Basically i'm trying to get the below month4 but in a macro format.
Been a while since i've done macro functions so this is a bit tricky.
data test;
input month $8.;
datalines ;
202210
202211
202201
202210
;
run;
data test2;
set test;
format month2 date9. month3 date9.;
test = cats(month,"01");
month2 = input(cats(month, "01"), yymmdd8.);
month3 = intnx("month",input(cats(month, "01"), yymmdd8.),-1);
month4=trim(substr(put(month3,yymmddn8.),1,6));
run;
I suspect this is what you are looking for, but your request is not very clear.
%let month=202212;
%let month4=%sysfunc(intnx(month,%sysfunc(inputn(&month,yymmn6)),1),yymmn6);
It uses the INPUTN() function to convert your YYYYMM string by reading it with the YYMMN6. informat. It then uses the INTNX() function to move to the beginning of the next month and returns the result formatted with YYMMN format so you get back another YYYYMM string of digits.
I have an sql table like this one
id | payment | date |
______|_____________|________________________|
obs1 | -20,10,13 | 21184,22765,22704 |
And so on (1M+ observation). I prepeared all the data for using finance() in SQL, so in SAS i just need to take them and pass to the function. I am confident, that the data i prepared will return right answer
The problem is that i can't find the most proper way to do caclulate the function on entire data. Right now i am going row by row in cycle and passing data to macro variables throught proc sql BUT i can't get string larger than 1000 characters, so my program isn't working.
I am running next function:
finance('XIRR', payment, date, 0.15);
Can you help me please? Thanks
The code i had before the answer. Worked unacceptable long!
%macro eir (input_data, cash_var, dt_var, output_data);
data rawdata;
set &input_data(dbmax_text=32000);
run;
proc sql noprint;
select count(*) into :n from rawdata ;
quit;
%let n = 100;
%do j=1 %to &n;
data x;
set rawdata(firstobs = &j obs= &j);
run;
proc sql noprint;
select &cash_var into: cf from x;
select &dt_var into: dt from x;
quit;
data x;
set x;
r= finance('xirr', &cf, &dt, 0.15);
drop &cash_var &dt_var;
run;
data out;
set %if &j>1 %then %do; out %end; x;
run;
%end;
proc append base = &output_data data=out;
run;
proc datasets nolist;
delete x out rawdata;
run;
%mend eir;
%eir(input_data = have, cash_var = pmt, dt_var = dt, output_data = ggg);
Took 20 minutes to calculate 50,000 rows
and now it's just
data want;
set have(dbmax_text=32000);
eir = input(resolve(catx(',','%sysfunc(finance(XIRR',pmt,dt,'0.15),hex16)')),hex16.);
run;
Took 6 minutes to calcuate 1,400,000 rows
Tom just saved our project =)
The FINANCE() function wants a list of values, not a character string. You could parse the string and convert the text back into numbers and pass those to the function. But if the lengths of the lists vary from observation to observation that will cause issues.
You could use the macro processor to help you. You can generate a call to %sysfunc(finance()) and read the generated string back into a numeric variable.
It also might work to pad the short lists with zero payments on the last recorded date.
Let's make some test data.
data have ;
infile cards dsd dlm='|' ;
length id $20 payment date $100 ;
input id payment date;
cards;
obs1 | -20,10,13 | 21184,22765,22704
obs2 | -20,10 | 21184,22765
;
Now let's try converting it two ways. One by creating numeric variables to pass to the FINANCE() function call and the other by generating %sysfunc(finance()) call so that we can make sure the %sysfunc() call is working properly.
data want;
set have ;
array v (3) _temporary_;
array d (3) _temporary_;
do i=1 to dim(v);
v(i)=coalesce(input(scan(payment,i,','),32.),0);
d(i)=input(scan(date,i,','),32.);
if missing(d(i)) and i>1 then d(i)=d(i-1);
end;
drop i;
value1=finance('XIRR',of v(*),of d(*),0.15);
value2=input(resolve(catx(',','%sysfunc(finance(XIRR',payment,date,'0.15),hex16)')),hex16.);
run;
Here's my best guess based on the limited details you've provided. I think you need to split out each date and payment into separate variables before you can call the finance function, e.g.:
data have;
infile datalines dlm='|';
input id :$8. amount :$20. date :$20.;
datalines;
obs1 | -20,10,13 | 21184,22765,22704
;
run;
data want;
set have;
array dates[3] d1-d3;
array amounts[3] a1-a3;
do i = 1 to 3;
amounts[i] = input(scan(amount, i, ','), 8.);
dates[i] = input(scan(date, i, ','), 8.);
end;
XIRR = finance('XIRR', of a1-a3, of d1-d3, 0.15);
run;
I suspect this will only work you have the same number of dates and payments in every row, otherwise you will run into array out of bounds issues or problems with the IRR calculation.
