image processing with parallel computing gives incorrect results

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Maria Kanwal
Maria Kanwal 2016-11-12
评论： Ahmet Cecen 2016-11-13
i need to read all images in a folder in groups of 3 and process them for that i am using a broadcast variable to store the names of all the images, i want to do it by using parallel computing but results are different from the results i get by using simple for loop. Here is the example code, any suggestion how can i get correct results?
inDir ='./abc/';
Imgs = dir([inDir '*.jpg']);
parfor indx=2:length(Imgs)-1
im1=imread([inDir Imgs(indx-1).name]);
im2=imread([inDir Imgs(indx).name]);
im3=imread([inDir Imgs(indx+1).name]);
% resultimg=someprocessing using im1,im2 and im3
% write resultimg
end

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回答（2 个）

Ahmet Cecen
Ahmet Cecen 2016-11-12
I am not very knowledgeable with the implementation of ParFor, but I would wager a guess the problem is your indexing: specifically the presence of indx-1 and indx+1.
Try to see if creating three lists of images and indexing them all with indx fixes the problem. Something like:
Imgs = dir([inDir '*.jpg']);
ImgsBefore = circshift(Imgs,1);
ImgsAfter = circshift(Imgs,-1)
parfor indx=2:length(Imgs)-1
im1=imread([inDir ImgsBefore (indx).name]);
im2=imread([inDir Imgs(indx).name]);
im3=imread([inDir ImgsAfter (indx).name]);
% resultimg=someprocessing using im1,im2 and im3
% write resultimg
end
As I mentioned, this is simply a guess and might not work. Just come back and let it known if it doesn't.
  2 个评论
Maria Kanwal
Maria Kanwal 2016-11-13
results are different than before but still do not match for loop results
Ahmet Cecen
Ahmet Cecen 2016-11-13
I have tried a basic operation of this form (a triple convolution) and I get consistent results with for and parfor. I have no idea what else can be the problem.

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Walter Roberson
Walter Roberson 2016-11-13
inDir ='./abc/';
Imgs = dir( fullfile(inDir, '*.jpg') );
numimg = length(Imgs);
parfor indx = 1 : numimg
im{indx} = imread( fullfile(inDir, Imgs(indx).name) );
end
parfor indx=2:length(Imgs)-1
im1 = im{indx-1};
im2 = im{indx};
im3 = im{indx+1};
% resultimg=someprocessing using im1,im2 and im3
% write resultimg
end
If the result is not the same as before then it might have to do with the processing you are doing.
  3 个评论
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Walter Roberson
Walter Roberson 2016-11-13
I think the differences you are observing have to do with the processing you are doing (which you did not show us.)

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