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How to read multiple jpg images in matlab

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i have 250 jpg images stored at in_dir = 'C:\Users\DJDJ\Desktop\pic';
these pictures are arranged as following and named : image1 , image2, image3.....image250
I want to read all these images and processing them starting by image1 after processing go to process image2 .and after image3.to..image250
then saved the resultant images after preproceesing in a specific file,for example at out_dir = 'C:\Users\DJDJ\Desktop\pic output';.
How can i do this in matlab?
thank you in advance

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Niels
Niels il 20 Gen 2017
Modificato: Niels il 20 Gen 2017
you can use a for loop and save the image data in a cell:
replace png by the file type of your images. result will be a 1x250 cell array
for i= 1:250
x{i}=imread(['image',num2str(i),'.png'])
end
>> save('image_Data','x')
  6 Commenti
Image Analyst
Image Analyst il 21 Gen 2017
You need to use exist() to verify that the file exists first. Also, splitting up long lines into multiple lines would make things simpler. Because of that, I can see that you totally ignored my answer to use the code in the FAQ and used your own code, thus causing your problems. I urge you to again look at the FAQ and see what it's doing that you aren't, and then the code will work.
djamaleddine djeldjli
djamaleddine djeldjli il 21 Gen 2017
Thank you very much for your assistance.Now,it's working well.
I have replaced line 11 by current_image = imread(fullfile(in_dir, fname));

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Più risposte (1)

Image Analyst
Image Analyst il 21 Gen 2017
You would use the first code sample in this entry of the FAQ: http://matlab.wikia.com/wiki/FAQ#How_can_I_process_a_sequence_of_files.3F
  2 Commenti
Maitham
Maitham il 21 Gen 2017
Hi Image Analyst. Sorry maybe my comment has not related to this topic. I need your help please with Strehl Ratio. I have a point source image and need to find and draw the Strehl Ratio for it. I would high appreciate it if you could help me please. Very kind regards Maitham
Image Analyst
Image Analyst il 21 Gen 2017
Read this link and then attach your data in a new question. We can't do anything without data to work on.

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