Hello,
I am trying to use the DATDIF function to retrieve the amount of actual days between two dates:
The dates I am working with are numeric and in the format eg. 170102 and 170320 (2nd Jan 2017 and 20th Mar 2017).
I am unsure of how to prep these dates such that when I input them into DATDIF it will be able to run properly.
Any help is much appreciated,
Declan
Is your value a numeric with the value of 170102. Or is it a date value of 20821 displayed with the SAS display format of yymmdd6.?
SAS data values which are required for any of the date functions to work are days offset from 1 Jan 1960. So if you date 170102 is supposed to be 2 Jan 2017 then the numeric value would have to be 20821.
BTW when working with SAS FORMAT has a very specific interpretation of a display, not the actual values.
If your value is numeric and ALL of them have 6 digits you can create a SAS date value for use with the date functions and date FORMATS with
newdatevar = input(put(datevar,6.),yymmdd6.);
in a data step where datevar is your existing date variable and newdatevar is the SAS date valued variable.
The appearance of the newdatevar would be set with a format, either at use or assign a default format at creation with a line like:
format newdatevar date9. ; <= displays date as 02JAN2017
format newdatevar mmddyy10. ; <= displays date as 01/02/2017 or any of many date format choices.
The SAS formats are very flexible in that you can use a different format in different procedures to create different groupings of your date variables without creating new variables.
A second hint: Using 2 digit years in this day an age is getting to be a poor practice. It is not obvious that 170102 is in 2017 or 1917.
And when the date in question is 010203 which is month, day and year is a complete guess.
Do something like this.
You can use the DATDIF function, but I prefer the INTCK function because it is much more flexible.
I included examples of both, yielding the same result 🙂
data have;
input date1:yymmdd6. date2:yymmdd6.;
datalines;
170102 170320
;
data want;
set have;
format date1 date2 date9.;
days_datdif = datdif(date1, date2, 'act/act');
days_intck = intck('day', date1, date2);
run;
Is your value a numeric with the value of 170102. Or is it a date value of 20821 displayed with the SAS display format of yymmdd6.?
SAS data values which are required for any of the date functions to work are days offset from 1 Jan 1960. So if you date 170102 is supposed to be 2 Jan 2017 then the numeric value would have to be 20821.
BTW when working with SAS FORMAT has a very specific interpretation of a display, not the actual values.
If your value is numeric and ALL of them have 6 digits you can create a SAS date value for use with the date functions and date FORMATS with
newdatevar = input(put(datevar,6.),yymmdd6.);
in a data step where datevar is your existing date variable and newdatevar is the SAS date valued variable.
The appearance of the newdatevar would be set with a format, either at use or assign a default format at creation with a line like:
format newdatevar date9. ; <= displays date as 02JAN2017
format newdatevar mmddyy10. ; <= displays date as 01/02/2017 or any of many date format choices.
The SAS formats are very flexible in that you can use a different format in different procedures to create different groupings of your date variables without creating new variables.
A second hint: Using 2 digit years in this day an age is getting to be a poor practice. It is not obvious that 170102 is in 2017 or 1917.
And when the date in question is 010203 which is month, day and year is a complete guess.
Days between = end date - start date.
You don't need datdif or intck for this.
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