Referring an input dataset in itself - Recursion using POINT option - SAS - sas

I have the below dataset wherein each employee is tied up to a manager position.
Now the employee's manager's employee ID needs to be found out using the manager position.
If in the immediate manager's position none is ACTIVE then we need to find the manager's manager position and find that any ACTIVE employee is tied to that posotion. This needs to be continued till an ACTIVE manager is found.
ID -> Employee ID
PSTN -> Employee Position code
MPSTN-> Manager Position code
STAT -> Employee Status (T - Term A - Active)
Input Dataset:
data input;
input id pstn mpstn stat$;
datalines;
1 10 30 A
2 20 30 T
3 30 40 T
6 30 40 T
4 40 50 A
7 40 50 T
5 50 50 A
;
run;
Output Dataset expected:
ID MGR_ID
1 4
2 4
3 4
6 4
4 5
7 5
5 5
I tried the RECURSIVE nature of the problem with POINT function.
It's working fine except for the recursive part - wherein searching for the next level active manager.
data output ;
set input;
flag = 1;
do I = 1 to last while (flag=1);
set input(rename=(pstn=pstn1 stat=stat1 mpstn=mpstn1 id=id1)) nobs=last
point=I;
if mpstn = pstn1 and stat1 = 'A' then
do;
MGRID = id1;
I=1;
flag=0;
end;
else flag=1;
end;
run;
Please help me with this.

You can use a hash map to loop through the input dataset to look up the values. Instead of recursively calling the lookup, I suggest putting the lookup into a do while() loop.
data input;
input id pstn mpstn stat$;
format stat $1.;
datalines;
1 10 30 A
2 20 30 T
3 30 40 T
6 30 40 T
4 40 50 A
7 40 50 T
5 50 50 A
;
run;
data out(keep=id mgr_id);
set input;
format pstn1 id1 mpstn1 best.
stat1 $1.;
if _n_ = 1 then do;
declare hash mgr(dataset:"input(rename=(pstn=pstn1 stat=stat1 mpstn=mpstn1 id=MGR_ID))");
rc = mgr.definekey("pstn1");
rc = mgr.definedata("MGR_ID");
rc = mgr.definedata("mpstn1");
rc = mgr.definedata("stat1");
rc = mgr.definedone();
end;
found = 0;
do while(^found);
pstn1 = mpstn;
rc = mgr.find();
if stat1 = "A" then do;
/*MGR Found*/
found = 1;
end;
else if rc then do;
/*RC^=0 when lookup fails*/
MGR_ID = .;
found = 1;
end;
else do;
mpstn = mpstn1;
end;
end;
run;

Related

recursive search in SAS [duplicate]

