Using regular expression to read part of the filename
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anton spektorov
am 1 Okt. 2018
Bearbeitet: Stephen23
am 2 Okt. 2018
Good afternoon,
I am attempting to use a regular expression feature to read part of the file that is generated by the microscope camera.
The file format is x_yyyymmdd_hhmmSSsss where small s is the milisecond and x is the number of the timelapsed photo.
I am trying to convert (truncate) say 5_20180927_161742655 to 20180927_161742 so that I can use this as a timestamp more easily.
Any help would be appreciated. Thank you.
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dpb
am 1 Okt. 2018
Bearbeitet: dpb
am 2 Okt. 2018
AMMENDED -- dpb
Don't need regular expressions for that purpose, just select the first N-3 characters in the file name...
S=15; % sublength of wanted string (15)
[~,TS]=fileparts(fname); % return the file name w/o path, extension
L=length(TS); % just get the overall length
TS=TS(L-(S+3)+1:L-3); % pick up between file and sss millisec portion
Alternatively, one can
L1=find(fn=='_',1)+1; % location just past first underscore
TS=TS(L1:L-3); % pick up between file and sss
NB: fileparts will return the name as either char array or cellstr depending on input form for fanme going in so may need to dereference a cell string if that is the form.
I'm sure there is a regexp expression that will work but I'm just not adept with its syntax, sorry...
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