https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9545#M694 I used a variation of this with the intnx function (which I never used before). Works great, thanks! Wed, 08 Jun 2011 19:07:00 GMT zekez 2011-06-08T19:07:00Z https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9542#M691 Hi!<BR /> <BR /> I am very new to SAS and am using EG. I need a formula, or code, that will generate the day of the year (example; for 7/1/2010, the value 181 would be generated). Any idea how to do this?<BR /> <BR /> Much thanks! Wed, 08 Jun 2011 16:23:26 GMT https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9542#M691 zekez 2011-06-08T16:23:26Z https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9543#M692 Here is one technique which uses INTNX function:<BR /> <BR /> 10 data _null_;<BR /> 11 dt = '01jul2011'd;<BR /> 12 x = dt - intnx('year',dt,0);<BR /> 13 put x= ;<BR /> 14 run;<BR /> <BR /> x=181<BR /> NOTE: DATA statement used (Total process time):<BR /> real time 0.00 seconds<BR /> cpu time 0.00 seconds<BR /> <BR /> <BR /> <BR /> Scott Barry<BR /> SBBWorks, Inc. Wed, 08 Jun 2011 16:36:47 GMT https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9543#M692 sbb 2011-06-08T16:36:47Z https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9544#M693 The JULDATE function gets you close. Use MOD to discard the year portion.<BR /> <BR /> [pre]<BR /> data _null_ ;<BR /> have = '01jul2010'd ;<BR /> want = mod(juldate(have),1000) ;<BR /> put want= ;<BR /> run ;<BR /> [/pre]<BR /> <BR /> &gt; Hi!<BR /> &gt; <BR /> &gt; I am very new to SAS and am using EG. I need a<BR /> &gt; formula, or code, that will generate the day of the<BR /> &gt; year (example; for 7/1/2010, the value 181 would be<BR /> &gt; generated). Any idea how to do this?<BR /> &gt; <BR /> &gt; Much thanks! Wed, 08 Jun 2011 17:14:38 GMT https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9544#M693 Howles 2011-06-08T17:14:38Z https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9545#M694 I used a variation of this with the intnx function (which I never used before). Works great, thanks! Wed, 08 Jun 2011 19:07:00 GMT https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9545#M694 zekez 2011-06-08T19:07:00Z https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9546#M695 Thanks all! One follow up question, any idea how to gererate days in a given year? EG: I want the output to be 365 for most years, and 366 for leap years.<BR /> Input 2003, output 365; Input 2004, output 366. Thu, 09 Jun 2011 15:13:44 GMT https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9546#M695 zekez 2011-06-09T15:13:44Z https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9547#M696 Same type of SAS assignment statement, again using INTNX for both the start and end dates, however also there is the INTCK function too. Have a peek at the DOC.<BR /> <BR /> Scott Barry<BR /> SBBWorks, Inc.<BR /> <BR /> Suggested Google advanced search argument, this topic / post:<BR /> <BR /> intck function site:sas.com<BR /> <BR /> calculate days in year site:sas.com Thu, 09 Jun 2011 15:23:45 GMT https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9547#M696 sbb 2011-06-09T15:23:45Z https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9548#M697 leap years are divisible by 4 but not 100 unless divisible by 400 Thu, 09 Jun 2011 16:24:08 GMT https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9548#M697 Peter_C 2011-06-09T16:24:08Z https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9549#M698 [pre]<BR /> <BR /> %let year=2004;<BR /> data _null_;<BR /> day=intck('day',"01jan&amp;year"d,"01jan%eval(&amp;year+1)"d);<BR /> put day=;<BR /> run;<BR /> [/pre]<BR /> <BR /> <BR /> Ksharp Fri, 10 Jun 2011 01:49:38 GMT https://communities.sas.com/t5/SAS-Procedures/Day-of-the-Year/m-p/9549#M698 Ksharp 2011-06-10T01:49:38Z