hello am trying access columns from library with specific date format and using year function on the columns in my macro code but it produces duplicate values... but the year function displays duplicate values and does not provide desired results. my code should return only the year from the input dates.
%macro dteyear(lib=,outdsn=);
proc sql noprint;
select distinct catx(".",libname,memname), name
into :dsns separated by " ", :varname separated by " "
from dictionary.columns
where libname = upcase("&lib") and format=('YYMMDD10.')
order by 1;
quit;
%put &dsns;
%put &varname;
%local olddsn curdsn curvbl i;
data &outdsn.;
set
%let olddsn=;
%do i=1 %to &sqlobs;
%let curdsn=%scan(&dsns,&i,%str( ));
%let curvbl=%scan(&varname,&i,%str( ));
%if &curdsn NE &olddsn
%then %do;
%if &olddsn NE
%then %do;
)
%end;
%let olddsn=&curdsn.;
&curdsn (keep=&curvbl
%end;
%else %do;
&curvbl
%end;
%end;
);
%do i=1 %to &sqlobs;
%scan(&varname,&i,%str( ))=year(&varname.);
%end;
run;
proc print data=&outdsn;run;
%MEND;
%dteyear(lib=dte3,outdsn=dtetst);
the input data is as follows
1975-12-04
1977-11-03
1989-09-15
1998-06-17
1999-05-31
2000-08-14
2001-03-11
2007-03-11
2007-12-28
2008-10-07
2009-12-03
duplicate output from my code is-->
Obs RFDTC
1 1965-05-19
2 1965-05-19
3 1965-05-19
4 1965-05-19
5 1965-05-19
6 1965-05-19
7 1965-05-19
8 1965-05-19
9 1965-05-19
10 1965-05-19
11 1965-05-19
12 1965-05-19
13 1965-05-19
The basic problem is that the YEAR() function returns a 4-digit number, and the variable's format is YYMMDD10., so the result is formatted as a SAS date very close to 1960 (SAS's beginning of all time).
What I did in the code below was change the format to 4.0, so it displays as a 4-digit number.
If you want to have access to the original date variable, you'll have to create a new variable for the year. I'll leave that to you.
There was an additional problem--that is, YEAR(&varname.) inserts the entire list of variables, not just the one you're working with. It works if there is only one date variable, but not if there are more than one. I fixed this, too.
%macro dteyear(lib=,outdsn=);
proc sql noprint;
select distinct catx(".",libname,memname), name
into :dsns separated by " ", :varname separated by " "
from dictionary.columns
where libname = upcase("&lib") and format=('YYMMDD10.')
order by 1;
quit;
%put &dsns;
%put &varname;
%local olddsn curdsn curvbl i;
data &outdsn.;
set
%let olddsn=;
%do i=1 %to &sqlobs;
%let curdsn=%scan(&dsns,&i,%str( ));
%let curvbl=%scan(&varname,&i,%str( ));
%if &curdsn NE &olddsn
%then %do;
%if &olddsn NE
%then %do;
)
%end;
%let olddsn=&curdsn.;
&curdsn (keep=&curvbl
%end;
%else %do;
&curvbl
%end;
%end;
);
%do i=1 %to &sqlobs;
%let curvbl=%scan(&varname,&i,%str( ));
&curvbl=year(&curvbl.);
format &curvbl 4.0;
%end;
run;
proc print data=&outdsn;run;
%MEND;
data have;
input datevar yymmdd10.;
format datevar yymmdd10.;
cards;
1975-12-04
1977-11-03
1989-09-15
1998-06-17
1999-05-31
2000-08-14
2001-03-11
2007-03-11
2007-12-28
2008-10-07
2009-12-03
run;
options mprint;
%dteyear(lib=work,outdsn=want)
The result, then, is:
Obs datevar
1 1975
2 1977
3 1989
4 1998
5 1999
6 2000
7 2001
8 2007
9 2007
10 2008
11 2009
To convert a date value to just a year you can use the YEAR() function, but you also need to change the format attached to the variable since you will have essentially divided the value stored in it by 365 to convert it from the number of days to the number of years.
