Lets say I have the following dates for the observations
data dates;
input obs date$11.;
cards;
1 06/10/1949
2 01/07/1952
3 02/10/1947
;
run;
But now I want to insert another column next to date called new date under the date9. format and this new date column is to be numeric.
I tried the following,
data newdata;
set dates;
newdate=input(date,date9.);
run;
But when I run this, the new date column seems to be empty
Your string values are not using a format that is compatible with the DATE. informat. They appear to be using either MMDDYY. or DDMMYY., but it is not possible to tell which from your example values.
data dates;
input obs datestr :$11.;
date1 = input(datestr,mmddyy10.);
date2 = input(datestr,ddmmyy10.);
format date1 date2 date9. ;
cards;
1 06/10/1949
2 01/07/1952
3 02/10/1947
;
results:
Obs obs datestr date1 date2
1 1 06/10/1949 10JUN1949 06OCT1949
2 2 01/07/1952 07JAN1952 01JUL1952
3 3 02/10/1947 10FEB1947 02OCT1947
Related
What am I trying to do is the following:
Have this table:
Table1
Item Date1 Date 2
1 6/1/2021 7/31/2021
2 7/4/2021 7/30/2021
3 6/20/2021 7/28/2021
....
My want table is the following:
Item Date
1 6/1/2021
1 6/6/2021
1 6/11/2021
1 6/16/2021
...
Basically I am trying to create a date by incrementing 5 days from the start date until the last date.
Something like this should get you started:
data want;
set have;
format date date1 date2 date9.;
do Date=date1 to date2 by 5;
Date = MIN(Date, Date2);
output;
end;
*keep Item Date;
run;
Date set having id and date .I want a date set with two duplicate id but condition is that one should be before 8th June and other should be after 8th June.
To take the first date and the first date after 2021-06-08 you can sort by ID and DATE and use LAG() to detect when you cross the date boundary.
data have ;
input id date :date. ;
format date date9.;
cards;
1 01jun2021
1 07jun2021
1 08jun2021
1 09jun2021
;
data want;
set have ;
by id date;
if first.id or ( (date<='08JUN2021'd) ne lag(date<='08JUN2021'd));
run;
results
Obs id date
1 1 01JUN2021
2 1 09JUN2021
I have a sas datebase with something like this:
id birthday Date1 Date2
1 12/4/01 12/4/13 12/3/14
2 12/3/01 12/6/13 12/2/14
3 12/9/01 12/4/03 12/9/14
4 12/8/13 12/3/14 12/10/16
And I want the data in this form:
id Date Datetype
1 12/4/01 birthday
1 12/4/13 1
1 12/3/14 2
2 12/3/01 birthday
2 12/6/13 1
2 12/2/14 2
3 12/9/01 birthday
3 12/4/03 1
3 12/9/14 2
4 12/8/13 birthday
4 12/3/14 1
4 12/10/16 2
Thanks by ur help, i'm on my second week using sas <3
Edit: thanks by remain me that i was not finding a sorting method.
Good day. The following should be what you are after. I did not come up with an easy way to rename the columns as they are not in beginning data.
/*Data generation for ease of testing*/
data begin;
input id birthday $ Date1 $ Date2 $;
cards;
1 12/4/01 12/4/13 12/3/14
2 12/3/01 12/6/13 12/2/14
3 12/9/01 12/4/03 12/9/14
4 12/8/13 12/3/14 12/10/16
; run;
/*The trick here is to use date: The colon means everything beginning with date, comparae with sql 'date%'*/
proc transpose data= begin out=trans;
by id;
var birthday date: ;
run;
/*Cleanup. Renaming the columns as you wanted.*/
data trans;
set trans;
rename _NAME_= Datetype COL1= Date;
run;
See more from Kent University site
Two steps
Pivot the data using Proc TRANSPOSE.
Change the names of the output columns and their labels with PROC DATASETS
Sample code
proc transpose
data=have
out=want
( keep=id _label_ col1)
;
by id;
var birthday date1 date2;
label birthday='birthday' date1='1' date2='2' ; * Trick to force values seen in pivot;
run;
proc datasets noprint lib=work;
modify want;
rename
_label_ = Datetype
col1 = Date
;
label
Datetype = 'Datetype'
;
run;
The column order in the TRANSPOSE output table is:
id variables
copy variables
_name_ and _label_
data based column names
The sample 'want' shows the data named columns before the _label_ / _name_ columns. The only way to change the underlying column order is to rewrite the data set. You can change how that order is perceived when viewed is by using an additional data view, or an output Proc that allows you to specify the specific order desired.
I have the following dataset:
DATA survey;
informat order_date date9. ;
INPUT id order_date ;
DATALINES;
1 11SEPT20016
2 12AUG2016
3 14JAN2016
;
RUN;
PROC PRINT data = survey;
format order_date date9.;
RUN;
What I would like to do now is classify the records based on their last visit. So what I want to do is:
Set a date (fe, 10SEPT 2016)
Classify all records that have a lastvisit > 30days as 1, Classify all records that have a lastvisit > 60days as 2 etc...
Any thoughts on how I need to program this?
You could build something like this (count the days between the dates, divide them by 30 and ceil them). Alternativly, if you want to use months and not 30 days, you can replace the first intck parameter with 'month' and remove the ceil and /30:
DATA survey;
informat order_date date9. ;
INPUT id order_date ;
DATALINES;
1 11SEP2016
2 12AUG2016
3 14JAN2016
4 09SEP2016
5 10AUG2016
;
RUN;
%let lastvisit=10SEP2016;
data result;
set survey;
days_30=ceil(intck('days', order_date,"&lastvisit"d)/30)-1;
run;
PROC PRINT data = result;
format order_date date9.;
RUN;
I have 4 columns in my SAS dataset as shown in first image below. I need to compare the dates of consecutive rows by ID. For each ID, if Date2 occurs before the next row's Date1 for the same ID, then keep the Bill amount. If Date2 occurs after the Date1 of the next row, delete the bill amount. So for each ID, only keep the bill where the Date2 is less than the next rows Date1. I have placed what the result set should look like at the bottom.
Result set should look like
You'll want to create a new variable that moves the next row's DATE1 up one row to make the comparison. Assuming your date variables are in a date format, use PROC EXPAND and make the comparison ensuring that you're not comparing the last value which will have a missing LEAD value:
DATA TEST;
INPUT ID: $3. DATE1: MMDDYY10. DATE2: MMDDYY10. BILL: 8.;
FORMAT DATE1 DATE2 MMDDYY10.;
DATALINES;
AA 07/23/2015 07/31/2015 34
AA 07/30/2015 08/10/2015 50
AA 08/12/2015 08/15/2015 18
BB 07/23/2015 07/24/2015 20
BB 07/30/2015 08/08/2015 20
BB 08/06/2015 08/08/2015 20
;
RUN;
PROC EXPAND DATA = TEST OUT=TEST1 METHOD=NONE;
BY ID;
CONVERT DATE1 = DATE1_LEAD / TRANSFORMOUT=(LEAD 1);
RUN;
DATA TEST2; SET TEST1;
IF DATE1_LEAD NE . AND DATE2 GT DATE1_LEAD THEN BILL=.;
RUN;
If you sort your data so that you are looking to the previous obs to compare your dates, you can use a the LAG Function in a DATA STEP.