days count each month wise

[BrahmanandaRao]

data a;                       
jan=intnx('month',01,0,'e');
feb=intnx('month',02,0,'e');
mar=intnx('month',03,0,'e');
apr=intnx('month',04,0,'e');
may=intnx('month',05,0,'e');
jun=intnx('month',06,0,'e');
jul=intnx('month',07,0,'e');
aug=intnx('month',08,0,'e');
sep=intnx('month',09,0,'e');
oct=intnx('month',10,0,'e');
nov=intnx('month',11,0,'e');
dec=intnx('month',12,0,'e');
run;    

How to count number of days each month wise

Regards,

anand

[PeterClemmensen · Accepted solution]

data a;
    do dt='01jan2019'd by 0 until (dt ge '31dec2019'd);
        dt=intnx('month', dt, 0, 'e');
        days=day(dt);
        output;
        dt=dt+1;
    end;
    format dt monname.;
run;

[BrahmanandaRao]

Hi draycut 

Code is running to long 

But  not correct output

[PeterClemmensen]

My code should not take too long. No data is being read, only 12 rows created.

For me to replicate the "correct" output, please post exactly what the correct output is.

[Kurt_Bremser]

---

@BrahmanandaRao wrote:

Hi draycut 

Code is running to long 

---

Can't be, @PeterClemmensen 's code takes just one hundredth of a second:

73         data a;
 74             do dt='01jan2019'd by 0 until (dt ge '31dec2019'd);
 75                 dt=intnx('month', dt, 0, 'e');
 76                 days=day(dt);
 77                 output;
 78                 dt=dt+1;
 79             end;
 80             format dt monname.;
 81         run;
 
 NOTE: The data set WORK.A has 12 observations and 2 variables.
 NOTE:  Verwendet wurde: DATA statement - (Gesamtverarbeitungszeit):
       real time           0.01 seconds
       cpu time            0.00 seconds

and the result looks fine to me:

                                                     Beob.           dt    days

                                                        1       January     31 
                                                        2      February     28 
                                                        3         March     31 
                                                        4         April     30 
                                                        5           May     31 
                                                        6          June     30 
                                                        7          July     31 
                                                        8        August     31 
                                                        9     September     30 
                                                       10       October     31 
                                                       11      November     30 
                                                       12      December     31 

[BrahmanandaRao]

Hi Draycut

Just i reopen my sas application and check your solution 

it's perfect output   

Thank You for your support 

Regards,

Anand

[yabwon]

Hi @BrahmanandaRao 

do you mean something like this:

data a;
  first_period = '1jan2018'd; 
  drop first_period;
  array Months jan feb mar apr may jun jul aug sep oct nov dec;

  do over Months; 
    Months = day(intnx('month',first_period, _i_-1, 'e'));
  end;
run; 

or a bit more general:

data first_period;
format first_period yymmdd10.;
first_period = '1jan2013'd; output;
first_period = '2feb2014'd; output;
first_period = '3mar2015'd; output;

first_period = '4apr2016'd; output;
first_period = '5may2017'd; output;
first_period = '6jun2018'd; output;

first_period = '7jul2019'd; output;
first_period = '8aug2020'd; output;
first_period = '9sep2021'd; output;

first_period = '10oct2022'd; output;
first_period = '11nov2023'd; output;
first_period = '12dec2024'd; output;
run;



data a;
  set first_period;
  array Months jan feb mar apr may jun jul aug sep oct nov dec;

  do over Months; 
    Months = day(intnx('month',first_period, _i_-month(first_period), 'e'));
  end;
run;  

All the best

Bart

[BrahmanandaRao]

Hi   Yabown

Thank you for your initiative