Oracle analytic functions - how to get the highest RANKing row?

Test data:

create table test (
  grp varchar2(16)
, mbr varchar2(16)
, reading1 number
, reading2 number
);

-- group A: 3 members, 1 duplicate set
-- group B: 2 members, 1 duplicate, one reading NULL
-- group C: 2 members, no repeats, no NULLs 
begin
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'A', 'x', '1.0', '2.0' ) ;
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'A', 'y', '1.1', '2.2' ) ;
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'A', 'z', '1.2', '2.4' ) ;
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'A', 'x', '1.0', '2.0' ) ;
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'A', 'y', '1.1', '2.2' ) ;
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'A', 'z', '1.2', '2.4' ) ;
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'B', 'y', '20.2', null ) ;
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'B', 'x', '20.4', '40.4' ) ;
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'B', 'y', '20.2', null ) ;
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'B', 'x', '20.4', '40.4' ) ;
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'C', 'r', '100.1', '200.2' ) ;
  insert into test ( grp, mbr, reading1, reading2 ) 
    values ( 'C', 's', '100.2', '200.4' ) ;
end;
/

See dbfiddle.

select * from test;
GRP  MBR  READING1  READING2  
A    x    1         2         
A    y    1.1       2.2       
A    z    1.2       2.4       
A    x    1         2         
A    y    1.1       2.2       
A    z    1.2       2.4       
B    y    20.2      NULL      
B    x    20.4      40.4      
B    y    20.2      NULL      
B    x    20.4      40.4      
C    r    100.1     200.2     
C    s    100.2     200.4

Problem:

Write a query that does all of the following:

{1} Find unique rows.

{2} Find the last 2 members (mbr) of each group (grp). Assumption: when the memb

I fail to see the point of a requirements such as avoiding WHERE rank_ =, but here it is, without RANK(), or hardcoding a constant (still, hardcoding is done by using FIRST_VALUE):

select distinct grp,
  first_value(result1) over (partition by grp order by mbr desc) as result1,
  first_value(result2) over (partition by grp order by mbr desc) as result2
from (
select
  grp, mbr, 
  reading1 - lag(reading2) over (partition by grp order by mbr) result1,
  reading2 - lag(reading1) over (partition by grp order by mbr) result2
from (select unique grp, mbr,
             nvl(reading1, 0) as reading1, nvl(reading2, 0) as reading2
      from test)
);

GRP                 RESULT1    RESULT2
---------------- ---------- ----------
A                        -1        1.3
B                     -20.2      -20.4
C                      -100      100.3

When this is easier to read in my opinion:

select grp, result1, result2 from (
select
  grp,
  reading1 - lag(reading2) over (partition by grp order by mbr) result1,
  reading2 - lag(reading1) over (partition by grp order by mbr) result2,
  rank() over (partition by grp order by mbr desc) as rank_
from (select unique grp, mbr,
             nvl(reading1, 0) as reading1, nvl(reading2, 0) as reading2
      from test)
) where rank_ = 1
;

StackExchange Database Administrators Q#199118, answer score: 3