Why this code does not return any thing (Busy Status)

Question

Ali Purse 2018-10-25

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/426066-why-this-code-does-not-return-any-thing-busy-status

评论： Adam 2018-10-25

      im=imread('file.jpg');
      fac= 30;
      [h,v]=size(im);
      for i=1:h
          for j=1:v
          im1( floor(1+(i-1)*fac), floor(1+(j-1)*fac),:)=im(i,j); %mapping pixels of given image to a new array so that zooming factor is satisfied. 
          end  
      end
      [h1,v1]=size(im1);
      disp('Unknown pixel values are being interpolated. Look at Figure1');
  %nearest neighbor interpolation
      for i=1:h1-1 
          for j=1:v1-1  
                 if(  (floor( 1+(    ceil((i-1)/fac+1)    -1)*fac )==i)  && ( floor( 1+(    ceil((j-1)/fac+1)    -1)*fac )==j )  ) % maped values from the original picture should not be recalculated.     
                 else  %for pixels that should be calculated    
                 if(floor( 1+(    ceil((i-1)/fac+1)    -1)*fac )==i) %on a defined vertical level
                     xo00=ceil((i-1)/fac+1);
                     x00=i;
                 else
                     xo00=floor((i-1)/fac+1); 
                     x00=floor(1+(xo00-1)*fac);
                 end
                 if( floor( 1+(    ceil((j-1)/fac+1)    -1)*fac )==j )%on a defined horizontal level
                     yo00=ceil((j-1)/fac+1);
                     y00=j;
                 else
                     yo00=floor((j-1)/fac+1);
                     y00=floor(1+(yo00-1)*fac);
                 end                  
                xo10=xo00+1;
                yo10=yo00;
                x10=floor(1+(xo10-1)*fac); %current coordinate of 10;
                y10=floor(1+(yo10-1)*fac);
                xo01=xo00;
                yo01=yo00+1;
                x01=floor(1+(xo01-1)*fac); %current coordinate of 01;
                y01=floor(1+(yo01-1)*fac);
                xo11=xo10;
                yo11=yo01;
                x11=x10; %current coordinate of 11;
                y11=y01;          
                xlen=x11-x00;
                ylen=y11-y00;
                x=i-x00;
                y=j-y00;
                dx=x/xlen; %converting coordinates of the pixel being calculated to (0*1)(0*1) domain.
                dy=y/ylen;          
                 if dx<=0.5 %if the pixel being calculated is near to the known upper pixel level. 
                     nx=0; 
                 else  %to known lower pixel level.
                     nx=1;
                 end
                 if dy<=0.5 %if the pixel being calculated is near to the known left pixel level.
                 ny=0;
                 else  %to known right pixel level.
                 ny=1; 
                 end  
                 if ((nx==0) &&(ny==0)) %assigning the upper left known pixel to the pixel being calculated.
                     h00=im1(x00,y00,:);
                     im1(i,j,:)=h00;
                 elseif ((nx==1) && (ny==0)) %assigning the lower left known pixel to the pixel being calculated.
                     h10=im1(x10,y10,:);
                     im1(i,j,:)=h10;
                 elseif((nx==0) && (ny==1)) %assigning the upper right known pixel to the pixel being calculated.
                      h01=im1( x01,y01,:);
                     im1(i,j,:)=h01;
                 elseif((nx==1)&&(ny==1)) %assigning the lower right known pixel to the pixel being calculated.
                     h11=im1(x11,y11,:);
                     im1(i,j,:)=h11;
                 end
             end
          end
        imshow(im1); %displaying image being interpolated.
      end
      disp('Zooming is complet. Look at Figure1');
      imshow(im1);
      imwrite(im1,'zoomed_pic.jpg');

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Adam 2018-10-25

I have no idea, just by looking at it. Use the Pause functionality if you are in a recent enough version of the software that has it. Otherwise step through in debug mode and find out where it is taking its time.

How long did you let it run? Double for loops are not all that fast, especially if your image is quite large.

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