rfdtc = year(rfdtc);
format rfdtc 4. ;
Your macro is attempting to read many variables from many datasets and generate a single output dataset. I am not sure the resulting dataset will be of much value to you since it will look like a checker board of missing values. Also if the same variable name appears in more than one input dataset you will get corrupted values because of applying the YEAR() function to value that has already been converted from a date value to a year value.
For example you could end up generating a data step like this:
data WANT ;
set ds1 (keep=datevar1)
ds1 (keep=datevar2)
ds2 (keep=datevar3)
ds3 (keep=datevar3)
;
datevar1=year(datevar1);
datevar2=year(datevar2);
datevar3=year(datevar3);
datevar3=year(datevar3);
format datevar1 datevar2 datevar3 datevar3 4.;
run;
Since both input datasets DS2 and DS3 have a variable named DATEVAR3 you will be applying the YEAR() function to the value twice. That will convert everything to the year 1965.
To eliminate the problem with running the YEAR() function on the same value multiple times and losing the actual year perhaps you just want to apply the YEAR. format instead of converting the stored value.
format datevar1 datevar2 datevar3 datevar4 year. ;
That would still leave the underlying different date values. If you really need to values to be identical perhaps you could convert the value to the first day of the year? You could use INTNX() function
datevar1 = intnx('year',datevar1,0,'b');
or the MDY() function
datevar1 = mdy(1,1,year(datevar1));
I have a data set with one row for each country and 100 columns (10 variables with 10 data years each).
For each variable I am trying to make a new data set with the three most recent data years for that variable for each country (which might not be successive).
This is what I have so far, but I know its wrong because of the nest loop, and its has same value for recent1 recent2 recent3 however I haven't figured out how to create recent1 recent2 recent3 without two loops.
%macro test();
data Maternal_care_recent;
set wb;
keep country MATERNAL_CARE_2004 -- MATERNAL_CARE_2013 recent_1 recent_2 recent_3;
%let rc = 1;
%do i = 2013 %to 2004 %by -1;
%do rc = 1 %to 3 %by 1;
%if MATERNAL_CARE_&i. ne . %then %do;
recent_&rc. = MATERNAL_CARE_&i.;
%end;
%end;
%end; run; %mend; %test();
You don't need to use a macro to do this - just some arrays:
data Maternal_care_recent;
set wb;
keep country MATERNAL_CARE_2004-MATERNAL_CARE_2013 recent_1 recent_2 recent_3;
array mc {*} MATERNAL_CARE_2004-MATERNAL_CARE_2013;
array recent {*} recent1-recent3;
do i = 2013 to 2004 by -1;
do rc = 1 to 3 by 1;
if mc[i] ne . then do;
recent[rc] = mc[i];
end;
end;
run;
Maybe I don't get your request, but according to your description:
"For each variable I am trying to make a new data set with the three most recent data years for that variable for each country (which might not be successive)" I created this sample dataset with dt1 and dt2 and 2 locations.
The output will be 2 datasets (and generally the number of the variables starting with DT) named DS1 and DS2 with 3 observations for each country, the first one for the first variable, the second one for the second variable.
This is the sample dataset:
data sample_ds;
length city $10 dt1 dt2 8.;
infile datalines dlm=',';
input city $ dt1 dt2;
datalines;
MS,5,0
MS,3,9
MS,3,9
MS,2,0
MS,1,8
MS,1,7
CA,6,1
CA,6,.
CA,6,.
