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marianob
Obsidian | Level 7

Hello guys!

 

I'm struggling to achieve this, but I can't find the solution.

What I need to do, is to count the records where the inquiry_date is older than the per_obs, for each per_obs, and make it accumulative for each id.
So, it must restart when the ID changes. (the per_obs is set at the beginning of the month):

 

This is the original table:

IDINQUIRY_DATEPERIOD
A9/12/20192019-12
A9/12/20192019-12
A9/12/20192019-12
A4/11/20192019-11
A19/10/20192019-10
A19/10/20192019-10
A8/10/20192019-10
A20/9/20192019-09
A21/8/20192019-08
A21/8/20192019-08
A28/7/20192019-07
A26/7/20192019-07
B23/8/20182018-08
B23/8/20182018-08
B9/8/20182018-08
B13/7/20182018-07
B25/6/20182018-06
B11/5/20182018-05
B18/4/20182018-04

 

This is what I need:

IDINQUIRY_DATEPERIODAccumulative Inquiry dates
A9/12/20192019-129
A9/12/20192019-129
A9/12/20192019-129
A4/11/20192019-118
A19/10/20192019-105
A19/10/20192019-105
A8/10/20192019-105
A20/9/20192019-094
A21/8/20192019-082
A21/8/20192019-082
A28/7/20192019-070
A26/7/20192019-070
B23/8/20182018-083
B23/8/20182018-083
B9/8/20182018-083
B13/7/20182018-073
B25/6/20182018-062
B11/5/20182018-051
B18/4/20182018-040

 

Any help would be awesome, thanks a lot!

 

 

1 ACCEPTED SOLUTION

Accepted Solutions
smantha
Lapis Lazuli | Level 10

data new;
informat date ddmmyy10.;
input id $ date period $;
format date mmddyy10.;
cards;;;
A 9/12/2019 2019-12
A 9/12/2019 2019-12
A 9/12/2019 2019-12
A 4/11/2019 2019-11
A 19/10/2019 2019-10
A 19/10/2019 2019-10
A 8/10/2019 2019-10
A 20/9/2019 2019-09
A 21/8/2019 2019-08
A 21/8/2019 2019-08
A 28/7/2019 2019-07
A 26/7/2019 2019-07
B 23/8/2018 2018-08
B 23/8/2018 2018-08
B 9/8/2018 2018-08
B 13/7/2018 2018-07
B 25/6/2018 2018-06
B 11/5/2018 2018-05
B 18/4/2018 2018-04
;;;
run;
proc sort data=new;
by id period;
run;

data new2;
set new;
by id period;
retain cur_queries 0 tot_prev_queries 0;
if first.period then do;

tot_prev_queries = tot_prev_queries+cur_queries;
cur_queries = 0;
end;
cur_queries = cur_queries +1;
if first.id then tot_prev_queries =0;
drop cur_queris;
run;

proc sort data=new2;
by id descending period ;
run;

data new;
informat date ddmmyy10.;
input id $ date period $;
format date mmddyy10.;
cards;;;
A	9/12/2019	2019-12
A	9/12/2019	2019-12
A	9/12/2019	2019-12
A	4/11/2019	2019-11
A	19/10/2019	2019-10
A	19/10/2019	2019-10
A	8/10/2019	2019-10
A	20/9/2019	2019-09
A	21/8/2019	2019-08
A	21/8/2019	2019-08
A	28/7/2019	2019-07
A	26/7/2019	2019-07
B	23/8/2018	2018-08
B	23/8/2018	2018-08
B	9/8/2018	2018-08
B	13/7/2018	2018-07
B	25/6/2018	2018-06
B	11/5/2018	2018-05
B	18/4/2018	2018-04
;;;
run;
proc sort data=new;
by id period;
run;

data new2;
set new;
by id period;
retain cur_queries 0 tot_prev_queries 0;
if first.period then  do;
   
   tot_prev_queries = tot_prev_queries+cur_queries;
   cur_queries = 0;
end;
cur_queries = cur_queries +1;
if first.id then tot_prev_queries =0;
drop cur_queris;
run;

proc sort data=new2;
by id descending period ;
run;

View solution in original post

14 REPLIES 14
yabwon
Onyx | Level 15

What's `per_obs`?

Bart

_______________
Polish SAS Users Group: www.polsug.com and communities.sas.com/polsug

"SAS Packages: the way to share" at SGF2020 Proceedings (the latest version), GitHub Repository, and YouTube Video.
Hands-on-Workshop: "Share your code with SAS Packages"
"My First SAS Package: A How-To" at SGF2021 Proceedings

SAS Ballot Ideas: one: SPF in SAS, two, and three
SAS Documentation



marianob
Obsidian | Level 7

Hi Bart!