I have two variables ID1 and ID2. They are both the same kinds of identifiers. When they appear in the same row of data it means they are in the same group. I want to make a group identifier for each ID. For example, I have
ID1 ID2
1 4
1 5
2 5
2 6
3 7
4 1
5 1
5 2
6 2
7 3
Then I would want
ID Group
1 1
2 1
3 2
4 1
5 1
6 1
7 2
Because 1,2,4,5,6 are paired by some combination in the original data they share a group. 3 and 7 are only paired with each other so they are a new group. I want to do this for ~20,000 rows. Every ID that is in ID1 is also in ID2 (more specifically if ID1=1 and ID2=2 for an observation, then there is another observation that is ID1=2 and ID2=1).
I've tried merging them back and forth but that doesn't work. I also tried call symput and trying to make a macro variable for each ID's group and then updating it as I move through rows, but I couldn't get that to work either.
I have used Haikuo Bian's answer as a starting point to develop a slightly more complex algorithm that seems to work for all the test cases I have tried so far. It could probably be optimised further, but it copes with 20000 rows in under a second on my PC while using only a few MB of memory. The input dataset does not need to be sorted in any particular order, but as written it assumes that every row is present at least once with id1 < id2.
Test cases:
/* Original test case */
data have;
input id1 id2;
cards;
1 4
1 5
2 5
2 6
3 7
4 1
5 1
5 2
6 2
7 3
;
run;
/* Revised test case - all in one group with connecting row right at the end */
data have;
input ID1 ID2;
/*Make sure each row has id1 < id2*/
if id1 > id2 then do;
t_id2 = id2;
id2 = id1;
id1 = t_id2;
end;
drop t_id2;
cards;
2 5
4 8
2 4
2 6
3 7
4 1
9 1
3 2
6 2
7 3
;
run;
/*Full scale test case*/
data have;
do _N_ = 1 to 20000;
call streaminit(1);
id1 = int(rand('uniform')*100000);
id2 = int(rand('uniform')*100000);
if id1 < id2 then output;
t_id2 = id2;
id2 = id1;
id1 = t_id2;
if id1 < id2 then output;
end;
drop t_id2;
run;
Code:
option fullstimer;
data _null_;
length id group 8;
declare hash h();
rc = h.definekey('id');
rc = h.definedata('id');
rc = h.definedata('group');
rc = h.definedone();
array ids(2) id1 id2;
array groups(2) group1 group2;
/*Initial group guesses (greedy algorithm)*/
do until (eof);
set have(where = (id1 < id2)) end = eof;
match = 0;
call missing(min_group);
do i = 1 to 2;
rc = h.find(key:ids[i]);
match + (rc=0);
if rc = 0 then min_group = min(group,min_group);
end;
/*If neither id was in a previously matched group, create a new one*/
if not(match) then do;
max_group + 1;
group = max_group;
end;
/*Otherwise, assign both to the matched group with the lowest number*/
else group = min_group;
do i = 1 to 2;
id = ids[i];
rc = h.replace();
end;
end;
/*We now need to work through the whole dataset multiple times
to deal with ids that were wrongly assigned to a separate group
at the end of the initial pass, so load the table into a
hash object + iterator*/
declare hash h2(dataset:'have(where = (id1 < id2))');
rc = h2.definekey('id1','id2');
rc = h2.definedata('id1','id2');
rc = h2.definedone();
declare hiter hi2('h2');
change_count = 1;
do while(change_count > 0);
change_count = 0;
rc = hi2.first();
do while(rc = 0);
/*Get the current group of each id from
the hash we made earlier*/
do i = 1 to 2;
rc = h.find(key:ids[i]);
groups[i] = group;
end;
/*If we find a row where the two ids have different groups,
move the id in the higher group to the lower group*/
if groups[1] < groups[2] then do;
id = ids[2];
group = groups[1];
rc = h.replace();
change_count + 1;
end;
else if groups[2] < groups[1] then do;
id = ids[1];
group = groups[2];
rc = h.replace();
change_count + 1;
end;
rc = hi2.next();
end;
pass + 1;
put pass= change_count=; /*For information only :)*/
end;
rc = h.output(dataset:'want');
run;
/*Renumber the groups sequentially*/
proc sort data = want;
by group id;
run;
data want;
set want;
by group;
if first.group then new_group + 1;
drop group;
rename new_group = group;
run;
/*Summarise by # of ids per group*/
proc sql;
select a.group, count(id) as FREQ
from want a
group by a.group
order by freq desc;
quit;
Interestingly, the suggested optimisation of not checking the group of id2 during the initial pass if id1 is already matched actually slows things down a little in this extended algorithm, because it means that more work has to be done in the subsequent passes if id2 is in a lower numbered group. E.g. output from a trial run I did earlier:
With 'optimisation':
pass=0 change_count=4696
pass=1 change_count=204
pass=2 change_count=23
pass=3 change_count=9
pass=4 change_count=2
pass=5 change_count=1
pass=6 change_count=0
NOTE: DATA statement used (Total process time):
real time 0.19 seconds
user cpu time 0.17 seconds
system cpu time 0.04 seconds
memory 9088.76k
OS Memory 35192.00k
Without:
pass=0 change_count=4637
pass=1 change_count=182
pass=2 change_count=23
pass=3 change_count=9
pass=4 change_count=2
pass=5 change_count=1
pass=6 change_count=0
NOTE: DATA statement used (Total process time):
real time 0.18 seconds
user cpu time 0.16 seconds
system cpu time 0.04 seconds
Please try the below code.
data have;
input ID1 ID2;
datalines;
1 4
1 5
2 5
2 6
3 7
4 1
5 1
5 2
6 2
7 3
;
run;
* Finding repeating in ID1;
proc sort data=have;by id1;run;
data want_1;
set have;
by id1;
attrib flagrepeat length=8.;
if not (first.id1 and last.id1) then flagrepeat=1;
else flagrepeat=0;
run;
* Finding repeating in ID2;
proc sort data=want_1;by id2;run;
data want_2;
set want_1;
by id2;
if not (first.id2 and last.id2) then flagrepeat=1;
run;
proc sort data=want_2 nodupkey;by id1 ;run;
data want(drop= ID2 flagrepeat rename=(ID1=ID));
set want_2;
attrib Group length=8.;
if(flagrepeat eq 1) then Group=1;
else Group=2;
run;
Hope this answer helps.
Like one commentator mentioned, Hash does seem to be a viable approach. In the following code, 'id' and 'group' is maintained in the Hash table, new 'group' is added only when no 'id' match is found for the entire row. Please note, 'do over' is an undocumented feature, it can be easily replaced with a little bit more coding.
data have;
input ID1 ID2;
cards;
1 4
1 5
2 5
2 6
3 7
4 1
5 1
5 2
6 2
7 3
;
data _null_;
if _n_=1 then
do;
declare hash h(ordered: 'a');
h.definekey('id');
h.definedata('id','group');
h.definedone();
call missing(id,group);
end;
set have end=last;
array ids id1 id2;
do over ids;
rc=sum(rc,h.find(key:ids)=0);
/*you can choose to 'leave' the loop here when first h.find(key:ids)=0 is met, for the sake of better efficiency*/
end;
if not rc > 0 then
group+1;
do over ids;
id=ids;
h.replace();
end;
if last then rc=h.output(dataset:'want');
run;

