I have a SAS data set grouped by clusters, as follows
data have;
input cluster date date9.;
cards;
1 1JAN2017
1 2JAN2017
1 7JAN2017
2 1JAN2017
2 3JAN2017
2 10JAN2017
;
run;
Within each cluster, I'd like to subtract a date from it's previous date, so I have the dataset below:
data want;
input cluster date date_diff;
cards;
1 1JAN2017 0
1 2JAN2017 1
1 7JAN2017 5
2 1JAN2017 0
2 3JAN2017 2
2 10JAN2017 7
;
run;
I think perhaps I should be using a lag function similar to what I have written below.
DATA test;
SET have;
BY cluster;
if first.cluster then do;
date_diff = date - lag(date);
END;
RUN;
Any advice would be appreciated! Thanks
I like dif for this (lag plus subtract in one function). You have the if first backwards, I think, but dif and lag have the same restriction - what they're really doing is building a queue, so the lag or dif statement cannot be conditionally executed for most use cases. Here I flip it around and calculate the dif, then set it to missing if on first.cluster.
I also encourage you to use missing, not 0, for the first.cluster dif.
DATA test;
SET have;
BY cluster;
date_diff = dif(date);
if first.cluster then call missing(date_diff);
RUN;
Conditional lags are tricky. It this case (and in many), you don't actually want a conditional lag. You want to compute the lag for every record, and then you can use it conditionally. One way is:
data want ;
set have ;
by cluster ;
lagdate=lag(date) ;
if first.cluster then date_diff=0 ;
else date_diff=date-lagdate ;
run ;
So you compute lagdate for every record. Then you can conditionally compute date_diff.
Related
So if I have identified a max value regarding a test result (Highest variable listed below), which occurred during one of the three dates that are being tested (testtime variables listed below), what I want to do is to create a new variable called Highesttime identifying the date when the test was given.
However, I am stuck in an array looping. SAS informs that "ERROR: Array subscript out of range at line x", guess there's something working regarding the logic? See codes below:
Example:
ID time1_a time_b time_c result_a result_b result_c Highest
001 1/1/22 1/2/22 1/3/22 3 2 4 4
002 12/1/21 12/23/21 1/5/22 6 1 2 6
003 12/22/21 1/6/22 2/2/22 5 5 7 7
...
data want;
set origin;
array testtime{3} time1_a time_b time_c;
array maxvalue{1} Highest;
array corr_time{1} Highesttime;
do i=1 to dim(testttime);
corr_time{i}=testttime{i=maxvalue{i}};
end;
run;
There is no need to make an array for HIGHEST since there is only one variable that you would put into that array. In that case just use the variable directly instead of trying to access it indirectly via an array reference.
First let's make an actual SAS dataset out of the listing you provided.
data have;
input ID (time_a time_b time_c) (:mmddyy.) result_a result_b result_c Highest ;
format time_a time_b time_c yymmdd10.;
cards;
001 1/1/22 1/2/22 1/3/22 3 2 4 4
002 12/1/21 12/23/21 1/5/22 6 1 2 6
003 12/22/21 1/6/22 2/2/22 5 5 7 7
;
If you want to loop then you need two arrays. One for times and the other for the values. Then you can loop until you find which index points to the highest value and use the same index into the other array.
data want ;
set have;
array times time_a time_b time_c ;
array results result_a result_b result_c;
do which_one=1 to dim(results) until (not missing(highest_time));
if results[which_one] = highest then highest_time=times[which_one];
end;
format highest_time yymmdd10.;
run;
Or you can avoid the looping by using the WHICHN() function to figure out which of three result variables is the first one that has that HIGHEST value. Then you can use that value as the index into the array of the TIME variables (which in your case have DATE instead of TIME or DATETIME values).
data want ;
set have;
which_one = whichn(highest, of result_a result_b result_c);
array times time_a time_b time_c ;
highest_time = times[which_one];
format highest_time yymmdd10.;
run;
Your code from this question was close, you just had the assignment backwards.
