Why doesn't Matlab recognize my data format?

Hello everyone,

thanks for reading. My point is that I have a time series from which I want to remove all the 29 of february. My time series is 10 years long.

Now, if I ask Matlab what's the format of my datetime, it replies me with:

ans =

    'dd/MM/uuuu'

and, in fact it is correct. But when I ask Matlab to find in my matrix the position of 29 of february I get this error:

Unable to convert the text '29/02/2018' to a datetime value because its
format was not recognized.

Does someone know why?

I specify that I also tried to do table2timetable before running the find function, but it doesn't work.

Thank you!

Answer 1

[Cris LaPierre]

The error is because 2018 was not a leap year. 2016 and 2020 were.

d1 = '29/02/2016';
datetime(d1)

ans = datetime
   29-Feb-2016

% The error you see
d2 = '29/02/2018';
datetime(d2)

Error using datetime
Could not recognize the date/time format of '29/02/2018'. You can specify a format using the 'InputFormat' parameter. If the date/time text contains day, month, or time zone names in a language
foreign to the 'en_US' locale, those might not be recognized. You can specify a different locale using the 'Locale' parameter.

[Jan]

You have answered this question 30 minutes before?! I did not see your answer before I typed my onw and even not after I sent it. Is it possible that I see the contents with a delay of 30 minutes, when I access the US server of mathworks.com from Europe?

[Cris LaPierre]

I wouldn't rule anything out. @Rena Berman might be able to answer more definitively.

Answer 2

[Jan]

a = datetime('28/02/2018', 'InputFormat', 'dd/MM/uuuu')

a = datetime
   28-Feb-2018

b = datetime('29/02/2018', 'InputFormat', 'dd/MM/uuuu')
% FAILS:
% Unable to convert the text '29/02/2018' to a datetime value because its
% format was not recognized.

There was no Februrary 29th in 2018.

a + days(1)   % 01-Mar-2018

ans = datetime
   01-Mar-2018

A workaround might be the old datenum format:

b = datetime(datevec('29/02/2018', 'dd/mm/yyyy'))

a = datetime
   01-Mar-2018

Ugly, but smarter.

[jjjSAN]

Thank you all. Finally was just this easy thing of leap years fortunately :)

Sorry