Filling in gaps between sequentially numbered records and updating a status indicator

In a summarized dataset, I have the status of an event at each hour after baseline in which it was recorded. I also have the last hour the event could have been recorded. I want to create a new dataset with one record for each hour from the first through the last hour, with the status for each record being the one from the last recorded status.
Here is an example dataset:
data new;
input hour status last_hour;
cards;
2 1 12
4 1 12
5 1 12
6 1 12
7 0 12
9 1 12
10 0 12
;
run;
In this case, the first recorded hour was the second, and the last recorded hour was the 10th. The last possible hour to record data was the 12th.
The final dataset should look like so:
0 . 12
1 . 12
2 1 12
3 1 12
4 1 12
5 1 12
6 1 12
7 0 12
8 0 12
9 1 12
10 0 12
11 0 12
12 0 12
I sort of have it working with this series of data steps, but I'm not sure if there's a cleaner way I'm not seeing.
data step1;
set new (keep=id hour);
by id;
do hour = 0 to last_hour;
output;
end;
run;
proc sort data=step1;
by id hour;
run;
proc sql;
create table step2 as
select distinct a.id, a.hour, b.status
from step1 as a
left join new as b
on a.id = b.id
and a.hour = b.hour
order by a.id, a.hour;
quit;
data step3;
set step2;
by id hour;
retain previous_status;
if first.id then do;
previous_status = .;
if status > . then previous_status = status;
end;
if not first.id then do;
if status = . and previous_status > . then status = previous_status;
if status > . then previous_status = status;
end;
run;
Seeing your code, it seems you left out of your question the fact that you also have id's. So this is a newer solution that deals with different id's. See further below for my first solution ignoring id's.
Since last_hour is always 12, I left it out of the have dataset. It will be added later on.
data have;
input id hour status;
cards;
1 2 1
1 4 1
1 5 1
1 6 1
1 7 0
1 9 1
1 10 0
2 2 1
2 4 1
2 5 1
2 6 1
2 7 0
2 9 1
2 10 0
;
Create a hours dataset, just containing numbers 0 thru 12;
data hours;
do i = 0 to 12;
hour = i;
output;
end;
drop i;
run;
Create a temporary dataset that will have the right number of rows (13 rows for every id, with valid hour values where they exist in the have table).
proc sql;
create table tmp as
select distinct t1.id, t2.hour, 12 as last_hour
from have as t1
cross join
(select hour from hours) as t2;
quit;
Then use merge and retain to fill in the missing hour column where appropriate.
data want;
merge have
tmp;
by id hour;
retain status_previous;
if not first.id then do;
if status ne . then status_previous = status;
else if status_previous ne . then status = status_previous;
end;
if last.id then status_previous = .;
drop status_previous;
run;
Previous solution (no id's)
If last_hour is always 12, then this should do it:
data have;
input hour status last_hour;
datalines;
2 1 12
4 1 12
5 1 12
6 1 12
7 0 12
9 1 12
10 0 12
;
data hours;
do i = 0 to 12;
hour = i;
last_hour = 12;
output;
end;
drop i;
run;
data want;
merge have
hours;
by hour;
retain status_previous;
if status ne . then status_previous = status;
else if status_previous ne . then status = status_previous;
drop status_previous;
run;

