2.erode(img, kernel1)
    print(erosion_dst)

首先我们创建一个10 x 10的图像,像素如下:

[[  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0 255 255 255 255   0   0   0]
 [  0   0   0 255 200 200 255   0   0   0]
 [  0   0   0 255 200 200 255   0   0   0]
 [  0   0   0 255 255 255 255   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]]

我们创建一个卷积核:

kernel1 = cv2.getStructuringElement(cv2.MORPH_RECT,(3,3))

getStructuringElement api

三个参数分别为卷积核的形状/大小/锚点位置. 默认锚点在矩阵的中心位置.

形状有三种

上面代码中我们创建的3 x 3矩形卷积核如下

用这个卷积核对原始图像做腐蚀后得到的矩阵如下

即矩阵有如下变化:

[[  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0 255 255 255 255   0   0   0]
 [  0   0   0 255 200 200 255   0   0   0]
 [  0   0   0 255 200 200 255   0   0   0]
 [  0   0   0 255 255 255 255   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]]
 
-->

[[  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0 200 200   0   0   0   0]
 [  0   0   0   0 200 200   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]]

我们考虑第三行第四列img[2,3,:]这个像素.当我们的卷积核矩阵的锚点位置与该像素重合时,我们取周边所有像素的最小值.最小值为0.所以该位置的像素值变为0. 其余位置的像素值同理可求.

我们稍微改一下我们的代码,然后再看一下不同卷积核作用下的不同结果,会理解的更清楚

import cv2
import numpy as np
def test1():
    img = np.zeros((10,10,1),np.uint8)
    img[3:7,3:7,:] = 255
    img[4:6,4:6,:] = 200
    
    kernel1 = cv2.getStructuringElement(cv2.MORPH_RECT,(3,3))
    print(kernel1)
    erosion_dst = cv2.erode(img, kernel1)
    print(erosion_dst)

def test2():
    img = np.zeros((10,10,1),np.uint8)
    img[3:7,3:7,:] = 255
    img[4:6,4:6,:] = 200
    img[2,4,:] = 100
    
    kernel1 = cv2.getStructuringElement(cv2.MORPH_RECT,(3,3))
    erosion_dst = cv2.erode(img, kernel1)
    print(erosion_dst)
    
    kernel2 = cv2.getStructuringElement(cv2.MORPH_CROSS,(3,3))
    erosion_dst2 = cv2.erode(img, kernel2)
    print(erosion_dst2)

test2()

我们把原始图像矩阵改为

[[  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0 100   0   0   0   0   0]
 [  0   0   0 255 255 255 255   0   0   0]
 [  0   0   0 255 200 200 255   0   0   0]
 [  0   0   0 255 200 200 255   0   0   0]
 [  0   0   0 255 255 255 255   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]
 [  0   0   0   0   0   0   0   0   0   0]]

用kernal1时,kernal1如下:

以第四行,第五列的像素为例,用卷积核的锚点与之对应,此时计算的是其周围八个像素的最小值,最小值为0.
所以我们得到的矩阵为

当我们用kernal2时,kernal2如下:

对第四行,第五列的像素,用卷积核的锚点与之对应,此时计算的不再是周围八个像素的最小值,而是其正上方,正下方,正左边,正右边的四个像素的最小值.该值为100.
所以我们得到的矩阵为

opencv示例

from __future__ import print_function
import cv2 as cv
import numpy as np
import argparse

erosion_size = 0
max_elem = 2
max_kernel_size = 21

title_trackbar_element_type = 'Element:\n 0: Rect \n 1: Cross \n 2: Ellipse'
title_trackbar_kernel_size = 'Kernel size:\n 2n +1'
title_erosion_window = 'Erosion Demo'
title_dilatation_window = 'Dilation Demo'

def erosion(val):
    erosion_size = cv.getTrackbarPos(title_trackbar_kernel_size, title_erosion_window)
    erosion_type = 0
    val_type = cv.getTrackbarPos(title_trackbar_element_type, title_erosion_window)
    if val_type == 0:
        erosion_type = cv.MORPH_RECT
    elif val_type == 1:
        erosion_type = cv.MORPH_CROSS
    elif val_type == 2:
        erosion_type = cv.MORPH_ELLIPSE
    element = cv.getStructuringElement(erosion_type, (2*erosion_size + 1, 2*erosion_size+1), (erosion_size, erosion_size))
    erosion_dst = cv.erode(src, element)
    cv.imshow(title_erosion_window, erosion_dst)
    
def dilatation(val):
    dilatation_size = cv.getTrackbarPos(title_trackbar_kernel_size, title_dilatation_window)
    dilatation_type = 0
    val_type = cv.getTrackbarPos(title_trackbar_element_type, title_dilatation_window)
    if val_type == 0:
        dilatation_type = cv.MORPH_RECT
    elif val_type == 1:
        dilatation_type = cv.MORPH_CROSS
    elif val_type == 2:
        dilatation_type = cv.MORPH_ELLIPSE
    element = cv.getStructuringElement(dilatation_type, (2*dilatation_size + 1, 2*dilatation_size+1), (dilatation_size, dilatation_size))
    dilatation_dst = cv.dilate(src, element)
    cv.imshow(title_dilatation_window, dilatation_dst)

src = cv.imread("/home/sc/disk/keepgoing/opencv_test/j.png")
cv.namedWindow(title_erosion_window)
cv.createTrackbar(title_trackbar_element_type, title_erosion_window , 0, max_elem, erosion)
cv.createTrackbar(title_trackbar_kernel_size, title_erosion_window , 0, max_kernel_size, erosion)

cv.namedWindow(title_dilatation_window)
cv.createTrackbar(title_trackbar_element_type, title_dilatation_window , 0, max_elem, dilatation)
cv.createTrackbar(title_trackbar_kernel_size, title_dilatation_window , 0, max_kernel_size, dilatation)

erosion(0)
dilatation(0)
cv.waitKey()

通过createTrackbar在窗口上创建两个bar,方便我们看不同种类不同大小的卷积核的影响.

cv.createTrackbar(title_trackbar_element_type, title_erosion_window , 0, max_elem, erosion)
cv.createTrackbar(title_trackbar_kernel_size, title_erosion_window , 0, max_kernel_size, erosion)

原始图片:

处理效果:

opencv实现

https://github.com/opencv/opencv/blob/master/modules/imgproc/src/morph.dispatch.cpp

posted @ 2019-10-14 16:52  core!  阅读(5580)  评论(1编辑  收藏  举报