I have a data in format 01 Jan 19.00.00 (datetime), and I want to extract the month name only from it.
Tried the below code but getting output in numbers i.e. 1 2 3 and so on. I want the output either in Jan Feb mar format or January February march format.
data want;
set detail;
month = month(datepart(BEGIN_DATE_TIME));
run;
You can use the MONNAME format.
data test;
dt = datetime();
monname = put(datepart(dt),MONNAME.);
put monname=;
run;
If you want "OCT" not "OCTOBER" you can add a 3 to the format (MONNAME3.).
If you are using the value in a report the better approach might be to use a date value formatted with MONNAME.
The values of a date formatted variable will be ordered properly when the variable is used in a CLASS or BY statement. If you had instead computed a new variable as the month name, the default ordering of values would be alphabetical.
data want;
set have;
begin_date = datepart(BEGIN_DATE_TIME);
format begin_date MONNAME3.;
run;
Related
I have a dataset with a column "Day" that has very specific time entries like the following. How can I transform these entries into month?
From
Day
31MAY2019:00:00:00
29MAY2020:00:00:00
30APR2021:00:00:00
To
Day
May 2019
May 2020
Apr 2021
You just need to use the datepart() function to get extract the date from the datetime value.
data want;
format datetime datetime20. date monyy7.;
datetime = '31MAY2019:00:00:00'dt;
date = datepart(datetime);
run;
Keep in mind this doesn't add the space in between the month and year and is still a number format. If you want this as a character you can do the following:
data want;
format datetime datetime20.;
datetime = '31MAY2019:00:00:00'dt;
date = put(datepart(datetime),monname3.)||' '||put(datepart(datetime),year.);
run;
I am new to SAS and I am struggling struggling with my code. I would love some help. Am I thinking about this the right way? I have a huge table and I want to extract that data from certain dates. My two dates: 1969-12-01 and 1948-01-01 my sample code:
data null;
call symput ('timenow',put (time(),time.));
call symput ('datenow',put (date(),date9.));
run;
title "The current time is timenow and the date is datenow";
proc print data=sashelp.buy;
run;
So first learn about your dataset. So for example run PROC CONTENTS.
proc contents data=sashelp.buy; run;
Which will show you that there is variable named DATE that has date values (number of days since 1960).
So to reference a specific date use a date literal. That is a quoted string in the style that the DATE informat can read followed by the letter D. You can then use a WHERE statement to filter the data.
data want;
set sashelp.buy;
where date = '31dec1969'd ;
run;
Which will not find any observations since that date does not appear in that dataset.
If you want to select for multiple dates you could either add more conditions using OR.
where (date = '31dec1969'd) or (date = '01jan1948'd);
You can also use the IN operator:
where date in ('31dec1969'd '01jan1948'd);
Note that if your variable contains datetime values (number of seconds) then to pick a specific date you would either need to use a range of datetime literals:
where datetime between '31dec1969:00:00'dt and '31dec1969:11:59:59'dt);
Or convert the number of seconds into number of days and compare to the date literal.
where datepart(datetime) = '31dec1969'd ;
Welcome to StackOverflow Sportsguy3090.
Here I make a dataset called sample with some sample dates. That dataset has a variable called name and another variable called date. Internally, SAS stores dates as the number of days until or after January 1st 1970. That is rough to look at. So I use the format statement to have the dates appear as a 10 character string with month/day/year.
data sample;
name = "Abe "; date = "01Dec1969"d; output;
name = "Betty"; date = "01Jan1948"d; output;
name = "Carl"; date = "06Jun1960"d; output;
name = "Doug"; date = "06Dec1969"d; output;
name = "Ed"; date = "01Jan1947"d; output;
format date mmddyy10.;
run;
The code below subsets the data and puts the good records into a new dataset called keepers. It only keeps the records that are in the date range (including the limit dates).
data keepers;
set sample;
where date between "01jan1948"d and "01Dec1969"d;
run;
I hope that helps.... if not send up another flare.
I created a day variable using the following code:
DAY=datepart(checkin_date_time); /*example of checkin_date_time 1/1/2014 4:44:00*/
format DAY DOWNAME.;
Sample Data:
ID checkin_date_time Admit_Type BED_ORDERED_TO_DISPO
1 1/1/2014 4:40:00 ICU 456
2 1/1/2014 5:64:00 Psych 146
3 1/1/2014 14:48:00 Acute 57
4 1/1/2014 20:34:00 ICU 952
5 1/2/2014 10:00:00 Psych 234
6 1/2/2014 3:48:00 Psych 846
7 1/2/2014 10:14:00 ICU 90
8 1/2/2014 22:27:00 ICU 148
I want to analyze some data using Proc Tab where day is one of the class variables and have the day of week appear in chronological order in the output; however, the output table begins with Tuesday. I would like it to start with Sunday. I've read over the the following page http://support.sas.com/resources/papers/proceedings11/085-2011.pdf and tried the proc format invalue code but it's producing a table that where the "day of week" = "21". Not quite sure where to go from here.
