I have monthly observations that I imported from a .csv file. I'd like to convert my date variable which is of str7 type to a %tm type.
My goal is to go, for instance, from 2013-1 of type str7 to 2013m1 of type %tm.
This is what I've tried so far, the name of my date variable is "date".
gen numdate = date(date, "ym")
format numdate %tm
But it does not work as
2014-1 gets converted to 3603m9.
I read this post but, as many of the examples of the stata dates guide, the data imported is composed of daily observations, but I have monthly observations.
The -monthly()- function solves the problem as said in the comments.
Related
I wanted to know, how can we define date format from given date
for example, i have date 20180423 then in sas I want to define format as 'yyyymmdd'
similarly , i have date given in data as 12022018 then i want to define as 'ddmmyyyy'
Please note that, date is provided to me in proper date, but i want to define format now.
Date given may be different in future
so I need to take care all of the date format through SAS
What I thought was given date 20180422
use substr function
data test;
a=20180422;
a=substr(a,1,4);
b=substr(a,5,1);
c=substr(a,7,1);
run;
but not sure.
If anyone can provide the solution,then it really helps me in my project work.
Thanks in Advance for help.
It sounds like you want to convert various values to a date. SAS stores dates as a number, being the number of days since 1st Jan 1960. It's then usual to format this number to display as a date, in whichever format is preferred.
When importing dates that's are already in a format, it is necessary to use the input function, along with an informat, to convert the formatted value to a SAS date. If the date values being read in are all in the same format, then the specific informat can be used. In your case, where different formats are used, you can use the anydtdte. informat which will convert most of the standard date formats to a SAS date.
The example below converts 3 different date formats to a SAS date, then displays the SAS date in the date9. format. I've printed both the unformatted and formatted new values to the log, just so you can see they are stored as numbers.
data _null_;
input date_in $20.;
date_out = input(date_in, anydtdte20.);
put date_in date_out date_out :date9.;
datalines;
20180422
12022018
27apr2018
;
run;
Use the input(a,anydtdte20.); this will convert any date to SAS date, then use the functions Year(), Month(), Day() to extract the data you want.
You will find this SAS Post very useful about dates and locales.
Solution:
I created a table with two rows; each row have a different date format YYYYMMDD & DDMMYYYY to show you how the code will handles different date formats, saved them to SAS date and broke them down to Year, Month & Day:
options DATESTYLE=DMY;
data have;
input a;
datalines;
20180422
12022018
;
run;
data test;
set have;
format date_a date9.;
date_a=input(a,anydtdte20.);
Year_a=year(date_a);
month_a=month(date_a);
day_a=day(date_a);
run;
Output:
a=20180422 date_a=22APR2018 Year_a=2018 month_a=4 day_a=22
a=12022018 date_a=12FEB2018 Year_a=2018 month_a=2 day_a=12
You can use an if condition inside a data step. Using If condition, check for the condition to be true (check date value satisfies the required criteria), then format the date using a put function.Put function can take a source as first argument and format as second argument , and return the formatted value. Different values of same column, can have different formats specified that way.
Something like this,
if a = 'date1CheckCondtion' then newA = put(a , dateformat1.);
if a = 'date2' then newA = put(a , dateformat2.);
You may then choose to get all values in a common format like this:
dateA=input(newA,mmddyy6.);
Feels like an obvious question, but Stata help hasn't yielded answers. Most Stata users are interested in converting a non-date variable into a date variable, but I want the opposite.
I have a date variable date, type long, format %tdCCYYNN. I'm trying to append it to a dataset in which the same variable date is type long and format %12.0g. To accurately do this, I need to convert date in the first dataset from %tdCCYYNN to %12.0g. When I do format %12.0g date, date values change to incorrect ones.
Let's say, in the first dataset, I have date=201204. I still want it to read 201204, just as a %12.0g variable. Is there a way to do this?
I +1 all the comments above by Nick and William and suggest you read help datetime. I have been using Stata for a few years and still frequently visit this help file. Stata's date/time functionality is fantastic and you will benefit from learning it earlier rather than later.
I would convert the other data to Stata date format. Really. But if you need to convert your %td date to an "integer YYYYNN" date, then pass through a temporary file. If you write your %td date to plain text, then it will keep the displayed format and you can read it back as an integer YYYYNN date.
// data that matches your decsription
clear
set obs 1
generate date = date("20120401", "YMD")
format date %tdCCYYNN
list
// write to tempfile as plain text
tempfile plainText
outsheet using "`plainText'"
// read back with dates as integers
preserve
tempfile StataData
insheet using "`plainText'", clear
rename date dateInteger
save "`StataData'"
restore
// merge to original data
merge 1:1 _n using "`StataData'"
list
describe
This yields the following.
. list
+---------------------------------+
| date dateIn~r _merge |
|---------------------------------|
1. | 201204 201204 matched (3) |
+---------------------------------+
. describe
Contains data
obs: 1
vars: 3
size: 7
-----------------------------------------------------------------------------------------------------
storage display value
variable name type format label variable label
-----------------------------------------------------------------------------------------------------
date int %tdCCYYNN
dateInteger long %12.0g
_merge byte %23.0g _merge
-----------------------------------------------------------------------------------------------------
Sorted by:
Note: Dataset has changed since last saved.
But I suggest you take advantage of Stata's date/time functionality.
