MATLAB Answers

Mary
2

Using regexp to capture parts of a filename

Asked by Mary
on 10 Jun 2013
I have a filename that is something like
Exp000_DD2CM000_PN000_block1_predecision
I need to extract different parts of this filename and use them to determine which path for a looping if statement to take.
For example:
I need it to load the file determine that file = Exp000_DD2CM000_PN000_block1_predecision experimentName = DD2CM000 participantName = PN000 block= 1 type = pre or post
if type = pre
...
elseif type = post
... code
end
right now I'm using
>> experimentName =regexp(str,'DD2CM(/d*)','match')
without any luck.
any ideas?
- ML

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3 Answers

Answer by Azzi Abdelmalek
on 10 Jun 2013
Edited by Azzi Abdelmalek
on 10 Jun 2013
 Accepted Answer

x='Exp000_DD2CM000_PN000_block1_predecision'
a=regexp(x,'_','split')
experimentName = a{2}
participantName = a{3}
block=regexp(a{4},'\d*')
type = a{5}

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this works wonderfully - Thanks!

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Answer by Jonathan Sullivan on 10 Jun 2013

Your slash is the wrong way. Try:
experimentName =regexp(str,'DD2CM(\d*)','match')

  1 Comment

my bad - I made this error while retyping. Thank you for the reply!

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Answer by Daniel Shub
on 10 Jun 2013
Edited by Daniel Shub
on 10 Jun 2013

Depending on how structured your file names are, it might be easier to skip the regexp part.
file = 'Exp000_DD2CM000_PN000_block1_predecision';
x = strfind(file, '_');
f = @(n,m,x)x((m(n)+1):(m(n+1)-1));
experimentName = f(2, [0, x, length(file)], file);
participantName = f(3, [0, x, length(file)], file);
block = f(4, [0, x, length(file)], file);
type = f(5, [0, x, length(file)], file);
Then you just want to throw out parts of each variable
experimentName(6:end)
participantName(3:end)
block(6:end)
type(1:3)
The structure of your data might more naturally fit
experimentName(end-2:end)
participantName(end-2:end)
block(end)
type(1:3)

  1 Comment

I'll give this a try! - Thank you :)

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