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Using datenum for fraction of a second

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Bruno Rodriguez
Bruno Rodriguez el 10 de Mzo. de 2017
Comentada: Walter Roberson el 11 de Mzo. de 2017
I am looking for a ways to convert some dates into datenum, given their special format. My dates look as follows:
'2017-02-28 21:36:51'
'2017-02-28 21:36:51.5'
'2017-02-28 21:36:52'
'2017-02-28 21:36:52.5'
'2017-02-28 21:36:53'
'2017-02-28 21:36:53.5'
etc. etc. etc.
In other words, the format of the dates alternates, such that every other entry has an additional 500 milliseconds at the end of it. How can I convert these to datenum simultaneously, given their different formats? (or at least, how can I convert dates with .5 to datenum?). I cannot skip or ignore any of this data.
Thanks
  1 comentario
Walter Roberson
Walter Roberson el 10 de Mzo. de 2017
Would datetime format instead of datenum format be acceptable?
datenum() tends to be more flexible in format conversion but sometimes datetime() is easier.

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Respuesta aceptada

Walter Roberson
Walter Roberson el 10 de Mzo. de 2017
dates = {'2017-02-28 21:36:51'
'2017-02-28 21:36:51.5'
'2017-02-28 21:36:52'
'2017-02-28 21:36:52.5'
'2017-02-28 21:36:53'
'2017-02-28 21:36:53.5'};
temp = regexprep(dates, ':\d\d$', '$&.0', 'lineanchors');
Now temp is in consistent format
  2 comentarios
Bruno Rodriguez
Bruno Rodriguez el 10 de Mzo. de 2017
Thank you, that certainly helps to make them consistent. However, I'm still unsure how to apply datenum to such a format. I seem to be getting two different datenum outputs for the first two cases above, but when I convert the second one back to datestr to double check, it still returns 21:36:51, even though it has a .5
Walter Roberson
Walter Roberson el 11 de Mzo. de 2017
datestr(datenum(temp), 'YYYY-mm-DD hh:MM:ss.fff')

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