Since I am new to SAS I need some help to understand how to combine the overlap date ranges into one row.I want to combine the overlap date ranges when they have matching Id. If the dates don’t overlap then I want to keep them as it is. IF they over lap by Matching Id and drug code Then it should combine into one line. Please look at the same ple data set which I have below and the expected results:
Current Data set:
ID Drug Code BEG_Date End_Date
1 100 1/1/2018 1/1/2019
1 100 1/1/2018 3/1/2018
1 100 2/1/2018 04/30/2018
1 90 4/1/2018 04/30/2018
1 100 5/1/2018 6/1/2018
1 98 6/1/2018 8/31/2018
1 100 9/1/2018 5/4/2019
Expected results:
ID Drug Code BEG_Date End_Date
1 100 1/1/2018 3/31/2018
1 90 4/1/2018 04/30/2018
1 100 5/1/2018 6/1/2018
1 98 6/2/2018 8/31/2018
1 100 9/1/2018 5/4/2019
I wrote some SAS code but I am combining the dates even when there is no overlap. I want to write some code which should work in SAS.
PROC SORT DATA=Want OUT=ONE;
BY PERSON_ID BEG_DATE DRUG_CODE END_DATE;
RUN;
data TWO (DROP=PERSON_ID2 DRUG_CODE2 BEG_DATE END_DATE
RENAME=(BEG2=BEG_DOS
END2=END_DOS));
SET ONE;
RETAIN BEG2 END2;
PERSON_ID2=LAG1(PERSON_ID);
DRUG_CODE2=LAG1(DRUG_CODE);
IF PERSON_ID2=PERSON_ID AND DRUG_CODE2=DRUG_CODE AND BEG_DATE LE(END2+1) THEN
DO;
BEG2=MIN(BEG_DATE,BEG2);
END2=MAX(END_DATE,END2);
END;
ELSE
DO;
SEG+1;
BEG2=BEG_DATE;
END2=END_DATE;
END;
FORMAT BEG2 END2 MMDDYY10.;
RUN;
DATA THREE(DROP=BEG_DOS END_DOS SEG);
RETAIN BEG_DATE END_DATE;
SET TWO;
BY PERSON_ID SEG;
FORMAT BEG_DATE END_DATE MMDDYY10.;
IF FIRST.SEG THEN
DO;
BEG_DATE=BEG_DOS;
END;
IF LAST.SEG THEN
DO;
END_DATE = END_DOS;
OUTPUT;
END;
RUN;
This is how I would do it. Create an obs for each ID DRUG and DATE. Flag the gaps and summarize by RUN.
data have;
input ID Drug_Code (BEG End)(:mmddyy.);
format BEG End mmddyyd10.;
cards;
1 100 1/1/2018 3/1/2018
1 100 2/1/2018 04/30/2018
1 90 4/1/2018 04/30/2018
1 90 6/1/2018 8/15/2018
1 100 5/1/2018 6/1/2018
1 98 6/1/2018 8/31/2018
1 100 9/1/2018 5/4/2019
;;;;
run;
proc print;
run;
/*1 100 1/1/2018 1/1/2019*/
data exv/ view=exv;
set have;
do date = beg to end;
output;
end;
drop beg end;
format date mmddyyd10.;
run;
proc sort data=exv out=ex nodupkey;
by id drug_code date;
run;
data breaksV / view=BreaksV;
set ex;
by id drug_code;
dif = dif(date);
if first.drug_code then do; dif=1; run=1; end;
if dif ne 1 then run+1;
run;
proc summary data=breaksV nway missing;
class id drug_code run;
var date;
output out=want(drop=_type_) min=Begin max=End;
run;
Proc print;
run;
Computing the extent range composed of overlapping segment ranges requires a good understanding of the range conditions (cases).
Consider the scenarios when sorted by start date (within any larger grouping set, G, such as id and drug)
Let [ and ] be endpoints of a range
# be date values (integers) within
Extent be the combined range that grows
Segment be the range in the current row
Case 1 - Growth. Within G Segment start before Extent end
Segment will either not contribute to Extent or extend it.
