How can I convert a character vector that includes date time and random text to datetime format?

I will like to convert a character vector which contains info like this '2018-05-19_07.11.16_test6.csv' to datetime format. The info I actually need is the date (2018-05-19) and time (07.11.16). How can I do this? Thank you!

[Guillaume]

Is the date and time format consistent?

Is the location of the random text consistent?

If so, what is the actual pattern?

Answer 1

[the cyclist]

Guessing at the pattern ...

s =  '2018-05-19_07.11.16_test6.csv'
idx = regexp(s,'_')
d = s(1:idx(1)-1)
t = s(idx(1)+1:idx(2)-1)

[vicsm]

It works! Thank you so much for your help!!

[Stephen23]

Simpler to use the 'split' option::

>> S =  '2018-05-19_07.11.16_test6.csv';
>> C = regexp(S,'_','split');
>> d = C{1}
d = 2018-05-19
>> t = C{2}
t = 07.11.16

[vicsm]

Also works well! Thank you!

Answer 2

[Adam Danz]

This fits the pattern in your example but it is not robust to pattern changes.

t = '2018-05-19_07.11.16_test6.csv'; 
dtchar = regexp(t, '\d+-\d+-\d+_\d+.\d+.\d+', 'match');    %date-time characters
dt = datetime(dtchar, 'InputFormat', 'yyyy-MM-dd_HH.mm.ss')

Answer 3

[Peter Perkins]

None of this will work unles the format is stable, and if it is, a simpler version would be

>> t = '2018-05-19_07.11.16_test6.csv';
>> dt = datetime(t(1:19), 'InputFormat', 'yyyy-MM-dd_HH.mm.ss')
dt = 
  datetime
   19-May-2018 07:11:16

If you are doing this for more than one file name, you presumably have those in a cell array of char row vectors, or (preferably) a string array. A simple modification

>> t = ["2018-05-19_07.11.16_test6.csv"; "2018-05-19_07.11.15_test77.csv"; "2018-05-19_07.11.18_test888.csv"];
>> dt = datetime(extractBefore(t,20), 'InputFormat', 'yyyy-MM-dd_HH.mm.ss')
dt = 
  3×1 datetime array
   19-May-2018 07:11:16
   19-May-2018 07:11:15
   19-May-2018 07:11:18