Problems trying to find fakes dates.

 idx =

        2005           2          28          18           0           0
        2005           2          28          19           0           0
        2005           2          28          20           0           0
        2005           2          28          21           0           0
        2005           2          28          22           0           0
        2005           2          28          23           0           0
        2005           3           1           0           0           0
        2005           3           1           1           0           0
        2005           3           1           2           0           0
        
>> time = datenum(idx)
   
time =

                 732371.75
          732371.791666667
          732371.833333333
                732371.875
          732371.916666667
          732371.958333333
                    732372
          732372.041666667
          732372.083333333
          
% looking for a date that is in the domain
>> find(time == datenum(2005,2,28,18,0,0))

ans =

     1
     
% looking for a date that is NOT in the domain     
>> find(time == datenum(2005,2,29,0,0,0))

ans =

     7   
     
% the output shuld be  0×1 empty double column vector.

          
% Why when u try to find fakes dates, the output is a value of an existing date in the domain?         

[Fangjun Jiang]

Good question. I remember using this trick to check if a year is a leap year.

I bet there is a rational for this. It could also be implented to report error if date inputs are "out of range".

datenum(2004,2,29)==datenum(2004,3,1)

ans = logical
   0

datenum(2005,2,29)==datenum(2005,3,1)

ans = logical
   1

datenum(2005,3,32)==datenum(2005,4,1)

ans = logical
   1

[Claudio Iturra]

Hi Fangjun, Thanks! I already add a line to my code that verify ramdoms date...basically, I'm evaluating parameters in the ocean during ramdon dates, but becouse Im stupid, I forget to add a line that controls february's days, thinking that the find(time==datenum()) will did not take fakes dates. Actually the code found those fakes dates during february and did not sent me any error, I just was cheking for bad dates and eventually found that! Maybe datenum needs to verify in some way corrects dates.

Big question here is if 2 elemets are not the same, why the find code sent you the closer one..

find(time == datenum(2005,2,29,0,0,0))

find(x == y)

(x == y)

Have a great day!

[Claudio Iturra]

......mmmmmm

find(time == datenum(2005,45,29,0,0,0)) %month 45 of the year 2005!

ans =

       59112

>> datevec(time(59112))

ans =

        2008           9          29           0           0           0

Answer 1

[the cyclist]

This behavior is due to "carryover" in date vectors and strings.

As recommended there, and many other places in the documentation, use of the newer datetime data type is encouraged.

[Claudio Iturra]

Hi, Same behavior using datetime....

%Same behavior using datetime

idx =

        2005           2          28          18           0           0
        2005           2          28          19           0           0
        2005           2          28          20           0           0
        2005           2          28          21           0           0
        2005           2          28          22           0           0
        2005           2          28          23           0           0
        2005           3           1           0           0           0
        2005           3           1           1           0           0
        2005           3           1           2           0           0

>> time = datetime(idx)

time = 

  9×1 datetime array

   28-Feb-2005 18:00:00
   28-Feb-2005 19:00:00
   28-Feb-2005 20:00:00
   28-Feb-2005 21:00:00
   28-Feb-2005 22:00:00
   28-Feb-2005 23:00:00
   01-Mar-2005 00:00:00
   01-Mar-2005 01:00:00
   01-Mar-2005 02:00:00
   
   
>> find(time == datetime(2005,2,29,0,0,0)) %fake date

ans =

     7

[the cyclist]

Sorry, I should not have implied that using datetime would solve the carryover issue. datetime also implement the carryover algorithm discussed on that page.