[####] Extent
+ [#] Segment range DOES NOT contribute
--------
[####] Extent (do not output a row, still growing)
or
[####] Extent
+ [#####] Segment range DOES contribute
--------
[#######] Extent (do not output a row, still growing)
Case 2 - Terminus. 3 possibilities:
Within G Segment start after Extent end,
Next G reached (different id/drug combination),
End of data reached.
#2 and #3 can be tested by checking the appropriate last. flag.
[####] Extent
+ ..[#] Segment beyond Extent (gap is 2)
--------
[####] output Extent
[#] reset Extent to Segment
You can adjust your rules for Segment being adjacent (gap=0) or close enough (gap < threshold) to mean an Extent is either expanded, or, output and reset to Segment.
Note: The situation is a little more (not shown) complicated for the real world cases of:
missing start means the Segment has an unknown start date (presume it to be epoch (0=01JAN1960, or some date that pre-dates all dates in the data or study)
missing end means the Segment is active today (end date is date when processing data)
Sample code:
data have;
call streaminit(42);
do id = 1 to 10;
do _n_ = 1 to 50;
drug = ceil(rand('UNIFORM', 10));
beg_date = intnx ('MONTH', '01JAN2008'D, rand('UNIFORM',20));
end_date = intnx ('DAY', beg_date, rand('UNIFORM',75));
OUTPUT;
end;
end;
format beg_date end_date yymmdd10.;
run;
proc sort data=have out=segments;
by id drug beg_date end_date;
run;
data want;
set segments;
by id drug beg_date end_date; * will error if incoming data is NOT sorted;
retain ext_beg ext_end;
retain gap_allowed 0; * set to 1 for contiguously adjacent segment ;
if first.drug then do;
ext_beg = beg_date;
ext_end = end_date;
segment_count = 0;
end;
if beg_date <= ext_end + gap_allowed then do;
ext_end = max (ext_end, end_date);
segment_count + 1;
end;
else do;
extent_id + 1;
OUTPUT;
ext_beg = beg_date;
ext_end = end_date;
segment_count = 1;
end;
if last.drug then do;
extent_id + 1;
OUTPUT;
* reset occurs implicitly;
* it will happen at first. logic when control returns to top of step;
end;
format ext_: yymmdd10.;
keep id drug ext_beg ext_end segment_count extent_id;
run;
Related
I have a dataset like below
Customer_ID Vistited_Date
1234 7-Feb-20
4567 7-Feb-20
9870 7-Feb-20
1234 14-Feb-20
7654 14-Feb-20
3421 14-Feb-20
I am trying find the cumulative unique count of customers by date, assuming my output will be like below
Cust_count Vistited_Date
3 7-Feb-20
2 14-Feb-20
7-Feb-2020 has 3 unique customers, whereas 14-Feb-2020 has only 2 hence customer 1234 has visited already.
Anyone knows how I could develop a data set in these conditions?
Sorry if my question is not clear enough, and I am available to give more details if necessary.
Thanks!
NOTE: #draycut's answer has the same logic but is faster, and I will explain why.
#draycut's code uses one hash method, add(), using the return code as test for conditional increment. My code uses check() to test for conditional increment and then add (which will never fail) to track. The one method approach can be perceived as being anywhere from 15% to 40% faster in performance (depending on number of groups, size of groups and id reuse rate)
You will need to track the IDs that have occurred in all prior groups, and exclude the tracked IDs from the current group count.
Tracking can be done with a hash, and conditional counting can be performed in a DOW loop over each group. A DOW loop places the SET statement inside an explicit DO.
Example:
data have;
input ID Date: date9.; format date date11.;
datalines;
1234 7-Feb-20
4567 7-Feb-20
9870 7-Feb-20
1234 14-Feb-20
7654 14-Feb-20
3421 14-Feb-20
;
data counts(keep=date count);
if _n_ = 1 then do;
declare hash tracker();
tracker.defineKey('id');
tracker.defineDone();
end;
do until (last.date);
set have;
by date;
if tracker.check() ne 0 then do;
count = sum(count, 1);
tracker.add();
end;
end;
run;
Raw performance benchmark - no disk io, cpu required to fill array before doing hashing, so those performance components are combined.