Note that an array method will assign the last date in the case of duplicate high results and WHICHN will report the first date so the answers are not identical unless you modify the loop to exit after the first maximum value is found.
With the changes suggested in the answer proposed:
data temp2_lead_f2022;
set temp_lead_f2022;
array _day {3} daybld_a daybld_b daybld_c;
array _month {3} mthbld_a mthbld_b mthbld_c;
array _dates {3} date1_a date2_b date3_c;
array _pblev{3} pblev_a pblev_b pblev_c;
do i = 1 to 3;
_dates{i} = mdy(_month{i}, _day{i}, 1990);
end;
maxlead= max(of _pblev(*));
do i=1 to 3;
if _pblev{i} = maxlead then max_date=_dates(i);
end;
*Using WHICHN to identify the maximum occurence;
max_first_index=whichn(maxlead, of _pblev(*));
max_date2 = _dates(max_first_index);
drop k;
format date1_a date2_b date3_c dob mmddyy8. ;
run;
I am editing my original question to simplify the problem statement:
I need to create a dataset that contains the principal paydown schedule of a security, which is split into 3 tranches. For each period for the security, I need to calculate the ending balances of principal owed for each tranche. For period 0 (i.e. starting period), I already have the balances owed. For subsequent periods, I need to take the balances from the previous periods and subtract the principal paid down in the current period. The same logic should continue through the last period.
In my SAS code, I am able to get period 1 to do the calculations correctly, but the balances from period 1 don't correctly make it into period 2, causing the calculation to break from that point onwards. I know lag or its placement is what is not working correctly. I am not able to figure out where to place it, or how to use retain (if not lag), such that my balances go from one row to the next.
%let n_t=3;
data xyz;
INFILE DATALINES DLM='#';
input ID $6. period PrincipalPaid best12.2;
datalines;
ABC123#00#0.0
ABC123#01#4.0
ABC123#02#3.92
ABC123#03#3.84
ABC123#04#3.76
ABC123#05#3.69
ABC123#06#3.62
ABC123#07#3.54
;run;
data xyz2;
set xyz;
by id;
if period=0 then do;
Bal1= 120;
Bal2= 8;
Bal3= 2;
end;
/*Code to push all starting balances from period 0 to 1*/
array prev_bal{&N_t.} prev_bal1-prev_bal&n_t.;
array bal{&N_t.} bal1-bal&n_t.;
do i=1 to &N_t.;
prev_bal{i}=lag(bal{i});
end;
/*code to calculate balances for periods >=1*/
if period>=1 then do;
array PrincipalPayDown{&N_t.} PrincipalPayDown1-PrincipalPayDown&N_t.;
do i = 1 to &N_t. ;
PrincipalPayDown{i}=round(PrincipalPaid*prev_bal{i}/sum(of prev_bal:),0.01);
bal{i}=max(prev_bal{i}-PrincipalPayDown{i},0);
end;
end;
drop i ;
run;
proc sql;
create table final as
select
id,period,PrincipalPaid,prev_bal1,prev_bal2,prev_bal3,
PrincipalPayDown1,PrincipalPayDown2,PrincipalPayDown3,Bal1,Bal2,Bal3
from xyz2;
quit;
I am also adding a picture of the final dataset with the correct output calculated in Excel. I want SAS to give me the same output for periods >=2.
Screenshot showing correct output in Excel
I am trying to create an array of strings and want to insert a value in it, if it does not exist already in the array.
I read somewhere that we can use 'IN' operator with Array. So, coded it as follows:
DATA WANT;
SET HAVE;
BY ID;
ARRAY R_PROS_SCRN_ID {2} $4. R_PROS_SCRN_ID_1 - R_PROS_SCRN_ID_2;
RETAIN R_PROS_SCRN_ID_1 - R_PROS_SCRN_ID_2;
IF NOT PROS_SCRN_ID IN R_PROS_SCRN_ID THEN DO;
DO I=1 to 2 ;
IF MISSING( R_PROS_SCRN_ID{i}) THEN DO;
R_PROS_SCRN_ID{i} = PROS_SCRN_ID;
LEAVE;
END;
END;
END;
IF LAST.ID THEN OUTPUT;
RUN;
In Array R_PROS_SCRN_ID, I want only the unique values from field PROS_SCRN_ID.
