sas time value
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From: peter zeng on 30 Oct 2009 16:37 I have two datetime values such as "13SEP99:19:45:04" and "13SEP99:23:05:36" I want (13SEP99:23:05:36)minus(13SEP99:19:45:04) so that I can get the actural numebrs Do everyone know how to do it here is my code format date date9. time time8. datetime datetime16.; datetime=(date*24*60*60)+ time; just learn SAS
From: Arthur Tabachneck on 30 Oct 2009 18:12 Peter, If you want the result shown in terms of hours you could just use: data have; start_time="13SEP99:19:45:04"dt; end_time="13SEP99:23:05:36"dt; run; data want; set have; diff=(end_time-start_time)/(60*60); run; HTH, Art ------------ On Oct 30, 4:37 pm, peter zeng <taifa.str...(a)gmail.com> wrote: > I have two datetime values such as "13SEP99:19:45:04" and > "13SEP99:23:05:36" > I want (13SEP99:23:05:36)minus(13SEP99:19:45:04) > so that I can get the actural numebrs > > Do everyone know how to do it > > here is my code > format date date9. time time8. datetime datetime16.; > datetime=(date*24*60*60)+ time; > > just learn SAS
From: Ruslan Kirhizau on 1 Nov 2009 11:52
Hi=2C =20 The answer depends on how your values are stored and/or read.=20 =20 If you have two SAS datetime values then you can just simply subtract one f= rom another and get difference in seconds as a result. SAS datetime value i= s numeric value representing number of seconds from midnight=2C January 1= =2C 1960.=20 SAS date value is number of days from January=2C 1=2C 1960.=20 SAS time value is number of seconds from midnight. It resets to 0 every mid= night.=20 All three counters (datetime=2C date and time) are numeric. =20 If you have your values stored as character strings=2C then you need to con= vert them to appropriate numeric date=2C time or datetime value first and p= erform calculations after that.=20 You can use input finction=2C for example=2C if variables character_string1= and character_string2 contain "13SEP99:19:45:04" and "13SEP99:23:05:36" := =20 num_dt1 =3D input (character_string1=2C datetime19.)=3B /*convert to the nu= mber of seconds from midnight 1st Jan 1960*/ = =20 num_dt2 =3D input (character_string2=2C datetime19.)=3B dif =3D num_dt2 - num_dt1 =3B /*difference in seconds*/ format dif time5. =3B =20 =20 =20 SAS date/time formats have no effect on performed calculations. Formats cou= ld be applied to the results for visual representation of the date or time.= =20 =20 Hope this helps a little =20 Thanks=20 =20 Ruslan =20 =20 =20 =20 =20 =20 =20 > Date: Fri=2C 30 Oct 2009 13:37:11 -0700 > From: taifa.strong(a)GMAIL.COM > Subject: sas time value > To: SAS-L(a)LISTSERV.UGA.EDU >=20 > I have two datetime values such as "13SEP99:19:45:04" and > "13SEP99:23:05:36" > I want (13SEP99:23:05:36)minus(13SEP99:19:45:04) > so that I can get the actural numebrs >=20 > Do everyone know how to do it >=20 > here is my code > format date date9. time time8. datetime datetime16.=3B > datetime=3D(date*24*60*60)+ time=3B >=20 > just learn SAS =20 _________________________________________________________________ Windows 7: It works the way you want. http://www.microsoft.com/Windows/windows-7/default.aspx?ocid=3DPID24727::T:= WLMTAGL:ON:WL:en-US:WWL_WIN_evergreen2:112009= |