How can I estimate the time required by textscan and the size of the output?
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Hello,
I am running Matlab 2013b on Windows 7. I have 8 GB RAM memory and a I set the swap file to 20 GB.
I am trying to read a relatively large txt file that is tab separated. The size of the file on the hard disk is a little over 2 GB. There are 6 columns and approx. 64 million rows in the file. The entries are mixed (strings and numbers with some missing values).
At this point I am using:
textscan(fid,repmat('%s',1,6),'delimiter','\t');
It is running for about 4 hours now using about 6.5 GB RAM.
1. I would like to know how can I estimate the time it takes to read the file and the size of the output.
2. After it is done I would like to extract the numerical values from the resulting cell matrix and save that to a .mat file. Any idea how long that would take?
3. Is there any better way of doing this? If I could extract from the file a matrix with the numerical values only (setting everything else to NaN) it would be great.
Thanks!
3 Comments
per isakson
on 31 Jul 2014
Edited: per isakson
on 31 Jul 2014
Strange! Could there be something using memory that the task manager doesn't report on? Is your system 64bit?
Accepted Answer
per isakson
on 31 Jul 2014
Edited: per isakson
on 31 Jul 2014
I made an experiment with R2013a, 64bit, Win7, 8GB RAM, 9GB page file size and a mechanical HD
- created a file with "6 columns and approx. 64 million rows"
- read a piece of the file with textscan after restart of Matlab
- Monitored the memory usage with the Windows Task Manager
Elapsed time is 192.597879 seconds.
>> cac{1}(1:3)
ans =
'Col1'
'Col1'
'Col1'
>> cac{2}(1:3)
ans =
1
1
1
>> cac{6}(1:3)
ans =
3
3
3
>> whos cac
Name Size Bytes Class Attributes
cac 1x6 1920000672 cell
where code is
fid = fopen('c:\tmp\test.txt');
M = cumsum(ones( 3, 64e6 ), 1 );
fprintf( 'Col1\t%4.1f\tCol2\t%4.1f\tCol3\t%4.1f\n', M )
fclose( fid );
tic
fid = fopen('c:\tmp\test.txt');
cac = textscan( fid, '%s%f%s%f%s%f', 5e6, 'Delimiter', '\t' );
fclose( fid );
toc
- start of Matlab
- running of experiment
Results
- Reading and parsing 5 million rows took three minutes and peaked at 4.8GB RAM usage
- 5 million rows produced a 2GB variable, cac, in Matlab.
- An experiment to read the entire file showed that speed decreased drastically when there was no more free physical RAM. (I killed the process.) 8GB RAM would allow effective reading of nearly ten million rows.
4 Comments
per isakson
on 31 Jul 2014
That's surprising results! It doesn't make sense to me. Here is another one that doesn't make sense.
%s%f%s%f%s%f
Elapsed time is 1.670506 seconds.
Elapsed time is 0.001105 seconds.
Elapsed time is 0.007156 seconds.
Elapsed time is 0.066814 seconds.
Elapsed time is 0.747540 seconds.
Elapsed time is 13.118503 seconds.
%s%s%s%s%s%s
Elapsed time is 0.754963 seconds.
Elapsed time is 0.001459 seconds.
Elapsed time is 0.009222 seconds.
Elapsed time is 0.090024 seconds.
Elapsed time is 1.193596 seconds.
Elapsed time is 36.568043 seconds.
>>
where
fid = fopen('c:\tmp\test.txt');
M = cumsum(ones( 3, 64e6 ), 1 );
fprintf( 'Col1\t%4.1f\tCol2\t%4.1f\tCol3\t%4.1f\n', M )
fclose( fid );
disp( '%s%f%s%f%s%f' )
for jj = 1 : 6
tic
fid = fopen('c:\tmp\test.txt');
cac = textscan( fid, '%s%f%s%f%s%f', 10^jj, 'Delimiter', '\t' );
fclose( fid );
toc
end
disp( '%s%s%s%s%s%s' )
for jj = 1 : 6
tic
fid = fopen('c:\tmp\test.txt');
cac = textscan( fid, '%s%s%s%s%s%s', 10^jj, 'Delimiter', '\t' );
fclose( fid );
toc
end
More Answers (1)
dpb
on 30 Jul 2014
Don't know that there is any metric to predict run time other than testing as it's so dependent upon the machine characteristics, not just size.
Two things I can think of to try --
a) Use the specific format for the data file -- strings for string, numeric for numbers. Skip ('%*s' for example to skip a string field) any fields that aren't mandatory. Use "'collectoutput',true" to gather the various types together. This will bypass a subsequent conversion step.
b) Use the feature of textscan to process the file in pieces -- say 1 to a few MB roughly per pass.
4 Comments
per isakson
on 31 Jul 2014
Edited: per isakson
on 31 Jul 2014
Yes, if it is a reasonable number of different string constants and you know them beforehand.
>> cac = textscan( 'char; 23.1; 16', '%f', 'Delimiter', ';' ...
, 'treatAsEmpty', {'char'} );
>> cac{:}
ans =
NaN
23.1000
16.0000
nan regardless of case is converted to "NaN"
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