The root performance is how fast can new items be added to the hash.
Simulate 3,000,000 'records', 1,000 groups of 3,000 dates, 10% id reuse (so the distinct ids will be ~2.7M).
%macro array_fill (top=3000000, n_group = 1000, overlap_factor=0.10);
%local group_size n_overlap index P Q;
%let group_size = %eval (&top / &n_group);
%if (&group_size < 1) %then %let group_size = 1;
%let n_overlap = %sysevalf (&group_size * &overlap_factor, floor);
%if &n_overlap < 0 %then %let n_overlap = 0;
%let top = %sysevalf (&group_size * &n_group);
P = 1;
Q = &group_size;
array ids(&top) _temporary_;
_n_ = 0;
do i = 1 to &n_group;
do j = P to Q;
_n_+1;
ids(_n_) = j;
end;
P = Q - &n_overlap;
Q = P + &group_size - 1;
end;
%mend;
options nomprint;
data _null_ (label='check then add');
length id 8;
declare hash h();
h.defineKey('id');
h.defineDone();
%array_fill;
do index = 1 to dim(ids);
id = ids(index);
if h.check() ne 0 then do;
count = sum(count,1);
h.add();
end;
end;
_n_ = h.num_items;
put 'num_items=' _n_ comma12.;
put index= comma12.;
stop;
run;
data _null_ (label='just add');
length id 8;
declare hash h();
h.defineKey('id');
h.defineDone();
%array_fill;
do index = 1 to dim(ids);
id = ids(index);
if h.add() = 0 then
count = sum(count,1);
end;
_n_ = h.num_items;
put 'num_items=' _n_ comma12.;
put index= comma12.;
stop;
run;
data have;
input Customer_ID Vistited_Date :anydtdte12.;
format Vistited_Date date9.;
datalines;
1234 7-Feb-2020
4567 7-Feb-2020
9870 7-Feb-2020
1234 14-Feb-2020
7654 14-Feb-2020
3421 14-Feb-2020
;
data want (drop=Customer_ID);
if _N_=1 then do;
declare hash h ();
h.definekey ('Customer_ID');
h.definedone ();
end;
do until (last.Vistited_Date);
set have;
by Vistited_Date;
if h.add() = 0 then Count = sum(Count, 1);
end;
run;
If your data is not sorted and you like the SQL maybe this solution is same good for you and it is very simple:
/* your example 3 rows */
data have;
input ID Date: date9.; format date date11.;
datalines;
1234 7-Feb-20
4567 7-Feb-20
9870 7-Feb-20
1234 14-Feb-20
7654 14-Feb-20
3421 14-Feb-20
1234 15-Feb-20
7654 15-Feb-20
1111 15-Feb-20
;
run;
/* simple set theory. Final dataset contains your final data like results
below*/
proc sql;
create table temp(where =(mindate=date)) as select
ID, date,min(date) as mindate from have
group by id;
create table final as select count(*) as customer_count,date from temp
group by date;
quit;
/* results:
customer_count Date
3 07.febr.20
2 14.febr.20
1 15.febr.20
*/
Another method cause I dont know hash so well. >_<
data have;
input ID Date: date9.; format date date11.;
datalines;
1234 7-Feb-20
4567 7-Feb-20
9870 7-Feb-20
1234 14-Feb-20
7654 14-Feb-20
3421 14-Feb-20
;
data want;
length Used $200.;
retain Used;
set have;
by Date;
if first.Date then count = .;
if not find(Used,cats(ID)) then do;
count + 1;
Used = catx(',',Used,ID);
end;
if last.Date;
put Date= count=;
run;
If you are not overly concerned with processing speed and want something simple:
proc sort data=have;
by id date;
** Get date of each customer's first unique visit **;
proc sort data=have out=first_visit nodupkey;
by id;
proc freq data=first_visit noprint;
tables date /out=want (keep=date count);
run;
During some data cleaning process, there is a need to compare the data between different rows. For example, if the rows have the same countryID and subjectID then keep the largest temperature:
CountryID SubjectID Temperature
1001 501 36
1001 501 38
1001 510 37
1013 501 36
1013 501 39
1095 532 36
In this case like this, I will use the lag() function as follows.
proc sort table;
by CountryID SubjectID descending Temperature;
run;
data table_laged;
set table;
CountryID_lag = lag(CountryID);
SubjectID_lag = lag(SubjectID);
Temperature_lag = lag(Temperature);
if CountryID = CountryID_lag and SubjectID = SubjectID_lag then do;
if Temperature < Temperature_lag then delete;
end;
drop CountryID_lag SubjectID_lag Temperature_lag;
run;
The code above may work.
