python 检测图片是否有马赛克

首先是canny边缘检测，将图片的边缘检测出来，参考博客

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

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