search last day calendar in several year

yearBegin=2015
yearEnd=2023
endMonth=9;

for anno=yearBegin:yearEnd
    for month=1:12
  if anno<=yearEnd and month<=endMonth
      
    bb=calendar(anno,mese)
   cond=bb(:,7) ;  
   idx=find(cond);
   gg=cond(idx) 
    hh=[anno,mese,gg,0,0,0]  
    RP(count)=datetime(hh);
  end 
 end
end
 

i want to do this:

example:

anno=2015;

endMonth=1;

gg=cond(idx)

cond =

3

10

17

24

31

0

gg =

3

10

17

24

31

RP(1)=3-gen-2015

RP(2)=10-gen-2015

RP(3)=17-gen-2015

RP(4)=24-gen-2015

RP(5)=31-gen-2015

next the second month ( mese=2)

RP(6)=7-feb-2015

RP(7)=14-feb-2015

RP(8)=21-feb-2015

RP(9)=28-feb-2015

...

this loop between 2015-2023 year for every month

Answer 1

[dpb]

Iffen I interpret the request correctly, you want the Saturdays from 2015 thru 2023...

dt=[datetime(2015,1,1):days(7):datetime(2023,12,31)].'; % datetime from first DOY to last DOY of years by week interval
RP=dateshift(dt,'dayofweek','Saturday');                % shift to following Saturday
RP.Format=['eeee, ' RP.Format];                         % format to see DOW string to be sure
[RP(1:5); RP(end-4:end)]                                % show first, last few for check...

ans = 10×1 datetime array
   Saturday, 03-Jan-2015
   Saturday, 10-Jan-2015
   Saturday, 17-Jan-2015
   Saturday, 24-Jan-2015
   Saturday, 31-Jan-2015
   Saturday, 02-Dec-2023
   Saturday, 09-Dec-2023
   Saturday, 16-Dec-2023
   Saturday, 23-Dec-2023
   Saturday, 30-Dec-2023

You don't need to keep the separate datetime array, of course, can build the output array in place; just did for showing...

numel(RP)

ans = 470

shows there are at total of 470 in the array.

[piero]

correct your code but

  10×1 datetime array

   Saturday, 03-Jan-2015
   Saturday, 10-Jan-2015
   Saturday, 17-Jan-2015
   Saturday, 24-Jan-2015
   Saturday, 31-Jan-2015
   Saturday, 02-Dec-2023
   Saturday, 09-Dec-2023
   Saturday, 16-Dec-2023
   Saturday, 23-Dec-2023
   Saturday, 30-Dec-2023
   
   
   i want year 2015/2016/2017...2023

[dpb]

They're all there, just displayed first/last five...go look at all of RP

[piero]

I posted the partial problem (because I can't post it completely) I have to do several searches including finding the first and third Saturday of the month... in this case what would the code look like?

[Dyuman Joshi]

Changing the days(7) to days(14) should work.

[piero]

thank you

[dpb]

Changing only to two-week difference won't be reliable; it would produce Saturdays two weeks apart, yes, but not only the first and third for months containing a total of five.

There is not a builtin option to return only the first and third; you'll have to select those by eliminating the ones not in first three weeks of a month or selecting the subset of first, third week in month. I added second answer showing the latter technique.

Answer 2

[dpb]

"...only changing to two-week difference won't be reliable; it would produce Saturdays two weeks apart, yes, but not only the first and third for months containing a total of five."

dt=[datetime(2015,1,1):days(7):datetime(2023,12,31)].'; % datetime from first DOY to last DOY of years by week interval
RP=dateshift(dt,'dayofweek','Saturday');                % shift to following Saturday
RP.Format=['eeee, ' RP.Format];                         % format to see DOW string to be sure
WKS=[1,3];                                              % weeks of month wanted
RP=RP(ismember(week(RP,'weekofmonth'),WKS));            % select the desired subset
[RP(1:5); RP(end-4:end)]                                % show first, last few for check...

ans = 10×1 datetime array
   Saturday, 03-Jan-2015
   Saturday, 17-Jan-2015
   Saturday, 07-Feb-2015
   Saturday, 21-Feb-2015
   Saturday, 07-Mar-2015
   Saturday, 21-Oct-2023
   Saturday, 04-Nov-2023
   Saturday, 18-Nov-2023
   Saturday, 02-Dec-2023
   Saturday, 16-Dec-2023

Adapt to fit; WKS here is a list of weeks of month desired to retain; most any other regular combination could be written with some perturbation of the above type of screening logic.