But I still want to know if there are any better ways to solve this kind of questions?
I think you complicate task. You can use proc sql and max function:
proc sql noprint;
create table table_laged as
select CountryID,SubjectID,max(Temperature)
from table
group by CountryID,SubjectID;
quit;
I don't know if you want it that way but you code would keep the highest temperatures
So when you have 2 1 3 for one subject if will keep 3. But when you have 1 4 3 4 4 it will keep 4 4 4. Better is to keep simple the first row for each subject which is the highest because of descending order.
proc sort data = table;
by CountryID SubjectID descending Temperature;
run;
data table_laged;
set table;
by CountryID SubjectID;
if first.SubjectID;
run;
You can use double DOW technique to:
Compute a measure over a group,
Apply the measure to items in the group.
The benefit of DOW looping is a single pass over the data set when incoming data is already grouped.
In this question, 1. is to identify the row in the group with the first highest temperature, and 2. is to select the row for output.
data want;
do _n_ = 1 by 1 until (last.SubjectId);
set have;
by CountryId SubjectId;
if temperature > _max_temp then do;
_max_temp = temperature;
_max_at_n = _n_;
end;
end;
do _n_ = 1 to _n_;
set have;
if _n_ = _max_at_n then OUTPUT;
end;
drop _:;
run;
The traditional procedural technique is Proc MEANS
data have;input
CountryID SubjectID Temperature; datalines;
1001 501 36
1001 501 38
1001 510 37
1013 501 36
1013 501 39
1095 532 36
run;
proc means noprint data=have;
by countryid subjectid;
output out=want(drop=_:) max(temperature)=temperature;
run;
If the data is disordered in CountryID and SubjectID going into the data step, a hash object can be used or SQL per #Aurieli.
I'm looking to take a variable observation's date and essentially keep rolling it forward by its specified repricing parameter until a target date
the dataset being used is:
data have;
input repricing_frequency date_of_last_repricing end_date;
datalines;
3 15399 21367
10 12265 21367
15 13879 21367
;
format date_of_last_repricing end_date date9.;
informat date_of_last_repricing end_date date9.;
run;
so the idea is that i'd keep applying the repricing frequency of either 3 months, 10 months or 15 months to the date_of_last_repricing until it is as close as it can be to the date "31DEC2017". Thanks in advance.
EDIT including my recent workings:
data want;
set have;
repricing_N = intck('Month',date_of_last_repricing,'31DEC2017'd,'continuous');
dateoflastrepricing = intnx('Month',date_of_last_repricing,repricing_N,'E');
format dateoflastrepricing date9.;
informat dateoflastrepricing date9.;
run;
The INTNX function will compute an incremented date value, and allows the resultant interval alignment to be specified (in your case the 'end' of the month n-months hence)
data have;
format date_of_last_repricing end_date date9.;
informat date_of_last_repricing end_date date9.;
* use 12. to read the raw date values in the datalines;
input repricing_frequency date_of_last_repricing: 12. end_date: 12.;
datalines;
3 15399 21367
10 12265 21367
15 13879 21367
;
run;
data want;
set have;
status = 'Original';
output;
* increment and iterate;
date_of_last_repricing = intnx('month',
date_of_last_repricing, repricing_frequency, 'end'
);
do while (date_of_last_repricing <= end_date);
status = 'Computed';
output;
date_of_last_repricing = intnx('month',
date_of_last_repricing, repricing_frequency, 'end'
);
end;
run;
If you want to compute only the nearest end date, as when iterating by repricing frequency, you do not have to iterate. You can divide the months apart by the frequency to get the number of iterations that would have occurred.
