how to read day of the year as datetime ?

I have vector of dates, it looks like this

2019001
2019002
.
.
.
2019364
2019365

2019 is year and next number is day of the year.

anybody tell me how to use datenum or datetime here !

I have around 3 years of dara.

Answer 1

[per isakson]

>> chr = sprintf( '%d', 2019364 );
>> sdn = datenum(str2double(chr(1:4)),1,0)+str2double(chr(5:end));
>> datetime( sdn, 'ConvertFrom', 'datenum' )
ans = 
  datetime
   30-Dec-2019 00:00:00
>> 

Added later

%%
d = [ 2019001; 2019002; 2019364; 2019365 ];
%%
chr = num2str( d );
chr = strjust( chr, 'left' );
sdn = datenum( str2num(chr(:,1:4)), 1, 0 ) + str2num(chr(:,5:end));
dt  = datetime( sdn, 'ConvertFrom', 'datenum' )

Outputs

dt = 
  4×1 datetime array
   01-Jan-2019 00:00:00
   02-Jan-2019 00:00:00
   30-Dec-2019 00:00:00
   31-Dec-2019 00:00:00

Answer 2

[madhan ravi]

 v = regexprep(""+vector,'(\d{4})(\d+)','00/$2/$1');
 Wanted =  datetime(datestr(v))

[madhan ravi]

 Days_in_a_year = double(regexprep(v, '(\d+)/(\d+)/(\d+)', '$2'))

[madhan ravi]

For older versions:

v = regexprep(sprintfc('%d',vector),'(\d{4})(\d+)','00/$2/$1'); % sprintfc() undocumented
Wanted =  datetime(datestr(v))

 Days_in_a_year = str2double(regexprep(v, '(\d+)/(\d+)/(\d+)', '$2'));

Answer 3

[Star Strider]

Another approach:

ydv = [2019001
       2019002
       2019364
       2019365];                                        % ‘YearDay’ Vector

Years_Days = datetime(num2str(fix(ydv/1000)), 'InputFormat','yyyy') + caldays(rem(ydv,1000)-1)

producing:

Years_Days = 

  4×1 datetime array

   01-Jan-2019
   02-Jan-2019
   30-Dec-2019
   31-Dec-2019

Use the 'Format' name-value pair to get the desired output format. The easiest way to do that (for one example) is simply:

Years_Days.Format = 'yyyy-MM-dd';

to get:

Years_Days = 

  4×1 datetime array

   2019-01-01
   2019-01-02
   2019-12-30
   2019-12-31

.