I have dataset with dates stored as strings in a format ddMonyy e.g. 19Dec16.
When converting the strings using date7. informat to SAS date, some years are interpreted as 19yy and some as 20yy.
Here is a sample code
data strDates;
infile cards;
input StringDate $;
cards;
31Dec99
01Jan00
19Dec16
31Dec25
01Jan26
;
run;
data convertTest;
set strDates;
format Date date9.;
Date=input(StringDate,date7.);
run;
Running the code today (19 Dec 2016) produces the following results
strDate date
31Dec99 31DEC1999
01Jan00 01JAN2000
19Dec16 19DEC2016
31Dec25 31DEC2025
01jan26 01JAN1926
Dates between 01Jan00 and 31Dec25 are assigned to years 2000-2025 while dates from 01Jan26-31Dec99 are treated as years 1926-1999
Question:
How is it determined if 2000 or 1900 is to added to the year? I suspect it is dependent on the runtime (calendar year when the code is run?) - but I was not able to find any reference to this in SAS documentation.
There is an option, YEARCUTOFF, which depending on your system and version probably has a value of either 20 or 26. See KB note 46368 for more information on the change.
It sounds like you're using SAS 9.4, which means the default is 26: anything from 0-25 will be '20xx' and anything from 26-99 will be '19xx'. You can change the YEARCUTOFF option if that value does not work for your data (or, construct the 4 digit year yourself).
Related
I have a date variable in ddmmyy8. a format such as 24/12/12, I want to change it to ddmmyy10. format, however, it automatically changes to 24/12/2012, which is not right, the date should be 24/12/1912.
Is there any way to tell SAS it is 1912, not 2012? or I have to minus 100 years manually?
How can you tell which dates should be 2012 and which should be 1912?
If your variable is really a SAS date variable then the century was already 2000, you just couldn't see it before because your were using a display format that didn't show the century. You can use intnx() to adjust the value.
want = intnx('year',have,-100,'same');
If you are converting strings into dates then you might be able to use YEARCUTOFF option to have it interpret 12 as 1912 instead of 2012.
857 %let optsave=%sysfunc(getoption(yearcutoff));
858 %put &=optsave;
OPTSAVE=1926
859 %put %sysfunc(inputn(24/12/12,ddmmyy8),date9);
24DEC2012
860 options yearcutoff=1912;
861 %put %sysfunc(inputn(24/12/12,ddmmyy8),date9);
24DEC1912
862 options yearcutoff=&optsave;
I want to assign the current year in a YY format to either a macro or data set variable.
I am able to use the automatic macro variables &sysdate or &sysdate9 to get the current date. However, extracting the year in a YY format is proving to be a nightmare. Below are some examples of what I've been trying.
There exists the YEARw. format. But when I try to use it I get errors or weird results. For instance, running
data _null_;
yy = year(input("&sysdate9.", year2.));
put yy=;
run;
produces the error
ERROR 48-59: The informat YEAR was not found or could not be loaded.
If I try to format the variable in the output, I get 1965 instead of the current year. The following
data _null_;
yy = year(input("&sysdate9.", date9.));
put yy= yy year2.;
run;
outputs
yy=2016 65
Please help.
This works to get you the 2-digit year number of the current year:
DATA _NULL_;
YEAR = PUT(TODAY(),YEAR2.);
PUT YEAR;
RUN;
/* Returns: 16 */
To breakdown what I am doing here:
I use TODAY() to get the current date as a DATE type. &SASDATE needs to be converted to a DATE, but also it is the date that the SAS session started. TODAY() is the current date.
PUT allows us to pass in a non-character (numeric/date) value, which is why it is used with TODAY() as opposed to INPUT.
I think it is worth exploring the issues here in more detail.
First, Formats are patterns for converting numeric values to a human readable format. That's what you want to do here: convert a date value to a human readable format, in this case to a year.
Informats, on the other hand, convert human readable information to numeric values. That's not what you're doing here; you have a value already.
Second, put matches with Formats, and input matches with informats, exclusively.
Third, you get close in your last try: but you misuse the year format. Formats are basically value mappings, so they map every possible numeric value in their range (sometimes "all values" is the range, sometimes not) to a display value (string). You need to know what kind of value is expected on the input. YEARw. expects a date value as input, not a year value: meaning input is "number of days from 1/1/1960", mapped to "year". So you cannot take a value you've already mapped to a year value and map it again with that method; it will not make any sense.
Let's look at it:
data _null_;
yy = year(input("&sysdate9.", date9.));
put yy= yy year2.;
run;
yy contains the result of the year function - 2016. Good so far. Now, you need the 2 digit year (16); you can get that through mod function, if you like, or put/substr/input:
data _null_;
yy = input(substr(put(year(input("&sysdate9.", date9.)),4.),3,2),2.);
put yy=;
run;
mod is probably easier though since it's a number. But of course you could've used year:
data _null_;
yy = put(input("&sysdate9.", date9.),year2.);
put yy=;
run;
Now, yy is character, so you could wrap that with input(...,2.) or leave it character depending on your purposes.
Finally - a use note on &sysdate9.. You can easily make this a date without input:
"&sysdate9."d
So:
yy = put("&sysdate9."d,year2.);
That's called a date literal (and "..."dt and "..."t also work for datetime,time). They require things in the standard SAS formats to work properly.