Thanks!
proc format;
invalue day_name
'Sunday'=1
'Monday'=2
'Tuesday'=3
'Wednesday'=4
'Thursday'=5
'Friday'=6
'Saturday'=7;
value day_names
1='Sunday'
2='Monday'
3='Tuesday'
4='Wednesday'
5='Thursday'
6='Friday'
7='Saturday';
run;
data Combined_day;
set Combined;
day_of_week = input(day,day_name.);
run;
proc tabulate data = Combined_day;
class Day Admit_Type;
var BED_ORDERED_TO_DISPO ;
format day_of_week day_names.;
table Day*Admit_Type, BED_ORDERED_TO_DISPO * (N Median);
run;
Fundamentally, you are confusing actual values with displayed values (i.e., formats). Specifically, datepart extracts the date portion out of a date/time field. Then, applying a format only changes how it is displayed not actual underlying value. So below DAY never contains the character values of 'WEDNESDAY' or 'THURSDAY' but original integer value (19724 and 19725).
DAY = datepart(checkin_date_time); // DATE VALUE
format DAY DOWNAME.; // FORMATTED DATE VALUE (SAME UNDERLYING DATE VALUE)
Consider actually assigning a column as weekday value using WEEKDAY function. Then apply your user-defined format for proc tabulate.
data Combined_day;
set Combined;
checkin_date = datepart(checkin_date_time); // NEW DATE VALUE (NO TIME)
format checkin_date date9.;
checkin_weekday = weekday(checkin_date); // NEW INTEGER VALUE OF WEEKDAY
run;
proc tabulate data = Combined_day;
class checkin_weekday Admit_Type;
var BED_ORDERED_TO_DISPO ;
format checkin_weekday day_names.; // APPLY USER DEFINED FORMAT
table checkin_weekday*Admit_Type, BED_ORDERED_TO_DISPO * (N Median);
run;
What I have
I have a column/variable named Release_Date with length 8 format 11. and informat 11.
The observation in that column are in the form YYYYMMDD SAS actually reads them as pure number
What I want
I want to convert those observation in DD/MM/YYYY format with format and informat as DDMMYY10.
Example
Release_Date is 20180612 then I want 12/06/2018
Any help with this regard will be appreciated
Convert to character
Read back in as date
make sure the variable is different since SAS does not allow you to change a variable type
releaseDate_NUM = input(put(Release_Date, 8. -l), yymmdd10.);
The following solution creates a simple dataset, then processes that dataset to add the converted date:
data myDates;
input intDate;
datalines;
20110101
20120202
;run;
data myDates;
set myDates;
format cvtDate mmddyy10.; * create new variable by specifying the format;
cvtDate = input(put(intDate,8.),yymmdd8.); * assign new variable value;
run;
Results should be:
intDate cvtDate
-------- ----------
20110101 01/01/2011
20120202 02/02/2012
I Have a file from excel that is in a short date format, but when SAS reads it in, it turns it into numbers in the 4000 range...when I try and convert this to an excel date with the below formula, it turns the year into 2077...is there a formula to ensure that this date remains in the original format on the read in, or avoid it turning into this 4000 range that is not at all close to the 2017 and 2018 year that my file is starting in. Does that make sense?
data change_date;
format Completed_Date mmddyy8. ;
set check;
completed_date = date_completed;
if 42005 => date_completed >=43466 and date_completed ^=. then
Completed_date = Date_Completed-21916; *commented out 12-21-17 Xalka
dates back to how they are expected;
run;
I am pretty sure this is a duplicate question, but I can't find it.
This is usually caused by mixing character and date values in the same column. This made SAS import the data as a character variable and it results in the actual dates being copied as character versions of the integers that Excel uses to store dates.
Frequently this is caused by entries that look like dates but are really character strings in the Excel file. The best way to fix it is to fix the Excel file so that the column only contains dates. Otherwise you just need to convert the strings to integers and adjust the values to account for the differences in index dates.
So if your values are in a SAS dataset named HAVE in the character variable DATESTRING then you could use this data step to create a new variable with an actual date value.
data want ;
set have ;
if indexc(datestring,'-/') then date=input(datestring,anydtdte32.);
else date = input(datestring,32.) + '01JAN1900'D -2;
format date yymmdd10. ;
run;
The minus 2 is because of difference in whether to start numbering with 1 or 0 and because Excel thinks 1900 was a leap year.
Excel and SAS have different default dates in back-end.
Day 0 in SAS is 1 January 1960 and Day 0 in Excel is 1 January 1900.
So, you will need to convert excel numeric date to sas date using the below formula.
SAS_date = Excel_date - 21916;
data dateExample;
informat dt mmddyy8.;
set dates;
SAS_date = dates - 21916;
dt=sas_Date;
format dt date9.;
run;