原理讲的很清晰，给原博主一个赞

边缘检测之后按照正方形检索来判定是否是马赛克内容

原理知晓了之后就很好做了

话说MATLAB转化为python的过程还是很有趣的

from PIL import Image

import numpy as np

import math

import warnings

#算法来源，博客https://www.cnblogs.com/techyan1990/p/7291771.html和https://blog.csdn.net/zhancf/article/details/49736823

highhold=200#高阈值

lowhold=40#低阈值

warnings.filterwarnings("ignore")

demo=Image.open("noise_check//23.jpg")

im=np.array(demo.convert('L'))#灰度化矩阵

print(im.shape)

print(im.dtype)

height=im.shape[0]#尺寸

width=im.shape[1]

gm=[[0 for i in range(width)]for j in range(height)]#梯度强度

gx=[[0 for i in range(width)]for j in range(height)]#梯度x

gy=[[0 for i in range(width)]for j in range(height)]#梯度y

theta=0#梯度方向角度360度

dirr=[[0 for i in range(width)]for j in range(height)]#0,1,2,3方位判定值

highorlow=[[0 for i in range(width)]for j in range(height)]#强边缘、弱边缘、忽略判定值2,1,0

rm=np.array([[0 for i in range(width)]for j in range(height)])#输出矩阵

#高斯滤波平滑，3x3

for i in range(1,height-1,1):

for j in range(1,width-1,1):

rm[i][j]=im[i-1][j-1]*0.0924+im[i-1][j]*0.1192+im[i-1][j+1]*0.0924+im[i][j-1]*0.1192+im[i][j]*0.1538+im[i][j+1]*0.1192+im[i+1][j-1]*0.0924+im[i+1][j]*0.1192+im[i+1][j+1]*0.0924

for i in range(1,height-1,1):#梯度强度和方向

for j in range(1,width-1,1):

gx[i][j]=-rm[i-1][j-1]+rm[i-1][j+1]-2*rm[i][j-1]+2*rm[i][j+1]-rm[i+1][j-1]+rm[i+1][j+1]

gy[i][j]=rm[i-1][j-1]+2*rm[i-1][j]+rm[i-1][j+1]-rm[i+1][j-1]-2*rm[i+1][j]-rm[i+1][j+1]

gm[i][j]=pow(gx[i][j]*gx[i][j]+gy[i][j]*gy[i][j],0.5)

theta=math.atan(gy[i][j]/gx[i][j])*180/3.1415926

if theta>=0 and theta<45:

dirr[i][j]=2

elif theta>=45 and theta<90:

dirr[i][j]=3

elif theta>=90 and theta<135:

dirr[i][j]=0

else:

dirr[i][j]=1

for i in range(1,height-1,1):#非极大值抑制,双阈值监测

for j in range(1,width-1,1):

NW=gm[i-1][j-1]

N=gm[i-1][j]

NE=gm[i-1][j+1]

W=gm[i][j-1]

E=gm[i][j+1]

SW=gm[i+1][j-1]

S=gm[i+1][j]

SE=gm[i+1][j+1]

if dirr[i][j]==0:

d=abs(gy[i][j]/gx[i][j])

gp1=(1-d)*E+d*NE

gp2=(1-d)*W+d*SW

elif dirr[i][j]==1:

d=abs(gx[i][j]/gy[i][j])

gp1=(1-d)*N+d*NE

gp2=(1-d)*S+d*SW

elif dirr[i][j]==2:

d=abs(gx[i][j]/gy[i][j])

gp1=(1-d)*N+d*NW

gp2=(1-d)*S+d*SE

elif dirr[i][j]==3:

d=abs(gy[i][j]/gx[i][j])

gp1=(1-d)*W+d*NW

gp2=(1-d)*E+d*SE

if gm[i][j]>=gp1 and gm[i][j]>=gp2:

if gm[i][j]>=highhold:

highorlow[i][j]=2

rm[i][j]=1

elif gm[i][j]>=lowhold:

highorlow[i][j]=1

else:

highorlow[i][j]=0

rm[i][j]=0

else:

highorlow[i][j]=0

rm[i][j]=0

for i in range(1,height-1,1):#抑制孤立低阈值点

for j in range(1,width-1,1):

if highorlow[i][j]==1 and (highorlow[i-1][j-1]==2 or highorlow[i-1][j]==2 or highorlow[i-1][j+1]==2 or highorlow[i][j-1]==2 or highorlow[i][j+1]==2 or highorlow[i+1][j-1]==2 or highorlow[i+1][j]==2 or highorlow[i+1][j+1]==2):

#highorlow[i][j]=2

rm[i][j]=1

#img=Image.fromarray(rm)#矩阵化为图片

#img.show()

#正方形法判定是否有马赛克

value=35

lowvalue=16

imgnumber=[0 for i in range(value)]

for i in range(1,height-1,1):#性价比高的8点判定法

for j in range(1,width-1,1):

for k in range(lowvalue,value):

count=0

if i+k-1>=height or j+k-1>=width:continue

if rm[i][j]!=0:count+=1#4个顶点

if rm[i+k-1][j]!=0:count+=1

if rm[i][j+k-1]!=0:count+=1

if rm[i+k-1][j+k-1]!=0:count+=1

e=(k-1)//2

if rm[i+e][j]!=0:count+=1

if rm[i][j+e]!=0:count+=1

if rm[i+e][j+k-1]!=0:count+=1

if rm[i+k-1][j+e]!=0:count+=1

if count>=6:

imgnumber[k]+=1

for i in range(lowvalue,value):

print("length:{} number:{}".format(i,imgnumber[i]))

结果图可以上一下了

可以看出在一定程度上能够检测出马赛克内容

原图

边缘图案

正方形数量

以上就是python 检测图片是否有马赛克的详细内容，更多关于python 检测图片马赛克的资料请关注其它相关文章！