How can I convert a character vector that includes date time and random text to datetime format?
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vicsm
el 16 de Nov. de 2018
Respondida: Peter Perkins
el 19 de Nov. de 2018
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!
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Guillaume
el 16 de Nov. de 2018
Is the date and time format consistent?
Is the location of the random text consistent?
If so, what is the actual pattern?
Respuesta aceptada
the cyclist
el 16 de Nov. de 2018
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)
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Más respuestas (2)
Adam Danz
el 16 de Nov. de 2018
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')
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Peter Perkins
el 19 de Nov. de 2018
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
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