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https://www.cnblogs.com/techyan1990/p/7291771.html

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 Imageimport numpy as npimport mathimport warnings#算法来源，博客https://www.cnblogs.com/techyan1990/p/7291771.html和https://blog.csdn.net/zhancf/article/details/49736823highhold=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xgy=[[0 for i in range(width)]for j in range(height)]#梯度ytheta=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2,1,0rm=np.array([[0 for i in range(width)]for j in range(height)])#输出矩阵#高斯滤波平滑，3x3for 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]*0.0924for 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]=1for 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]=0for 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=35lowvalue=16imgnumber=[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]+=1for 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好吧啦网其它相关文章！