Obtain all possible pairs from a list

[ilikesas]

Hi,

I have the following data:

data have;
input element$ 1-2 qt;
datalines;
A 1
B 2
Cc3
D 4
Ee5
;
run;

I want to obtain all possible pairs like this:

A B

A Cc

A D

A Ee

B Cc

B D

...

Thanks!

[novinosrin]

Will this help? 

data have;
input element$ 1-2 qt;
datalines;
A 1
B 2
Cc3
D 4
Ee5
;
run;

proc sql;
select max(qt) into :m
from have;
quit;

data want;
if _n_=1 then do;
if 0 then set have;
  dcl hash H (dataset:'have') ;
   h.definekey  ("qt") ;
   h.definedata ("element") ;
   h.definedone () ;
end;
do i=1 to &m-1;
do j=i+1 to &m;
	rc=h.find(key:i);
	elementa=element;
	rc=h.find(key:j);
	elementb=element;
output;
end;
end;
keep elementa elementb;
run;

[ilikesas]

Hi novinsorin,

I am not familiar with hash, is it possible to get the solution in a macro form similar to this:

proc sql;
select element
into :element_list separated by ' '
from have
;
quit;
proc sql;
select count(element)
into :n
from have
;
quit;

%put &element_list;
%put &n;

%macro m1;
data want;
%do i=1 %to &n;
%let var1=%scan(&element_list,&i);
var1=&var1;

%end;
%mend m1;
%m1

Thank you

[novinosrin]

I  have a class to attend at my college in 5 mins,. If you are familiar with proc format cntrl in/cntrl out look up formatted dataset, you could replace my hash with formats which essentially does the same. Or i can look into it in 3 hours

Alternatively, you can filter the Cartesian product given by  paigemiller to only output the pairs with a where condition

[novinosrin]

data have;
input element$ 1-2 qt;
datalines;
A 1
B 2
Cc3
D 4
Ee5
;
run;
DATA fmtDataset;
 SET have;
 RETAIN fmtname 'fmtcode' type 'N';
 RENAME qt = Start;
 LABEL =element;
RUN; 
PROC FORMAT CNTLIN=fmtDataset;
RUN; 
proc sql;
select max(qt) into :m
from have;
quit;

data want;
do i=1 to &m-1;
do j=i+1 to &m;
elementa=put(i,fmtcode.); 
elementb=put(j,fmtcode.); 
output;
end;
end;
run;

[PaigeMiller]

proc sql;
    create table want as select a.element,b.element as element1 from have as a,have as b;
quit;

[ilikesas]

Hi PaigeMiller,

this code is quite simple, and I didn't even think of it!

But this code has repetitions, that is, since I have 5 elements it gives me all the pairs twice and also the pair of each element with itself

5sq = 25 pairs,

but if I want the unique non-same pairs I should get (25-5)/2 = 10 pairs

Thanks! 

[ballardw · Accepted solution]

---

@ilikesas wrote:

Hi PaigeMiller,

this code is quite simple, and I didn't even think of it!

But this code has repetitions, that is, since I have 5 elements it gives me all the pairs twice and also the pair of each element with itself

5sq = 25 pairs,

but if I want the unique non-same pairs I should get (25-5)/2 = 10 pairs

Thanks! 

---

Minor modification of @PaigeMiller's code;

proc sql;
   create table want as
   select distinct a.element,b.element as elementb
   from have as a, have as b
   where a.element<b.element;
quit;

You did not request unique solutions or ordered, so add the distinct to get the one group and the where to select the ordered pair.

[Reeza]

---

@ilikesas wrote:

Hi PaigeMiller,

this code is quite simple, and I didn't even think of it!

But this code has repetitions, that is, since I have 5 elements it gives me all the pairs twice and also the pair of each element with itself

5sq = 25 pairs,

but if I want the unique non-same pairs I should get (25-5)/2 = 10 pairs

Thanks! 

---

That's not all possible, that's all possible unique combinations, irrespective of order.

Another method is using ALLCOMB().

See this example here. 

[ilikesas]

These combinations include duplicates and same-element combinations, whereas I don't want these duplicates and same element combinations - how would I in this case delete the duplicates and same value combs?

[Reeza]

Sorry, I forgot the link.

%let n=5;              /* total number of items */          
%let k=2;              /* number of items per row */
/* generate combinations of n items taken k at a time */
data Comb(keep=c1-c&k);
array c[&k] (&k.*0);   /* initialize array to 0 */
array list[&n] _temporary_ (1, 2, 3, 4, 5); /*values in the array to set into groups of 2*/
ncomb = comb(&n, &k);  /* number of combinations */
do j = 1 to ncomb;
   rc = lexcombi(&n, &k, of c[*]);
   
   do i=1 to dim(c);
        c(i)=list(c(i));
   end;
   output;
end;
run;

Edit: and forgot it again:

https://blogs.sas.com/content/iml/2014/07/28/lexicographic-combinations.html 

[ChrisNZ]

The solution by @ballardw yields 10 records as wanted 

1. Please accept it (or @Reeza's solution if you prefer) as the answer to your question

2. To save everyone's time, do try to ask the full question right away rather than adding conditions as you go 

[PaigeMiller]

---

@ilikesas wrote:

Hi PaigeMiller,

this code is quite simple, and I didn't even think of it!

But this code has repetitions, that is, since I have 5 elements it gives me all the pairs twice and also the pair of each element with itself

5sq = 25 pairs,

but if I want the unique non-same pairs I should get (25-5)/2 = 10 pairs

Thanks! 

---

As @ballardw (and probably others) have pointed out ... you can then reduce the number of pairs to what you want via any number of methods in SAS (SQL or data step) ... but your original problem did not ask for "unique non-same pairs", it asked for "all possible pairs" (in fact, you asked for "all possible pairs" twice, once in the title and once in the text of your original message).

[Ksharp]

data have;
input element$ 1-2 qt;
datalines;
A 1
B 2
Cc3
D 4
Ee5
;

data want;
 set have end=last nobs=nobs;
 array x{999} $ 32 _temporary_;
 x{_n_}=element;
 if last then do;
   do i=1 to nobs-1;
     var1=x{i};
     do j=i+1 to nobs;
       var2=x{j};output;
	 end;
   end;
 end;
 keep var1 var2;
run;