I did some searching, and did not find a definitve date validation method in MATLAB. The topic has come up a few times in this forum. It seems to me that this thread is your best bet, which seems to have settled on this code:

function valid = valid_date(year, month, date)
if(nargin ~= 3)
    valid = false;
elseif ((~isscalar(year)||(mod(year,1)~=0) || year<0))
    valid = false;
elseif ((~isscalar(month))||(mod(month,1)~=0) || (month<=0) || (month>12))
    valid = false;
elseif ((~isscalar(date))||(mod(date,1)~=0) || (date<=0))
    valid = false;
elseif(any(month==[1:2:7,8:2:12])&& date>31)
    valid = false;
elseif (any(month==[4,6,9,11]) && date>30)
    valid = false;
elseif month==2 && date>(28+(mod(year,400)==0 || (mod(year,100)~=0 && mod(year,4)==0)))
    valid=false;    
else
    valid = true;    
end

Answer 2

[Eric Sofen]

The reason datetime(2005,2,29) works is that we want to support constructing a vector of datetimes:

datetime(2023,1,1:100)

ans = 1×100 datetime array
   01-Jan-2023   02-Jan-2023   03-Jan-2023   04-Jan-2023   05-Jan-2023   06-Jan-2023   07-Jan-2023   08-Jan-2023   09-Jan-2023   10-Jan-2023   11-Jan-2023   12-Jan-2023   13-Jan-2023   14-Jan-2023   15-Jan-2023   16-Jan-2023   17-Jan-2023   18-Jan-2023   19-Jan-2023   20-Jan-2023   21-Jan-2023   22-Jan-2023   23-Jan-2023   24-Jan-2023   25-Jan-2023   26-Jan-2023   27-Jan-2023   28-Jan-2023   29-Jan-2023   30-Jan-2023   31-Jan-2023   01-Feb-2023   02-Feb-2023   03-Feb-2023   04-Feb-2023   05-Feb-2023   06-Feb-2023   07-Feb-2023   08-Feb-2023   09-Feb-2023   10-Feb-2023   11-Feb-2023   12-Feb-2023   13-Feb-2023   14-Feb-2023   15-Feb-2023   16-Feb-2023   17-Feb-2023   18-Feb-2023   19-Feb-2023   20-Feb-2023   21-Feb-2023   22-Feb-2023   23-Feb-2023   24-Feb-2023   25-Feb-2023   26-Feb-2023   27-Feb-2023   28-Feb-2023   01-Mar-2023   02-Mar-2023   03-Mar-2023   04-Mar-2023   05-Mar-2023   06-Mar-2023   07-Mar-2023   08-Mar-2023   09-Mar-2023   10-Mar-2023   11-Mar-2023   12-Mar-2023   13-Mar-2023   14-Mar-2023   15-Mar-2023   16-Mar-2023   17-Mar-2023   18-Mar-2023   19-Mar-2023   20-Mar-2023   21-Mar-2023   22-Mar-2023   23-Mar-2023   24-Mar-2023   25-Mar-2023   26-Mar-2023   27-Mar-2023   28-Mar-2023   29-Mar-2023   30-Mar-2023   31-Mar-2023   01-Apr-2023   02-Apr-2023   03-Apr-2023   04-Apr-2023   05-Apr-2023   06-Apr-2023   07-Apr-2023   08-Apr-2023   09-Apr-2023   10-Apr-2023

Note that this is only for constructing datetime from numeric inputs. For text timestamps, it needs to be a valid date/time on the calendar:

% This errors:
% datetime("2005-02-29")
% Error using datetime
% Could not recognize the date/time format of '2005-02-29'. 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.

In your actual workflow, do you have access to the original Y/M/D data before constructing a datetime from it? If so, you could compare the month/day resulting from constructing the datetime to the original to check for spurious data with values that datetime wrapped to the next month:

ymd = [2005 2 28; 2005 2 29; 2005 3 1]

ymd = 3×3

2005 2 28 2005 2 29 2005 3 1

d = datetime(ymd);
find(d.Month == ymd(:,2))

ans = 2×1

1 3

I'd be hesitant to construct a full date validation scheme as @the cyclist laid out. Yes, it's doable, but many over the years have fallen into various traps around leap years and Daylight Saving Time!