It is throwing error:
NOTE: Invalid numeric data, PROS_SCRN_ID='MED' , at line 17352 column 201.
I think it is because I did not initialize the Array before comparing and hence it is considering it as Numeric Array. But, I have specified the format as $4. Why is it throwing error?
Also, I am not sure if this is the best way get unique values in an Array. Is there any better way to implement this?
Your code appears to be collecting unique values by group, pivoting from a tall data structure to a wide data structure.
One of the clearest DATA step ways is to use what we call DOW loop in which SET is within the loop. This sample code presumes no more than 10 unique satellite values per group. (The by variables can be thought of as key variables, and all other variables would be satellites)
data have;
input user_id screen_id ;
datalines;
1 1
1 2
1 1
1 1
1 1
1 3
2 1
2 1
2 1
3 0
4 1
4 2
4 3
5 11
5 11
5 11
5 5
5 1
5 5
5 6
5 1
run;
data want;
_index = 0;
do until (last.user_id);
set have;
by user_id;
array ids screen_id1-screen_id10;
if screen_id not in ids then do;
_index + 1;
ids(_index) = screen_id;
end;
end;
drop _index screen_id;
run;
One of the clearest procedural ways is to select the unique values and transpose them.
proc sql;
create view uniqueScreenByUser as
select distinct user_id, screen_id
from have
order by user_id
;
proc transpose data=uniqueScreenByUser prefix=screen_id out=wantWide(drop=_name_);
by user_id;
var screen_id;
run;
I have a datset sort of like this
obs| foo | bar | more
1 | 111 | 11 | 9
2 | 9 | 2 | 2
........
I need to throw out the 4 largest and 4 smallest of foo (later then I would do a similar thing with bar) basically to proceed but I'm unsure the most effective way to do this. I know there are functions smallest and largest but I don't understand how I can use them to get the smallest 4 or largest 4 from an already made dataset. I guess alternatively I could just remove the min and max 4 times but that sounds needlessly tedious/time consuming. Is there a better way?
PROC RANK will do this for you pretty easily. If you know the total count of observations, it's trivial - it's slightly harder if you don't.
proc rank data=sashelp.class out=class_ranks(where=(height_r>4 and weight_r>4));
ranks height_r weight_r;
var height weight;
run;
That removes any observation that is in the 4 smallest heights or weights, for example. The largest 4 would require knowing the maximum rank, or doing a second processing step.
data class_final;
set class_ranks nobs=nobs;
if height_r lt (nobs-3) and weight_r lt (nobs-3);
run;
Of course if you're just removing the values then do it all in the data step and call missing the variable if the condition is met rather than deleting the observation.
You are going to need to make at least 2 passes through your dataset however you do this - one to find out what the top and bottom 4 values are, and one to exclude those observations.
You can use proc univariate to get the top and bottom 5 values, and then use the output from that to create a where filter for a subsequent data step. Here's an example:
ods _all_ close;
ods output extremeobs = extremeobs;
proc univariate data = sashelp.cars;
var MSRP INVOICE;
run;
ods listing;
data _null_;
do _N_ = 1 by 1 until (last.varname);
set extremeobs;
by varname notsorted;
if _n_ = 2 then call symput(cats(varname,'_top4'),high);
if _n_ = 4 then call symput(cats(varname,'_bottom4'),low);
end;
run;
data cars_filtered;
set sashelp.cars(where = ( &MSRP_BOTTOM4 < MSRP < &MSRP_TOP4
and &INVOICE_BOTTOM4 < INVOICE < &INVOICE_TOP4
)
);
run;
If there are multiple observations that tie for 4th largest / smallest this will filter out all of them.
You can use proc sql to place the number of distinct values of foo into a macro var (includes null values as distinct).
