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differnce between 8.2 and 9.1.3
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NEW FORMATS:
The new length for numeric format names is 32. For
character names it is 31. Format names can be associated
with a data set. The only difference between SAS V9 and
previous versions of data sets is the existence of a format
name longer than the traditional 8 bytes.
Version 9 data sets that use 8 byte or shorter format names
can be read by Version 8.
NEW INFORMATS
The new length for numeric informat names is 31 and for
character names is 30. Informat names can be associated
with a data set. The only difference between SAS V9 and
previous versions of data sets is the existence of a format
name longer than the traditional 8 bytes. Version 9 data
sets that use 8 byte or shorter format names can be read by
Version 8.
FIND THE MATTER IN BELOW LINK:
http://www2.sas.com/proceedings/sugi30/266-30.pdf
PLEASE CORRECT IF ERROR IDENTIFIED.
| Is This Answer Correct ? | 0 Yes | 1 No |
How would you code a merge that will keep only the observations that have matches from both sets.
5 Answers Accenture, Bank Of America,
if the Id has more then two transcatiion then show the first observation, IF Id has only two observation then It show both the observation
/* This is example of age caluculate wihtout to display perfect days and years in output window */ data age; retain dob "12jun2003"d now "24may2011"d; age1=now-dob; age=(now-dob)/365.25; years=int(age); days1=round((age-years)*365.25); months=month(now)-1; if days1 gt 30 and months in(12,10,8,6,4,2)then do; month1=days1/30.4375; month=int(days1/30.4375); if day(now)=1 then days=1; else days=round((month1-month)*30.4375)+1; drop days1 month1 month; end; else if days1 gt 30 and months in (1,3,5,7,9,11)then do; month1=days1/30.4375; month=int(days1/30.4375); if day(now)=1 then days=1; else days=round((month1-month) *30.4375); drop days1 month1 month; end; drop age age1; proc print data=age; format dob now date.; run;
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