And as pointed out in Nicarus' answer, today() is a bit better than &sysdate9 since it is guaranteed to be today. If you're running this in batch or restart your session daily, this won't matter, but it will if you have a long-running session.
Apply the year function to the date variable
Convert to string
Take last 2 digits
EDIT: change input to PUT
Year = substr(put(year(today()), 4.), 3);
I have a problem in Stata with the format of the dates. I believe it is a very simple question but I can't see how to fix it.
I have a csv file (file.csv) that looks like
v1 v2
01/01/2000 1.1
01/02/2000 1.2
01/03/2000 1.3
...
01/12/2000 1.12
01/02/2001 1.1
...
01/12/2001 1.12
The form of v1 is dd/mm/yyyy.
I import the file in Stata using import delimited ...file.csv
v1 is a string variable, v2 is a float.
I want to transform v1 in a monthly date that Stata can read.
My attempts:
1)
gen Time = date(v1, "DMY")
format Time %tm
which gives me
Time
3177m7
3180m2
3182m7
...
that looks wrong.
2) In alternative
gen v1_1=v1
replace v1_1 = substr(v1_1,4,length(v1_1))
gen Time_1 = date(v1_1, "MY")
format Time_1 %tm
which gives exactly the same result.
And if I type
tsset Time, format(%tm)
it tells me that there are gaps but there are no gaps in the data.
Could you help me to understand what I'm doing wrong?
Stata has wonderful documentation on dates and times, which you should read from beginning to end if you plan on using time-related variables. Reading this documentation will not only solve your current problem, but will potentially prevent costly errors in the future. The section related to your question is titled "SIF-to-SIF conversion." SIF means "Stata internal form."
To explain your current issue:
Stata stores dates as numbers; you interpret them as "dates" when you assign a format. Consider the following:
set obs 1
gen dt = date("01/01/2003", "DMY")
list dt
// 15706
So that date is assigned the value 15706. Let's format it to look like a day:
format dt %td
list
// 01jan2003
Now let's format it to be a month:
format dt %tm
list
// 3268m11
Notice that dt is just a number that you can format and use like a day or month. To get a "month number" from a "day number", do the following:
gen mt = mofd(dt) // mofd = month of day
format mt %tm
list
// dt mt
// 3268m11 2003m1
The variable mt now equals 516. January 2003 is 516 months from January 1960. Stata's "epoch time" is January 1, 1960 00:00:00.000. Date variables are stored as days since the epoch time, and datetime variables are stored as miliseconds since the epoch time. A month variable can be stored as months since the epoch time (that's how the %tm formatting determines which month to show).
Hi I have a date conversion problem in SAS,
I imported an excel file which has the following dates.,
2012-01-09
2011-01-31
2010-06-28
2005-06-10
2012-09-19
2012-09-19
2007-06-12
2012-09-20
2004-11-01
2007-03-27
2008-06-23
2006-04-20
2012-09-20
2010-07-14
after I imported the dates have changed like this
40917
40574
40357
38513
41171
41171
39245
41172
38292
39168
39622
38827
41172
40373
I have used the input function to convert the dates but it gives a strange result.,
the code I used.,
want_date=input(have_date, anydtdte12.);
informat want_date date9.; format have_date date9.;run;
I get very stange and out of the World dates., any idea how can I convert these?
You can encourage SAS to convert the data as date during the import, although this isn't necessarily a panacea.
proc import file=whatever out=whatever dbms=excel replace;
dbdsopts=(dbSasType=( datevar=date ) );
run;
where datevar is your date column name. This tells SAS to expect this to be a date and to try to convert it.
See So Your Data Are in Excel for more information, or the documentation.
From : http://www2.sas.com/proceedings/sugi29/068-29.pdf
Times are counted internally in SAS as seconds since midnight and
date/time combinations are calculated as the number of seconds since
midnight 1 January 1960.
Excel also uses simple numerical values for dates and times
internally. For the date values the difference with the SAS date is
only the anchor point. Excel uses 1 January 1900 as day one.
So add a constant.
EXAMPLES:
SAS_date = Excel_date - 21916;
SAS_time = Excel_time * 86400;
SAS_date_time = (Excel_date_time - 21916) * 86400;
As Justin wrote you need to correct for the different zero date (SAS vs. Excel).
Then you just need to apply a format (if you want to get a date variable to do calculations):
want_date = have_date-21916;
format want_date date9.;
Or convert it to a string:
want_date = put(have_date-21916, date9.);
In either case you can choose the date format you prefer.
I'm trying to convert a SAS string of the form "MO-YR" (e.g. "Jan-04") to a SAS date.
Unfortunately, I don't think there is a date format in SAS that takes that form, so I can't just use an input statement like this date = input(datestring, sasformat).
I've been using this site to find date formats: http://v8doc.sas.com/sashtml/lrcon/zenid-63.htm
Thanks,
Michael
MONYY seems to work for me.
data _null_;
input #1 mydate monyy6.;
put mydate= date9.;
datalines;
Jan-04
Dec-12
;;;;
run;
Puts the correct values to the log.