Delete the group that none of its observation contain the certain value in SAS

I want to delete the whole group that none of its observation has NUM=14
So something likes this:
Original DATA
ID NUM
1 14
1 12
1 10
2 13
2 11
2 10
3 14
3 10
Since none of the ID=2 contain NUM=14, I delete group 2.
And it should looks like this:
ID NUM
1 14
1 12
1 10
3 14
3 10
This is what I have so far, but it doesn't seem to work.
data originaldat;
set newdat;
by ID;
If first.ID then do;
IF NUM EQ 14 then Score = 100;
Else Score = 10;
end;
else SCORE+1;
run;
data newdat;
set newdat;
If score LT 50 then delete;
run;
An approach using proc sql would be:
proc sql;
create table newdat as
select *
from originaldat
where ID in (
select ID
from originaldat
where NUM = 14
);
quit;
The sub query selects the IDs for groups that contain an observation where NUM = 14. The where clause then limits the selected data to only these groups.
The equivalent data step approach would be:
/* Get all the groups that contain an observation where N = 14 */
data keepGroups;
set originaldat;
if NUM = 14;
keep ID;
run;
/* Sort both data sets to ensure the data step merge works as expected */
proc sort data = originaldat;
by ID;
run;
/* Make sure there are no duplicates values in the groups to be kept */
proc sort data = keepGroups nodupkey;
by ID;
run;
/*
Merge the original data with the groups to keep and only keep records
where an observation exists in the groups to keep dataset
*/
data newdat;
merge
originaldat
keepGroups (in = k);
by ID;
if k;
run;
In both datasets the subsetting if statement is used to only output observations when the condition is met. In the second case k is a temporary variable with value 1(true) when a value is read from keepGroups an 0(false) otherwise.
You're sort of getting at a DoW loop here, but not quite doing it right. The problem (Assuming the DATA/SET names are mistyped and not actually wrong in your program) is the first data step doesn't append that 100 to every row - only to the 14 row. What you need is one 'line' per ID value with a keep/no keep decision.
You can either do this by doing your first data step, but RETAIN score, and only output one row per ID. Your code would actually work, based on 14 being the first row, if you just fixed your data/set typo; but it only works when 14 is the first row.
data originaldat;
input ID NUM ;
datalines;
1 14
1 12
1 10
2 13
2 11
2 10
3 14
3 10
;;;;
run;
data has_fourteen;
set originaldat;
by ID;
retain keep;
If first.ID then keep=0;
if num=14 then keep=1;
if last.id then output;
run;
data newdata;
merge originaldat has_fourteen;
by id;
if keep=1;
run;
That works by merging the value from a 1-per-ID to the whole dataset.
A double DoW also works.
data newdata;
keep=0;
do _n_=1 by 1 until (last.id);
set originaldat;
by id;
if num=14 then keep=1;
end;
do _n_=1 by 1 until (last.id);
set originaldat;
by id;
if keep=1 then output;
end;
run;
This works because it iterates over the dataset twice; for each ID, it iterates once through all records, looking for a 14, if it finds one then setting keep to 1. Then it reads all records again for that ID, and keeps if keep=1. Then it goes on to the next set of records by ID.
data in;
input id num;
cards;
1 14
1 12
1 10
2 16
2 13
3 14
3 67
;
/* To find out the list of groups which contains num=14, use below SQL */
proc sql;
select distinct id into :lst separated by ','
from in
where num = 14;
quit;
/* If you want to create a new data set with only groups containing num=14 then use following data step */
data out;
set in;
where id in (&lst.);
run;