data want2;
set have;
nearest_end_month = intnx('month', end_date, 0, 'end');
if nearest_end_month > end_date then nearest_end_month = intnx('month', nearest_end_month, -1, 'end');
months_apart = intck('month', date_of_last_repricing, nearest_end_month);
iterations_apart = floor(months_apart / repricing_frequency);
iteration_months = iterations_apart * repricing_frequency;
nearest_end_date = intnx('month', date_of_last_repricing, iteration_months, 'end');
format nearest: date9.;
run;
proc sql;
select id, max(date_of_last_repricing) as nearest_end_date format=date9. from want group by id;
select id, nearest_end_date from want2;
quit;
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;
ESN is an id column that has multiple observations per esn, so repeated values of esn occur. For a given esn, I want to find the earliest service start date (and call it first), and I want to find the proper end date (called last) the if/then statement for how "last" is chosen is correct, but I get the following errors when I run the code below:
340 first = min(of start(*));
---
71
ERROR 71-185: The MIN function call does not have enough arguments.
here is the code I used
data three_1; /*first and last date created ?? used to ignore ? in data*/
set three;
format first MMDDYY10. last MMDDYY10.;
by esn;
array start(*) service_start_date;
array stop(*) service_end_date entry_date_est ;
do i=1 to dim(start);
first = min(of start(*));
end;
do i=1 to dim(stop);
if esn_status = 'Cancelled' then last = min(input(service_end_date, MMDDYY10.), input(entry_date_est, MMDDYY10.));
else last = max(input(service_end_date, MMDDYY10.), input(entry_date_est, MMDDYY10.));
end;
run;
"esn" "service_start_date" "service_end_date" "entry_date_est" "esn_status"
1 10/12/2010 01/01/2100 10/12/2012 cancelled
1 05/02/2009 02/12/2010 10/09/2012 cancelled
1 04/05/2011 03/04/2100 10/02/2012 cancelled
the results should be first= 05/02/2009 and last=10/12/2012
Arrays and the min(), max(), etc. functions operate horizontally across rows of a data set, not vertically across multiple records.
Assuming esn_status is constant for a given esn, then you need to sort your input by esn and service_start_date. You can use a data step to collect the values you want.
data three; /*thanks Joe for the data step to create the example data*/
length esn_status $10;
format service_start_date service_end_date entry_date_est MMDDYY10.;
input esn (service_start_date service_end_date entry_date_est) (:mmddyy10.) esn_status $;
datalines;
1 10/12/2010 01/01/2100 10/12/2012 cancelled
1 05/02/2009 02/12/2010 10/09/2012 cancelled
1 04/05/2011 03/04/2100 10/02/2012 cancelled
;;;;
run;
proc sort data=three;
by esn service_start_date;
run;
data three_1(keep=esn esn_status start last);
set three;
format start last date9.;
by esn;
retain start last;
if first.esn then do;
start = service_start_date;
last = service_end_date;
end;
if esn_status = "cancelled" then
last = min(last,service_end_date,entry_date_est);
else
last = max(last,service_end_date,entry_date_est);
if last.esn then
output;
run;
A DoW loop will get to where you want, or you could do it in SQL. Your desired results don't match with the actual results as far as I can tell, so you may need to make some adjustments. You'd need a second WANT dataset for the non-cancelled folks, I don't think there's an easy way to put it in one data step.
data have;
length esn_status $10;
format service_start_date service_end_date entry_date_est MMDDYY10.;
input esn (service_start_date service_end_date entry_date_est) (:mmddyy10.) esn_status $;
datalines;
1 10/12/2010 01/01/2100 10/12/2012 cancelled
1 05/02/2009 02/12/2010 10/09/2012 cancelled
1 04/05/2011 03/04/2100 10/02/2012 cancelled
;;;;
run;
data want_cancelled;
first = 99999;
last = 99999;
do _n_ = 1 by 1 until (last.esn);
set have(where=(esn_status='cancelled'));
by esn;
first = min(first,service_start_date);
last = min(last,service_end_date,entry_date_est);
end;
output;
keep first last esn;
format first last mmddyy10.;
run;