Removing Leap day from my timetable

I need to remove leap days of all years from my timetable. I mean all values belong to 29 feb. my timetable is a hourly time table.

[Adam Danz]

That data you attached does not contain a timetable or any kind of datatime data.

Answer 1

[Adam Danz]

TT is the timetable and TT.Time is the time column.

isLeapDay is a logical vector identiying the rows of TT that belong to a leap day.

isLeapDay = month(TT.Time)==2 & day(TT.Time)==29; 
TT(isLeapDay,:) = [];  % remove leap days. 

[Hamed Hedayatnia]

I use this code and get this error:

Unable to use a value of type 'datetime' as an index.

[Hamed Hedayatnia]

this is a part of my timetable

[Adam Danz]

Could you show me your code and explain the variables in the code? I'm not using datetime as an index so I suspect there is an implementation error.

[Hamed Hedayatnia]

I made a timetable:

Data_m = table2timetable(Data,'RowTimes',timeSamples_1);

then change your code to with my table name

isLeapDay = month(Data_m.Time)==2 & day(Data_m.Time)==29;

Data_m(isLeapDay,:) = []; % remove leap days.

after applying i got this error

[Hamed Hedayatnia]

isLeapDay = month(TT.Time)==2 & day(TT.Time)==29;

I think this creates an index

[Adam Danz]

Yes, that creates an index; not a datetime vector. The error message you shared "Unable to use a value of type 'datetime' as an index." indicates that the error is caused by something else.

I just applied the answer to the data you shared and it works without error.

Please provide the full copy-pasted error message and include the line of code that the error message points to. The error message is not associated with the code from my answer.

Also, what version of Matlab are you using?

[Hamed Hedayatnia]

Thanks Adam, I applied it in anouther file and it works great.

[Adam Danz]

Sounds like you worked it out! Glad I could help.