SUAVe Open Problem (May 8 th 2018 meeting)

Transcription

1 SUAVe Open Problem (May 8 th 2018 meeting) For this Open Problem, you are given a list of people and the set of (ICD9) diagnoses coded for each person by one or more physicians within a 12 month period. Each person has between one and five distinct ICD9 codes. The problem dataset contains two columns: Name and ICD9. There is one row for each ICD9 code that applies to each person. Your challenge is to find matched pairs of two people each, where both people have the exact same set of ICD9 codes. That is, the full set of ICD9 codes for both persons must be identical. A matched pair could consist of two people who have the same single diagnosis, up to two people who have the same five ICD9 codes. People without a match can be dropped. Here are a couple of sample matches, drawn from the data below: Dora Ted (both have exactly these two codes only: 401 & 53081) Susan Dawn (both have exactly this list of codes: 493, 709, 7099, V048) When there are more than two people with the same set of diagnoses, it is not necessary to identify all pairings among those people (unless you want to). For example, if k people have the exact same codes, it is sufficient to identify the k 1 pairs that include one particular person in the set, or instead a list of k 2 mutually exclusive pairs of people, rather than all ( k ) pairs. However you handle this scenario, describe your approach. 2 To load the problem data into a SAS dataset, you can run the following code. data op_data; infile cards4; input name $ icd9 cards4; Aiden 250 Alfred 311 Alfred 595 Alfred 7099 Alfred 7832 Brian 493 Brian 709 Brian 7099 Brian V048 Brianna 250 Charles 244 Charles 250 Cheryl 7099 Cheryl 7832 Cynthia V25 Dawn 493 Dawn 709 Dawn 7099 Dawn V048 Dora 401 Dora Dylan 401 Dylan Dylan 709 Dylan 715 Dylan V048 Eric V25 Ethan 493 Ethan 7099 Hannah 401 Hannah Hannah 709 Hannah 715 Hannah V048 Helen 244 Janet 311 Janet Janet 595 Janet 7099 Janet 7832 John 244 John 250 Mary 244 Mary 250 Matthew 250 Michael 244 Norbert 7099 Norbert 7832 Robert 401 Robert Samantha 250 Sean 493 Sean 7099 Sean 7832 Sophia 493 Sophia 7099 Steven 595 Steven 715 Steven V048 Susan 493 Susan 709 Susan 7099 Susan V048 Ted 401 Ted Violet 401 Violet Willliam 311 Willliam 595 Willliam 7099 Willliam 7832 ;;;; 1

2 Solution 1. David Mah, BC Ministry of Health * put the diag codes into separate columns; data TMP1; set OP_DATA; if first.name then ct=1; else ct+1; proc sort data=tmp1; by name ct icd9; data TMP; set TMP1; format diag1-diag5 $5.; array diag(5) diag1-diag5; do i=1 to 5; if i=ct then do; diag(i) = icd9; keep name diag1-diag5; * combine the diag codes into one row per person; proc sort data=tmp; by name diag1-diag5; data UPDT; update TMP (obs=0) TMP; data COMBINE; set UPDT; diag=diag1 diag2 diag3 diag4 diag5; d2=compress(diag,''); drop diag1-diag5 diag; proc sort data=combine; by d2 name; data RETDAT; set COMBINE; by d2 name; format diagcd $20.; if first.d2 then do; retain d2; diagcd=d2; * count the number of distinct people in the data and put the number in a global var; create table PEEPS as select count(*) as peepcnt from RETDAT; 2

3 data NULL; set PEEPS; %global nums; call symput('nums',peepcnt); * create separate datasets for each person; %macro DODS (y); %let x=&y.; %do z=1 %to &x.; data DS&z.; set RETDAT; if _n_ = &z. then output DS&z.; % %m %DODS(&nums.); * match datasets; * checks for matching pairs of datasets; * deletes empty datasets; %macro MATCHDS (y); %global f; *number of successful matches; %let a=1; %let z=1; %do %until (&z.=&y.); %let z1=&z.; %let z2=%eval(&z1.+1); create table MATCH&z1. as select a.name, b.name as name2, a.diagcd, b.d2 as diagcd2 from DS&z1. a, DS&z2. b where a.d2=b.d2; * check for empty dataset looking at the number of observations in each dataset; data _null_; call symput('obscnt',0); set MATCH&z1.; call symput('obscnt',_n_); stop; * deletes empty datasets; %If &obscnt=0 %then %do; %put "There are no records in the dataset MATCH&z."; proc delete data=match&z1. (gennum=all); * delete dataset; % %else %do; create table SUCCESS&a. as select * from MATCH&z1.; %let a=%eval(&a.+1); %let f=%eval(&a.-1); % %let z=%eval(&z1.+1); % 3

4 %m %MATCHDS (&nums.); * combine all non empty datasets into one final dataset; data FINAL; set SUCCESS1-SUCCESS&f.; Solution 2. Arthur Tabachneck, Analyst Finder, Inc. proc transpose data=op_data out=op_data_wide (drop=_:) prefix=icd9_; var icd9; /* Sort table in order of icd9 codes */ proc sort data=op_data_wide; by icd9_1-icd9_5; /* Select mutually exclusive pairs for first person with matching codes */ data want (keep=pair); set op_data_wide; by icd9_1-icd9_5; retain first; if not (first.icd9_5 and last.icd9_5) then do; if first.icd9_5 then first=name; else do; pair=catx('+',first,name); Solution 3. Peter Ott, BC Ministry of FLNRORD data convert; *convert codes to a wide character string; length string $ 30; set op_data; retain string; if first.name then do; string=''; string=catx(",",string,icd9); if last.name then keep name string; 4

5 data pairs; *all possible unique matching pairs; set convert; pos=_n_; do k=pos+1 to nobs; *note starting index of loop; set convert(rename=(name=cname string=cstring)) point=k nobs=nobs; if string=cstring then do; pair1=name; pair2=cname; keep pair1 pair2 string; Solution 4. Robert Ellsworth, Ellsworth Stewart Consulting /*** The approach is to build a key with all the codes a person has (max 5). I sort by the name so I can create a compound key as I scan through the rows. I sort by the code so the codes will be in the same order for everyone. Join the table to itself using my constructed key to get all the matches. I use the a.name < b.name so I don t get a person matching themselves or the same match in a different order. ***/ data op_unique; retain icd9by5; length icd9by5 $25; set op_data; if first.name then icd9by5 = icd9; else icd9by5 = trim(icd9by5) "," icd9; if last.name ; create table matches as select a.name as name_1, b.name as name_2, a.icd9by5 from op_unique a join op_unique b on a.icd9by5 = b.icd9by5 and a.name < b.name order by a.icd9by5, a.name, b.name; proc print data =matches noobs; var name_1 name_2 icd9by5; 5