Calculate maximum difference between grouped rows

I have the following data where people in households are sorted by age (oldest to youngest):
data houses;
input HouseID PersonID Age;
datalines;
1 1 25
1 2 20
2 1 32
2 2 16
2 3 14
2 4 12
3 1 44
3 2 42
3 3 10
3 4 5
;
run;
I would like to calculate for each household the maximum age difference between consecutively aged people. So this example would give values of 5 (=25-20), 16 (=32-16) and 32 (=42-10) for households 1, 2 and 3 consecutively.
I could do this using lots of merges (i.e. extract person 1, merge with extract of person 2, and so on), but as there can be upto 20+ people in a household I'm looking for a much more direct method.
Here's a two pass solution. Same first step as the two solutions above, sort by age. In the second step keep track of max_diff per row, at the last record of HouseID output the results. This results in only two passes through the data.
proc sort data=houses; by houseid age;run;
data want;
set houses;
by houseID;
retain max_diff 0;
diff = dif1(age)*-1;
if first.HouseID then do;
diff = .; max_diff=.;
end;
if diff>max_diff then max_diff=diff;
if last.houseID then output;
keep houseID max_diff;
run;
proc sort data=houses; by houseid personid age;run;
data _t1;
set houses;
diff = dif1(age) * (-1);
if personid = 1 then diff = .;
run;
proc sql;
create table want as
select houseid, max(diff) as Max_Diff
from _t1
group by houseid;
proc sort data = house;
by houseid descending age;
run;
data house;
set house;
by houseid;
lag_age = lag1(age);
if first.houseid then age_diff = 0;
age_diff = lag_age - age;
run;
proc sql;
select houseid,max(age_diff) as max_age_diff
from house
group by houseid;
quit;
Working:
First sort the data set using houseid and descending Age.
Second data step will calculate difference between current age value (in PDV) and previous age value in PDV. Then, using sql procedure, we can get the max age difference for each houseid.
Just throwing one more into the mix. This one is a condensed version of Reeza's response.
/* No need to sort by PersonID as age is the only concern */
proc sort data = houses;
by HouseID Age;
run;
data want;
set houses;
by HouseID;
/* Keep the diff when a new row is loaded */
retain diff;
/* Only replace the diff if it is larger than previous */
diff = max(diff, abs(dif(Age)));
/* Reset diff for each new house */
if first.HouseID then diff = 0;
/* Only output the final diff for each house */
if last.HouseID;
keep HouseID diff;
run;
Here is an example using FIRST. and LAST. with one pass (after sort) through the data.
data houses;
input HouseID PersonID Age;
datalines;
1 1 25
1 2 20
2 1 32
2 2 16
2 3 14
2 4 12
3 1 44
3 2 42
3 3 10
3 4 5
;
run;
Proc sort data=HOUSES;
by houseid descending age ;
run;
Data WANT(keep=houseid max_diff);
format houseid max_diff;
retain max_diff age1 age2;
Set HOUSES;
by houseid descending age ;
if first.houseid and last.houseid then do;
max_diff=0;
output;
end;
else if first.houseid then do;
call missing(max_diff,age1,age2);
age1=age;
end;
else if not(first.houseid or last.houseid) then do;
age2=age;
temp=age1-age2;
if temp>max_diff then max_diff=temp;
age1=age;
end;
else if last.houseid then do;
age2=age;
temp=age1-age2;
if temp>max_diff then max_diff=temp;
output;
end;
Run;