 

Sorry, PER_OB is the same as PERIOD.

PaigeMiller
Diamond | Level 26

Please explain further the calculations desired.

 

Why is this one a zero

A 28/7/2019 2019-07 0

 

and this one a two?

A 21/8/2019 2019-08 2
--
Paige Miller
marianob
Obsidian | Level 7

Paige,

 

The first one is 0 because the ID A doesn't have any inquiry_date older than the PERIOD/PER_OBS. (It has two on the same month, 28/7/2019 and 26/7/2019 , but I don't have to consider inquiry_dates on the same period).

 

Meanwhile, the 2019-08 has two inquiry_dates from the historical data, 28/7/2019 and 26/7/2019, so the counter is marking 2.

 

Thanks!

novinosrin
Tourmaline | Level 20

Hi @marianob  Please try


data have;
input ID $	INQUIRY_DATE :ddmmyy10.	_PERIOD :$7.;
period=input(catx('-',_period,'01'),yymmdd10.);
format period yymmdd10. INQUIRY_DATE ddmmyy10.;
cards;
A	9/12/2019	2019-12
A	9/12/2019	2019-12
A	9/12/2019	2019-12
A	4/11/2019	2019-11
A	19/10/2019	2019-10
A	19/10/2019	2019-10
A	8/10/2019	2019-10
A	20/9/2019	2019-09
A	21/8/2019	2019-08
A	21/8/2019	2019-08
A	28/7/2019	2019-07
A	26/7/2019	2019-07
B	23/8/2018	2018-08
B	23/8/2018	2018-08
B	9/8/2018	2018-08
B	13/7/2018	2018-07
B	25/6/2018	2018-06
B	11/5/2018	2018-05
B	18/4/2018	2018-04
;


data want ;
 if _n_=1 then do;
  dcl hash H (multidata:'y') ;
  h.definekey  ("_INQUIRY_DATE") ;
  h.definedone () ;
  dcl hiter hi('h');
 end;
 do until(last.id);
  set have;
  by id;
  _INQUIRY_DATE=INQUIRY_DATE;
  h.add();
 end;
 do until(last.id);
  set have;
  by id INQUIRY_DATE period notsorted;
  if first.period then do;
   count=0;
   do while(hi.next()=0);
    count=sum(_INQUIRY_DATE<period,count);
   end;
  end;
  output;
 end;
 h.clear();
run;
ID INQUIRY_DATE _PERIOD period _INQUIRY_DATE count
A 09/12/2019 2019-12 2019-12-01 21841 9
A 09/12/2019 2019-12 2019-12-01 21841 9
A 09/12/2019 2019-12 2019-12-01 21841 9
A 04/11/2019 2019-11 2019-11-01 21841 8
A 19/10/2019 2019-10 2019-10-01 21841 5
A 19/10/2019 2019-10 2019-10-01 21841 5
A 08/10/2019 2019-10 2019-10-01 21841 5
A 20/09/2019 2019-09 2019-09-01 21841 4
A 21/08/2019 2019-08 2019-08-01 21841 2
A 21/08/2019 2019-08 2019-08-01 21841 2
A 28/07/2019 2019-07 2019-07-01 21841 0
A 26/07/2019 2019-07 2019-07-01 21841 0
B 23/08/2018 2018-08 2018-08-01 21292 4
B 23/08/2018 2018-08 2018-08-01 21292 4
B 09/08/2018 2018-08 2018-08-01 21292 4
B 13/07/2018 2018-07 2018-07-01 21292 3
B 25/06/2018 2018-06 2018-06-01 21292 2
B 11/05/2018 2018-05 2018-05-01 21292 1
B 18/04/2018 2018-04 2018-04-01 21292 0
yabwon
Onyx | Level 15

Double DoW-loop and Hash Table in one, I love it! 😎

 

Bart

_______________
Polish SAS Users Group: www.polsug.com and communities.sas.com/polsug

"SAS Packages: the way to share" at SGF2020 Proceedings (the latest version), GitHub Repository, and YouTube Video.
Hands-on-Workshop: "Share your code with SAS Packages"
"My First SAS Package: A How-To" at SGF2021 Proceedings

