how to make unzip faster in matlab?

i have 50000 .zip files..i need to unzip and process with the files inside those zip files. i coded to unzip in a loop.but its very slow and taked 2 hr to complete unzip of all 5000 files.any way to speedup the unzipping in matlab?

回答(5 个)

I had the same issue. Using an external program (7-zip) solved the issue for me. Extracting and repacking 1000+ zip files was > 27x faster than the Matlab unzip function.
Is used it like this to unpack:
sys_cmd_unzip = @(z,d) ['"C:\Program Files\7-Zip\7z.exe" e -y -r -o"' d '" "' z '" *'];
cellfun(@(z,d) system(sys_cmd_unzip(z,d)), all_zip_files, new_zip_file_dirs,'uniformoutput',false)
and like this to repack:
sys_cmd_unzip = @(zd) ['"C:\Program Files\7-Zip\7z.exe" a -y -sdel -tzip -r "' zd '.zip" "' zd '/*";'];
cellfun(@(zfd) system(sys_cmd_unzip(zfd)),new_zip_file_dirs,'uniformoutput',false)
cellfun(@(zfd) rmdir(zfd), new_zip_file_dirs) % remove directory where files where stored.
for the command line options of 7zip, see help of 7z.exe.
You can use external software with " ! ", if you use unzip, then, call it this way:
! unzip your_file.zip

1 个评论

thanks... this is applicable for one zip file..but how to use this(!) for 5500 files in a loop..it showing error?

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Walter Roberson
Walter Roberson 2013-8-28

1 个投票

Have you considered calling upon an external unzipping program by using system() ?

1 个评论

this s my code of unzipping for N number of files.(nearly 5500).it took 3hr to finish all process, how can i use system() to access external program..help me
zip_files=dir('*.zip');
for k=1:numel(zip_files)
unzip(zip_files(k).name,'output');
T ='.\Extracted Files.\*.tim';
R='.\Extracted Files.\*.run';
movefile(T,'.\T Files');
movefile(R,'.\R Files');
end

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You can try something like this:
my_zips = dir('*.zip');
for k=1:numel(my_zips)
tmp_str = horzcat('unzip ', my_zips(k).name);
system(tmp_str);
end
This will unzip all your zip files.

2 个评论

his code takes more time to unzip all my files.can you suggest more specific way to reduce unzipping times for 55000 zip files
Sometimes you can not reduce any more the time taken to perform some tasks. 5000 files are a whole lot of files to unzip. You say it takes 2 hours to perform the job, that means you are unzipping a file every 1.44 seconds, which is not a long time to unzip a file of small size.

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I understand the post is somewhat old but it may be useful for people looking to speed up their zip extraction. If you don't specifically need every single file in the archive at once, you can use something like:
zip_file = 'foo.zip'
destination_folder = 'C:\foo\bar\'
system(['unzip ' '"' zip_file '"' ' file_inside_to_extract.txt' ' -d ' '"' ...
destination_folder '"'])
The system unzip command (for Windows at least) allows you to pull out specific files from the archive.

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提问:

2013-8-28

回答:

2017-8-7

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