Clearly you need to peruse the datetime documentation and related links closely; the list of functions at the top of the doc page is invaluable aid in learning of available features...

[piero]

yes thanks...in fact I saw that the suggestion Dyuman Joshi was incorrect

[dpb]

The way to make use of the two-week differential in this case would be to generate only the first Saturday of the month initial, then add two calendar weeks to each of those...

[piero]

another questions:

i want to find end of month

i try it:

dt=eomday(datetime(2015,1,1):datetime(2023,12,31))

Error using eomday

Y must be an integer year number.

[Walter Roberson]

[y,m,~] = ymd(datetime(2015,1,1):calmonths(1):datetime(2023,12,31));
dt = eomday(y,  m)

dt = 1×108

31 28 31 30 31 30 31 31 30 31 30 31 31 29 31 30 31 30 31 31 30 31 30 31 31 28 31 30 31 30

[piero]

thank but can i receive output in this format:

Saturday, 03-Jan-2015
   Saturday, 17-Jan-2015
   Saturday, 07-Feb-2015
   Saturday, 21-Feb-2015
   Saturday, 07-Mar-2015

[Walter Roberson]

dateshift(datetime(2015,1,1):calmonths(1):datetime(2023,12,31), 'end', 'month').'

ans = 108×1 datetime array
   31-Jan-2015
   28-Feb-2015
   31-Mar-2015
   30-Apr-2015
   31-May-2015
   30-Jun-2015
   31-Jul-2015
   31-Aug-2015
   30-Sep-2015
   31-Oct-2015
   30-Nov-2015
   31-Dec-2015
   31-Jan-2016
   29-Feb-2016
   31-Mar-2016
   30-Apr-2016
   31-May-2016
   30-Jun-2016
   31-Jul-2016
   31-Aug-2016
   30-Sep-2016
   31-Oct-2016
   30-Nov-2016
   31-Dec-2016
   31-Jan-2017
   28-Feb-2017
   31-Mar-2017
   30-Apr-2017
   31-May-2017
   30-Jun-2017

[piero]

Thanks, it's very kind of you

[Steven Lord]

To determine the last day of the month, you could subtract 1 day from the first day of the next month.

dt = datetime(2023, (1:12).' + 1, 1) - caldays(1) % or

dt = 12×1 datetime array
   31-Jan-2023
   28-Feb-2023
   31-Mar-2023
   30-Apr-2023
   31-May-2023
   30-Jun-2023
   31-Jul-2023
   31-Aug-2023
   30-Sep-2023
   31-Oct-2023
   30-Nov-2023
   31-Dec-2023

dt2 = datetime(2023, (1:12).' + 1, 0) % Using the concept of "day 0" of the month

dt2 = 12×1 datetime array
   31-Jan-2023
   28-Feb-2023
   31-Mar-2023
   30-Apr-2023
   31-May-2023
   30-Jun-2023
   31-Jul-2023
   31-Aug-2023
   30-Sep-2023
   31-Oct-2023
   30-Nov-2023
   31-Dec-2023

Or if you have a vector of dates that aren't necessarily the first day of the month, you could dateshift your dates to the end of the month. Take some random data:

data = datetime(2023, randi(12, 5, 1), randi(28, 5, 1))

data = 5×1 datetime array
   20-Jan-2023
   03-May-2023
   22-Apr-2023
   19-Jun-2023
   09-Mar-2023

and shift the dates to the end of the month.

dt3 = dateshift(data, 'end', 'month')

dt3 = 5×1 datetime array
   31-Jan-2023
   31-May-2023
   30-Apr-2023
   30-Jun-2023
   31-Mar-2023

[piero]

thank