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How do I get the date out of a filename using regexp?

조회 수: 20 (최근 30일)
Shayma Al Ali
Shayma Al Ali 2019년 11월 21일
답변: Stephen23 2019년 11월 21일
Currently, I have a list of files that are saved as the following 'xxx_2014_06_03_00_00_01'. I'd like to extract the date in the file name using regexp. Currently, my code is:
fdate=regexp(fList(i).name,'_/d+','once')
dt=datetime(fdate,'InputFormat','dd-MM-yyyy HH:mm:ss')
However, whenever I test it out, fdate is always returned as empty. I'm extremely confused about regexp and how to use it.
Thanks!

채택된 답변

Star Strider
Star Strider 2019년 11월 21일
There are likely several ways to approach this. I would do it a bit differently:
str = 'xxx_2014_06_03_00_00_01';
fdate = regexp(str, '\d*','match');
datev = str2double(fdate);
dt=datetime(datev)
producing (here):
dt =
datetime
03-Jun-2014 00:00:01
I left these as separate lines to demonstrate how it works. The str2double call could be inserted into the datetime call.

추가 답변 (2개)

Jeremy
Jeremy 2019년 11월 21일
편집: Jeremy 2019년 11월 21일
str = 'xxx_2014_06_03_00_00_01';
id = regexp(str,'\d')
Returns:
id =
5 6 7 8 10 11 13 14 16 17 19 20 22 23
These are the indices of str that contain numeric digits. If you know that the 'xxx' part of the filename will not contain numeric digits, then you could then go ahead and say
year = str2double(str(id(1:4)));
month = str2double(str(id(5:6)));
etc.
Alternatively, you could use 'strtok' and pass it the underscore character as a delimiter to get a string of the datetime and convert those to integers. This is likely less efficient, though.

Stephen23
Stephen23 2019년 11월 21일
In one line without intermediate indices or double vector:
>> str = 'xxx_2014_06_03_00_00_01';
>> dtm = datetime(regexp(str,'(\d+_?){6}','match','once'),'InputFormat','yyyy_MM_dd_HH_mm_ss')
dtm =
03-Jun-2014 00:00:01

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