How to separate a portion of filename from a file

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How to separate a portion of filename from a file like I have the file 'scrubbed.MOD_D3_AOD_550.20020112.nc' I just want to extract the '20020112' part

Accepted Answer

Adam Danz
Adam Danz on 8 Sep 2019
Edited: Adam Danz on 8 Sep 2019
[~, fname] = fileparts('scrubbed.MOD_D3_AOD_550.20020112.nc');
[~,tok] = regexp(fname,'.(\d+)$','match','tokens');
str = tok{1}{1};
  4 Comments
Adam Danz
Adam Danz on 8 Sep 2019
Edited: Adam Danz on 8 Sep 2019
Glad I could help. The other answers here reminded me to make clear the assumption in my answer that the string of interest is always at the end of the filename (ignoring the final file extension) and is preceeded by a decimal point.

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More Answers (3)

Stephen23
Stephen23 on 8 Sep 2019
Simpler:
>> str = 'scrubbed.MOD_D3_AOD_550.20020112.nc';
>> out = regexp(str,'\d{8}','match','once')
out = 20020112
  2 Comments
Adam Danz
Adam Danz on 8 Sep 2019
It is simpler and assumes that the string of interest will always have 8 digits and that will be the only sub-string with 8 digits.

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Image Analyst
Image Analyst on 8 Sep 2019
Try strsplit():
parts = strsplit('scrubbed.MOD_D3_AOD_550.20020112.nc', '.') % Separate in between dots.
yourNumber = parts{end-1} % Take the next to the last one.
  2 Comments
Adam Danz
Adam Danz on 8 Sep 2019
Edited: Adam Danz on 8 Sep 2019
This is also simpler than my answer if the assumptions are true that the string of interest is the 2nd to last segment surrounded by decimal points.
S Roy
S Roy on 8 Sep 2019
Thanks. Now I know many different ways to do it.

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madhan ravi
madhan ravi on 8 Sep 2019
regexp('scrubbed.MOD_D3_AOD_550.20020112.nc',...
'\d*(?=\.nc)','match','once')

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