SAS Ballot Ideas: one: SPF in SAS, two, and three
SAS Documentation



novinosrin
Tourmaline | Level 20

Again, my initial kudos goes to Krakow and Marcin 🙂 ( to whom and Poland I owe my start 🙂 )Btw @yabwon  My colleagues love your packages albeit they are complaining that it's not accesible in work professional machines. Can you make it secure for us to access and download if you don't mind. I was gonna reach out to you offline for this request

yabwon
Onyx | Level 15
Just define "secure for us to access and download" and I gladly help.
Bart
_______________
Polish SAS Users Group: www.polsug.com and communities.sas.com/polsug

"SAS Packages: the way to share" at SGF2020 Proceedings (the latest version), GitHub Repository, and YouTube Video.
Hands-on-Workshop: "Share your code with SAS Packages"
"My First SAS Package: A How-To" at SGF2021 Proceedings

SAS Ballot Ideas: one: SPF in SAS, two, and three
SAS Documentation



novinosrin
Tourmaline | Level 20

Alright, let me explain over linkedin later and give you official email. Thank you for your attention!

marianob
Obsidian | Level 7

It works perfectly! Thanks a lot for your time!

smantha
Lapis Lazuli | Level 10

data new;
informat date ddmmyy10.;
input id $ date period $;
format date mmddyy10.;
cards;;;
A 9/12/2019 2019-12
A 9/12/2019 2019-12
A 9/12/2019 2019-12
A 4/11/2019 2019-11
A 19/10/2019 2019-10
A 19/10/2019 2019-10
A 8/10/2019 2019-10
A 20/9/2019 2019-09
A 21/8/2019 2019-08
A 21/8/2019 2019-08
A 28/7/2019 2019-07
A 26/7/2019 2019-07
B 23/8/2018 2018-08
B 23/8/2018 2018-08
B 9/8/2018 2018-08
B 13/7/2018 2018-07
B 25/6/2018 2018-06
B 11/5/2018 2018-05
B 18/4/2018 2018-04
;;;
run;
proc sort data=new;
by id period;
run;

data new2;
set new;
by id period;
retain cur_queries 0 tot_prev_queries 0;
if first.period then do;

tot_prev_queries = tot_prev_queries+cur_queries;
cur_queries = 0;
end;
cur_queries = cur_queries +1;
if first.id then tot_prev_queries =0;
drop cur_queris;
run;

proc sort data=new2;
by id descending period ;
run;

data new;
informat date ddmmyy10.;
input id $ date period $;
format date mmddyy10.;
cards;;;
A	9/12/2019	2019-12
A	9/12/2019	2019-12
A	9/12/2019	2019-12
A	4/11/2019	2019-11
A	19/10/2019	2019-10
A	19/10/2019	2019-10
A	8/10/2019	2019-10
A	20/9/2019	2019-09
A	21/8/2019	2019-08
A	21/8/2019	2019-08
A	28/7/2019	2019-07
A	26/7/2019	2019-07
B	23/8/2018	2018-08
B	23/8/2018	2018-08
B	9/8/2018	2018-08
B	13/7/2018	2018-07
B	25/6/2018	2018-06
B	11/5/2018	2018-05
B	18/4/2018	2018-04
;;;
run;
proc sort data=new;
by id period;
run;

data new2;
set new;
by id period;
retain cur_queries 0 tot_prev_queries 0;
if first.period then  do;
   
   tot_prev_queries = tot_prev_queries+cur_queries;
   cur_queries = 0;
end;
cur_queries = cur_queries +1;
if first.id then tot_prev_queries =0;
drop cur_queris;
run;

proc sort data=new2;
by id descending period ;
run;
marianob
Obsidian | Level 7

It works just great as @novinosrin code.

Thanks both for helping me out!

Ksharp
Super User
data new;
infile cards expandtabs;
informat date ddmmyy10.;
input id $ date period $;
format date mmddyy10.;  
cards;;;
A	9/12/2019	2019-12
A	9/12/2019	2019-12
A	9/12/2019	2019-12
A	4/11/2019	2019-11
A	19/10/2019	2019-10
A	19/10/2019	2019-10
A	8/10/2019	2019-10
A	20/9/2019	2019-09
A	21/8/2019	2019-08
A	21/8/2019	2019-08
A	28/7/2019	2019-07
A	26/7/2019	2019-07
B	23/8/2018	2018-08
B	23/8/2018	2018-08
B	9/8/2018	2018-08
B	13/7/2018	2018-07
B	25/6/2018	2018-06
B	11/5/2018	2018-05
B	18/4/2018	2018-04
;;;
run;
proc sql;
create table want as
select *,(select count(*) from new 
 where id=a.id and date<input(a.period,anydtdte.)) as count
 from new as a;
quit;
marianob
Obsidian | Level 7

Thanks, Ksharp! Your code worked great too!

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