I want to change a date format in SAS to another date informat using the query builder.
I thought it should be as simple as:
INPUT(t1.In_date, MMDDYY10.)
But this returns the following error:
ERROR: INPUT function requires a character argument.
So I played about with it & tried a few things for example:
INPUT("t1.In_date", MMDDYY10.)
This doesn't create an error message but only produces blanks.
I've googled the error message but I can't see a way to solve this problem using the query builder.
The informat of t1.In_date is DATETIME18.
If anyone has any suggestion I would be grateful.
You've got a few problems here.
First, informat is the wrong terminology. Your date variables are dates (numbers) with formats; so, you want to change their format. informat is converting a text string to a number; format is changing how a number (or text string) is displayed.
Second, you have a datetime. Datetimes are stored as the number of seconds from 1/1/1960, while Dates are the number of days. That means that they will be not compatible format-wise. You have to use datepart to convert a datetime to a date (basically, this means dividing by 86400).
So, what you want to do:
put(datepart(t1.in_date),mmddyy10.)
I am trying to see if a variable falls into a boundary of dates.
I hate a DATE1 already in MMDDYY10.
I use the following code
DATA GIANT;
SET GIANT;
UPPER_BOUND= intnx('week', DATE1, 2);
run;
it gives me back something in Num 8.
I want to restore it to MMDDYY10. so that I can compare it to my other dates.
Two Questions:
How can I convert a NUMERIC of length 8 into a date?
Why does intnx ... designed to work with dates return a numeric and not something in the same format?
I tried to convert it like this:
DATA GIANT;
SET GIANT;
UP_DATE=INPUT(PUT(UPPER_BOUND, 8.), MMDDYY10.);
FORMAT UP_DOS MMDDYY10.;
run;
but now it all comes up as null.
SAS Dates are always numeric (# of days since 1/1/1960). Date formats are simply a way of making that numeric readable. INTNX returns a numeric because that's all a date is; it's up to you to apply a date format to the new variable.
In your case it's very simple. You almost got it right in your attempt, but you don't need the input/put business.
data giant;
set giant;
upper_bound=intnx('week',Date1,2);
format upper_bound MMDDYY10.;
run;
INPUT converts human readable text into a value (usually a number). PUT converts a value into human readable text. PUT(INPUT(...)) is commonly used to convert a formatted value into a different kind of formatted value (for example, to convert the string "1/1/1960" to "01JAN1960"); INPUT(PUT(...)) is not very commonly used unless you are parsing the string that PUT created (such as, to read just a particular date element or something like that). Both change the type (from numeric to character in PUT or other way in INPUT) in most cases and certainly change the actual stored value.
Applying a format to a numeric column leaves the column as a numeric (which is usually good) but tells SAS how to display that numeric so you can understand it (also usually good). So underneath the value is 19857 but what is displayed is 05/14/2014.
I have imported some data into SAS from some Excel spreadsheets sent to me. When I view the output from the imported table, the date appears as "01APR2014" and maintains chronological order. When I view the column properties the type is "Date" and the length is 8. Both the format and informat are DATE9.
I need to be able to convert this date to week-year and month-year, but no matter what I try I always get Jan, 1960.
Using proc sql, I used the below to get the week-year,
"(put(datepart(a.fnlz_date),weeku3.))|| "-" ||(put(datepart(a.fnlz_date),year.)) as FNLZD_WK_YR,"
but all I got was "W00-1960". I've used the formula above successfully many times before with SAS datetime values.
For month-yr, using proc sql, I tried
"datepart(a.fnlz_date) as DT_FNLZD format=monyy.,"
but the only value returned is "JAN60".
I also tried using SUBSTR, but got an error saying it requires a character argument, so SAS must see it as a number at least.
My question; does anyone know a way to get the week-yr and/or month-yr from this format? If so, how? I'm not opposed to using a data step, but I haven't been able to get that to work either.
Thanks in advance for any help or insight provided.
datepart converts datetimes to dates. Not helpful here.
If you're just displaying this, then you have a few options, particularly for month. You can just change the format of the variable (This changes what's displayed, but not the underlying value; consider this a value label).
When you use this like this (again, it looks like you got most of the way there):
proc sql;
select datevar format=monyy5. from table;
quit;
Just don't include that datepart function call as that's not appropriate unless you have a datetime. (Date=# of days since 1/1/1960, Datetime = # of seconds since 1/1/1960:00:00:00).
That will display it with MONYY5. format, which would be MAY10 for May, 2010. You have some other format options, see the documentation on formats by category for more details.
I can't think of a Week format that matches what you want (there are week formats, like WEEKW., as you clearly found, but I don't know that they do exactly what you want. So, if you want to build one yourself, you can either build a custom picture format, or you can make a string.
Building a custom picture format isn't too hard; see the documentation on Picture formats or google SAS Date Picture Format.
proc format;
picture weekyear (default=8)
low-high = 'W%0U-%Y' (datatype=date) ;
quit;
Now you can use that as a normal format.
To get at the week/etc. to build values, you can also use functions week(), month(), etc., if that's easier.
Since the data was already in a date format, I only needed to drop the DATEPART function that only works with datetime values. So, for month-yr,
"a.fnlz_date as fnlz_mnth format=monyy.,"
gives me the results I'm looking for.
Cheers!