CA,2,8
CA,1,5
CA,0,4
;
This is the sample macro:
%macro help(ds=);
data vars(keep=dt:); set &ds; if _n_ not >0; run;
%let op = %sysfunc(open(vars));
%let nvrs = %sysfunc(attrn(&op,nvars));
%let cl = %sysfunc(close(&op));
%do idx=1 %to &nvrs.;
proc sort data=&ds(keep=city dt&idx.) out=ds&idx.(where=(dt&idx. ne .)) nodupkey; by city DESCENDING dt&idx.; run;
data ds&idx.; set ds&idx.;
retain cnt;
by city DESCENDING dt&idx.;
if first.city then cnt=0; else cnt=cnt+1;
run;
data ds&idx.(drop=cnt); set ds&idx.(where=(cnt<3)); rename dt&idx.=act&idx.; run;
%end;
%mend;
You will run this macro with:
%help(ds=sample_ds);
In the first statement of the macro I select the variables on which I want to iterate:
data vars(keep=dt:); set &ds; if _n_ not >0; run;
Work on this if you want to make this work for your code, or simply rename your variables as DT1 DT2...
Let me know if it is correct for you.
When writing macro code, always keep in mind what has to be done when. SAS processes your code stepwise.
Before your sas code is even compiled, your macro variables are resolved and your macro code is executed
Then the resulting SAS Base code is compiled
Finally the code is executed.
When you write %if MATERNAL_CARE_&i. ne . %then %do, this is macro code interpreded before compilation.
At that time MATERNAL_CARE_&i. is not a variable but a text string containing a macro variable.
The first time you run trhough your %do i = 2013 %to 2004 by -1, it is filled in as MATERNAL_CARE_2013, the second as MATERNAL_CARE_2012., etc.
Then the macro %if statement is interpreted, and as the text string MATERNAL_CARE_1 is not equal to a dot, it is evaluated to FALSE
and recent_&rc. = MATERNAL_CARE_&i. is not included in the code to pass to your compiler.
You can see that if you run your code with option mprint;
The resolution;
options mprint;
%macro test();
data Maternal_care_recent;
set wb;
keep country MATERNAL_CARE_: recent_:;
** The : acts as a wild card here **;
%do i = 2013 %to 2004 %by -1;
if MATERNAL_CARE_&i. ne . then do;
%do rc = 1 %to 3 %by 1;
recent_&rc. = MATERNAL_CARE_&i.;
%end;
end;
%end;
run;
%mend;
%test();
Now, before compilation of if MATERNAL_CARE_&i. ne . then do, only the &i. is evalueated and if MATERNAL_CARE_2013 ne . then do is passed to the compiler.
The compiler will see this as a test if the SAS variable MATERNAL_CARE_1 has value missing, and that is just what you wanted;
Remark:
It is not essential that I moved the if statement above the ``. It is just more efficient because the condition is then evaluated less often.
It is however essential that you close your %ifs and %dos with an %end and your ifs and dos with an end;
Remark:
you do not need %let rc = 1, because %do rc = 1 to 3 already initialises &rc.;
For completeness SAS is compiled stepwise:
The next PROC or data step and its macro code are only considered when the preveous one is executed.
That is why you can write macro variables from a data step or sql select into that will influence the code you compile in your next step,
somehting you can not do for instance with C++ pre compilation;
Thanks everyone. Found a hybrid solution from a few solutions posted.
data sample_ds;
infile datalines dlm=',';
input country $ maternal_2004 maternal_2005
maternal_2006 maternal_2007 maternal_2008 maternal_2009 maternal_2010 maternal_2011 maternal_2012 maternal_2013;
datalines;
MS,5,0,5,0,5,.,5,.,5,.
MW,3,9,5,0,5,0,5,.,5,0
WE,3,9,5,0,5,.,.,.,.,0
HU,2,0,5,.,5,.,5,0,5,0
MI,1,8,5,0,5,0,5,.,5,0
HJ,1,7,5,0,5,0,.,0,.,0
CJ,6,1,5,0,5,0,5,0,5,0
CN,6,1,.,5,0,5,0,5,0,5
CE,6,5,0,5,0,.,0,5,.,8
CT,2,5,0,5,0,5,0,5,0,9
CW,1,5,0,5,0,5,.,.,0,7
CH,0,5,0,5,0,.,0,.,0,5
;
%macro test(var);
data &var._recent;
set sample_ds;
keep country &var._1 &var._2 &var._3;
array mc {*} &var._2004-&var._2013;
array recent {*} &var._1-&var._25;
count=1;
do i = 10 to 1 by -1;
if mc[i] ne . then do;
recent[count] = mc[i];
count=count+1;
end;
end;
run;
%mend;