In you data step you can use first.foo and the macro var to selectively output only those that are no the smallest or largest 4 values.
proc sql noprint;
select count(distinct foo) + count(distinct case when foo is null then 1 end)
into :distinct_obs from have;
quit;
proc sort data = have; by foo; run;
data want;
set have;
by foo;
if first.foo then count+1;
if 4 < count < (&distinct_obs. - 3) then output;
drop count;
run;
I also found a way to do it that seems to work with IML (I'm practicing by trying to redo things different ways). I knew my maximum number of observations and had already sorted it by order of the variable of interest.
PROC IML;
EDIT data_set;
DELETE point {{1, 2, 3, 4,51, 52, 53, 54};
PURGE;
close data_set;
run;
I've not used IML very much but I stumbled upon this while reading documentation. Thank you to everyone who answered my question!
I'm working in SAS and I'm trying to sum all observations, leaving out one each time.
For example, if I have:
Count Name Grade
1 Sam 90
2 Adam 100
3 John 80
4 Max 60
5 Andrea 70
I want to output a value for Sam that is the sum of all grades but his own, and a value for Adam that is a sum of all grades but his own - etc.
Any ideas? Thanks!
You can do it in a single proc sql instead, using key word calculated:
data have;
input Count Name $ Grade;
datalines;
1 Sam 90
2 Adam 100
3 John 80
4 Max 60
5 Andrea 70
;;;;
run;
proc sql;
create table want as
select *, sum(grade) as all_grades, calculated all_grades-grade as minus_grade
from have;
quit;
Here's a nearly one pass solution (it will be about the same speed as a one pass solution if the dataset fits in the read buffer). I actually calculate the mean here instead of just the sum, as I feel that's a more interesting result (and the sum is of course the mean without the division).
data have;
input Count Name $ Grade;
datalines;
1 Sam 90
2 Adam 100
3 John 80
4 Max 60
5 Andrea 70
;;;;
run;
data want;
retain grademean;
if _n_=1 then do;
do _n_ = 1 to nobs_have;
set have(keep=grade) point=_n_ nobs=nobs_have;
gradesum+grade;
end;
grademean=gradesum/nobs_have;
end;
set have;
grade_noti = ((grademean*nobs_have)-grade)/(nobs_have-1);
run;
Calculate the mean, then for each record subtract the portion that record contributed to the mean. This is a super useful technique for stat testing when you want to compare a record to the rest of the population, and you have a complicated class combination where you'd rather do the mean first. In those cases you use PROC MEANS first and then merge it on, then do this subtraction.
proc sql;
create table temp as select
sum(grade) as all_grades
from orig_data;
quit;
proc sql;
create table temp2 as select
a.count,
a.name,
a.grade,
(b.all_grades-a.grade) as sum_other_grades
from orig_data a
left join temp b;
quit;
Haven't tested it but the above should work. It creates a new dataset temp that has the sum of all grades and merges that back to create a new table with the sum of all grades less the current students grade as sum_other_grades.
This solution performs takes each observation of your starting dataset, and then loops through the same dataset summing up grade values for any records with different names, so beginning with 'Sam', we only add the oth_g variable when we find names that are NOT 'Sam':
data want;
set have;
oth_g=0;
do i=1 to n;
set have
(keep=name grade rename=(name=name_loop grade=grade_loop))
nobs=n point=i;
if name^=name_loop then oth_g+grade_loop;
end;
drop grade_loop name_loop i n;
run;
This is a slight modification to the answer #Reese provided above.
proc sql;
create table want as
select *,
(select sum(grade) from have) as all_grades,
calculated all_grades - grade as minus_grade
from have;
quit;
I've rearranged it this way to avoid the below message being printed to the log:
NOTE: The query requires remerging summary statistics back with the original data.
If you see the above message, it almost always means that you have made a mistake. If you actually did mean to remerge summary stats back with the original data, you should do so explicitly (like I have done above by refactoring #reese 's query.
Personally I think the refactored version is also easier to understand.