How to speed up loading of .mat files

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Faisal Ahmed
Faisal Ahmed el 14 de Mzo. de 2017
Comentada: David el 16 de Ag. de 2023
I have around 200K .mat files which I need to analyze. It will take me a lot of time if I load each file to access a particular field of interest. I'll highly appreciate your good advice.

Respuestas (3)

Jan
Jan el 14 de Mzo. de 2017
Store the files on a SSD.
  7 comentarios
Steven Lord
Steven Lord el 15 de Sept. de 2017
You may be able to use the matfile function to access your data on disk.
Jan
Jan el 15 de Sept. de 2017
@Andre: What does "Saving the file is just as bad" exactly mean? Storing a single array might be more efficient with fwrite in a binary format.
I thought of publishing an alternative save command, which uses 7zip (optimize for output size and time for reading, but slow writing) or minilzo (fast, but no powerful compression) for a compression. Unfortunately the details are critical: nested struct arrays containing function handles and user-defined objects, brrr. I cannot decide if I should implement a feature for extracting parts of the file (some variables or a slices of large arrays). It is easy to optimize such a tool for a specific purpose, but then it can never compete with the established, flexible and massively tested MAT format. Therefore I still use binary files without compression.
x = rand(1, 2e8); % 1.6GB data
tic;
f = fopen('test.dat', 'w');
fwrite(f, x, 'double');
fclose(f);
toc
Elapsed time is 7.466022 seconds.
About 210 MB/s with an old hard disk.

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Jan
Jan el 15 de Sept. de 2017
See: http://www.mathworks.com/matlabcentral/fileexchange/47698-savezip : This saves an array into a ZIP or GZIP file.

David
David el 16 de Ag. de 2023
Editada: David el 16 de Ag. de 2023
If the mat files are:
  • large
  • have a lot of variables or nested variables or structs, most of which you dont need
  • saved as version 7.3 or later
Then it might be worth bypassing 'load' entirely and taking advantage of the fact that .mat v7.3 is just a HDF5.
Load the variable inside the files you want directly, without bothering with the variables you dont want to load. Itll load insanely fast, regardless of size.
Say you have a .mat file with path/to/my/file.mat with variables 'var1', 'var2', 'var3.a.b.c.d', and you just want var2 .
myVarName = 'var2';
myFile = fullfile('path','to','my','file.mat');
function argOut = quickLoad(myFile, myVarName)
% Get the location of the variable in the file using hdf syntax
% / by itself is the root of the file, then variables names come after
% Note: also works very nicely for nested structures where a.b.c.d.e would
% have varName as /a/b/c/d/e.
h5loc = ['/' myVarName]; % Always /, not like windows/linux filesep
% Open the file using H5F.
fid = H5F.open(pathToMatfile);
% Open the file using H5D.
dsetid = H5D.open(fid,h5loc);
% Load in the dataset
argOut = H5D.read(dsetid,h5Loc); % All done
% Clean up
H5D.close(dsetid);
H5F.close(fid);
Should include some input checking (existence of the variable without using 'whos' which is very slow), but that can be another post.
Try it out... itll make you happy.
myData = quickLoad(myFile, myVarName)
  2 comentarios
David
David el 16 de Ag. de 2023
Check if this file is actually v7.3 using:
tf = H5F.is_hdf(myFilePath);
Can check existence of the variable or dataset in another thread.
David
David el 16 de Ag. de 2023
This will also work for nested structures, which can be handy:
myVarName = 'var3/a/b/c/d';
d = quickLoad(myFile, myVarName);

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