Detect the difference b/w ages greater than some value using SAS

I am trying to detect groups which contain the difference between first age and second age are greater than 5. For example, if I have the following data, the difference between age in grp=1 is 39 so I want to output that group in a separate data set. Same goes for grp 4.
id grp age sex
1 1 60 M
2 1 21 M
3 2 30 M
4 2 25 F
5 3 45 F
6 3 30 F
7 3 18 M
8 4 32 M
9 4 18 M
10 4 16 M
My initial idea was to sort them by grp and then get the absolute value between ages using something like if first.grp then do;. But I don't know how to get the absolute value between first age and second age by group or actually I don't know how should I start this.
Thanks in advance.
Here's one way that I think works.
data have;
input id $ grp $ age sex $;
datalines;
1 1 60 M
2 1 21 M
3 2 30 M
4 2 25 F
5 3 45 F
6 3 30 F
7 3 18 M
8 4 32 M
9 4 18 M
10 4 16 M
;
proc sort data=have ;
by grp descending age;
run;
data temp(keep=grp);
retain old;
set have;
by grp descending age;
if first.grp then old=age;
if last.grp then do;
diff=old-age;
if diff>5 then output ;
end;
run;
Data want;
merge temp(in=a) have(in=b);
by grp ;
if a and b;
run;
I would use PROC TRANSPOSE so the values in each group can easily be compared. For example:
data groups1;
input id $ grp age sex $;
datalines;
1 1 60 M
2 1 21 M
3 2 30 M
4 2 25 F
5 3 45 F
6 3 30 F
7 3 18 M
8 4 32 M
9 4 18 M
10 4 16 M
;
run;
proc sort data=groups1;
by grp; /* This maintains age order */
run;
proc transpose data=groups1 out=groups2;
by grp;
var age;
run;
With the transposed data you can do whatever comparison you like (I can't tell from your question what exactly you want, so I just compare first two ages):
/* With all ages of a particular group in a single row, it is easy to compare */
data outgroups1(keep=grp);
set groups2;
if abs(col1-col2)>5 then output;
run;
In this instance this would be my preferred method for creating a separate data set for each group that satisfies whatever condition is applied (generate and include code dynamically):
/* A separate data set per GRP value in OUTGROUPS1 */
filename dynacode catalog "work.dynacode.mycode.source";
data _null_;
set outgroups1;
file dynacode;
put "data grp" grp ";";
put " set groups1(where=(grp=" grp "));";
put "run;" /;
run;
%inc dynacode;
If you are after the difference between just the 1st and 2nd ages, then the following code is a fairly straightforward way of extracting these. It reads though the dataset to identify the groups, then uses the direct access method, POINT=, to extract the relevant records. I put in an extra condition, grp=lag(grp) just in case you have any groups with only 1 record.
data want;
set have;
by grp;
if first.grp then do;
num_grp=0;
outflag=0;
end;
outflag+ifn(lag(first.grp)=1 and grp=lag(grp) and abs(dif(age))>5,1,0) /* set flag to determine if group meets criteria */;
if not first.grp then num_grp+1; /* count number of records in group */
if last.grp and outflag=1 then do i=_n_-num_grp to _n_;
set have point=i; /* extract required group records */
drop num_grp outflag;
output;
end;
run;
Here's an SQL approach (using CarolinaJay's code to create the dataset):
data groups1;
input id grp age sex $;
datalines;
1 1 60 M
2 1 21 M
3 2 30 M
4 2 25 F
5 3 45 F
6 3 30 F
7 3 18 M
8 4 32 M
9 4 18 M
10 4 16 M
;
run;
proc sql noprint;
create table xx as
select a.*
from groups1 a
where grp in (select b.grp
from groups1 b
join groups1 c on c.id = b.id+1
and c.grp = b.grp
and abs(c.age - b.age) > 5
left join groups1 d on d.id = b.id-1
and d.grp = b.grp
where d.id eq .
)
;
quit;
The join on C finds all occurrences where the subsequent record in the same group has an absolute value > 5. The join on D (and the where clause) makes sure we only consider the results from the C join if the record is the very first record in the group.
data have;
input id $ grp $ age sex $;
datalines;
1 1 60 M
2 1 21 M
3 2 30 M
4 2 25 F
5 3 45 F
6 3 30 F
7 3 18 M
8 4 32 M
9 4 18 M
10 4 16 M
;
data want;
do i = 1 by 1 until(last.grp);
set have;
by grp notsorted;
if first.grp then cnt = 0;
cnt + 1;
if cnt = 1 then age1 = age;
if cnt = 2 then age2 = age;
diff = sum( age1, -age2 );
end;
do until(last.grp);
set have;
by grp;
if diff > 5 then output;
end;
run;