MESH Data Tree:

example:
Hallux;A01.378.610.250.300.792.380

where:
A01 is Body Regions
A01.378 is Extremities
A01.378.610 is Lower Extremity
A01.378.610.250 is Foot
A01.378.610.250.300 is Forefoot, Human
A01.378.610.250.300.792 is Toes

CREATE OR REPLACE FUNCTION mesh_split(text) RETURNS text[]
AS $$
return [split('\.',$_[0])];
$$ LANGUAGE plperlu;

arancia=# select mesh_split('A01.378.610.250.300.792.380');
          mesh_split
-------------------------------
 {A01,378,610,250,300,792,380}
(1 row)


/*
   Is it a real array?
   If it is, why can I not use index to access its items?
 */

arancia=# select mesh_split('A01.378.610.250.300.792.380')[1];
ERROR:  syntax error at or near "["
LINE 1: select mesh_split('A01.378.610.250.300.792.380')[1];
                                                        ^
/*
    but it is an array, it behaves as it is.
 */
arancia=> select array_length(mesh_split('A01.378.610.250.300.792.380'),1);
 array_length
--------------
            7
(1 row)

/* How to get access to its items then?
 */


Another problem related:

arancia=> select * from meshtree where code = ANY mesh_split('A01.378.610.250.300.792.380');
ERROR:  syntax error at or near "mesh_split"
LINE 1: select * from meshtree where code = ANY mesh_split('A01.378....
                                                ^

select * from meshtree, unnest(mesh_split('A01.378.610.250.300.792.380')) as c where c=meshtree.code; parent | id | code | description
--------+-------+------+-------------------------------------------------------------------
     10 |    11 | 300  | Dehydroepiandrosterone Sulfate
     33 |    34 | 250  | Cymarine
     48 |    49 | 250  | Cymarine
     61 |    62 | 250  | Dihydrotachysterol
     66 |    68 | 300  | Calcitriol
     65 |    69 | 250  | Calcifediol
     92 |    93 | 380  | Glycodeoxycholic Acid
     98 |    99 | 250  | Finasteride
    111 |   117 | 300  | Chenodeoxycholic Acid
    145 |   146 | 300  | Dehydroepiandrosterone Sulfate
    180 |   182 | 250  | Ethinyl Estradiol-Norgestrel Combination
    190 |   191 | 250  | Desoximetasone
[..]
        | 18638 | A01  | Body Regions
[..]
    190 |   192 | 300  | Dexamethasone Isonicotinate
    195 |   196 | 250  | Clobetasol
    199 |   200 | 300  | Fluocinonide
    206 |   207 | 250  | Diflucortolone
    266 |   267 | 300  | Dexamethasone Isonicotinate
    281 |   282 | 250  | Diflucortolone
    290 |   293 | 250  | Dehydrocholesterols
    305 |   306 | 250  | Dihydrotachysterol
    312 |   314 | 300  | Calcitriol
    311 |   315 | 250  | Calcifediol
    320 |   321 | 250  | Cholestanol
    328 |   330 | 300  | Calcitriol
[..]
  52135 | 52136 | 250  | Eye Injuries
  52136 | 52137 | 250  | Eye Burns
  52149 | 52155 | 300  | Hematoma, Epidural, Cranial
  52181 | 52196 | 300  | Gallbladder Emptying
  52269 | 52277 | 300  | Caplan Syndrome
  52360 | 52368 | 300  | Caplan Syndrome
  52428 | 52442 | 380  | Hemothorax
  52476 | 52491 | 610  | Pneumonia
  52534 | 52535 | 380  | Legionnaires' Disease
(2204 rows)

I really want to write better similar query:

arancia=> with recursive t(id,parent,codeparts,idx,last,descriptions) as (
  SELECT
id, parent, mesh_split('A01.378.610.250.300.792.380'), 1, array_length(mesh_split('A01.378.610.250.300.792.380'),1), ARRAY[description]
    FROM meshtree WHERE code='A01'
    UNION ALL
SELECT m.id, m.parent, t.codeparts, idx+1, last, descriptions || ARRAY[description]
    FROM meshtree AS m JOIN t ON (t.id=m.parent)
   WHERE idx<=last AND m.code=t.codeparts[idx+1])
 SELECT t.* FROM t;
id | parent | codeparts | idx | last | descriptions
-------+--------+-------------------------------+-----+------+--------...
18638 | | {A01,378,610,250,300,792,380} | 1 | 7 | {"Body Regions"} 18675 | 18638 | {A01,378,610,250,300,792,380} | 2 | 7 | {"Body Regions",Extremities} 18676 | 18675 | {A01,378,610,250,300,792,380} | 3 | 7 | {"Body Regions",Extremities,"Lower Extremity"} 18679 | 18676 | {A01,378,610,250,300,792,380} | 4 | 7 | {"Body Regions",Extremities,"Lower Extremity",Foot} 18682 | 18679 | {A01,378,610,250,300,792,380} | 5 | 7 | {"Body Regions",Extremities,"Lower Extremity",Foot,"Forefoot, Human"} 18683 | 18682 | {A01,378,610,250,300,792,380} | 6 | 7 | {"Body Regions",Extremities,"Lower Extremity",Foot,"Forefoot, Human",Toes} 18684 | 18683 | {A01,378,610,250,300,792,380} | 7 | 7 | {"Body Regions",Extremities,"Lower Extremity",Foot,"Forefoot, Human",Toes,Hallux}
(7 rows)

explain analyze with recursive t(id,parent,codeparts,idx,last,descriptions) as ( select id,parent,mesh_split('A01.378.610.250.300.792.380'),1,array_length(mesh_split('A01.378.610.250.300.792.380'),1),ARRAY[description] from meshtree where code='A01'
  union all
select m.id,m.parent,t.codeparts,idx+1,last,descriptions || ARRAY[description] from meshtree as m join t on (t.id=m.parent) where idx<=last and m.code=t.codeparts[idx+1]) select t.* from t;

QUERY PLAN
------------------------------------------------------------------------------------------------------------------------------------------------------
CTE Scan on t (cost=6336.53..6337.17 rows=32 width=80) (actual time=4.850..9.453 rows=7 loops=1)
   CTE t
-> Recursive Union (cost=0.00..6336.53 rows=32 width=99) (actual time=4.839..9.397 rows=7 loops=1) -> Index Scan using meshtree_id_code on meshtree (cost=0.00..1030.38 rows=22 width=27) (actual time=4.828..8.895 rows=1 loops=1)
                 Index Cond: (code = 'A01'::text)
-> Nested Loop (cost=0.00..530.55 rows=1 width=99) (actual time=0.051..0.061 rows=1 loops=7) -> WorkTable Scan on t (cost=0.00..4.95 rows=73 width=76) (actual time=0.005..0.008 rows=1 loops=7)
                       Filter: (idx <= last)
-> Index Scan using meshtree_parent_code on meshtree m (cost=0.00..7.18 rows=1 width=31) (actual time=0.031..0.034 rows=1 loops=7) Index Cond: ((m.parent = t.id) AND (m.code = t.codeparts[(t.idx + 1)]))
 Total runtime: 9.758 ms
(11 rows)


PostgreSQL rocks!


Thank you in advance,              \ferz

<<attachment: nonsolosoft.vcf>>

-- 
Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
To make changes to your subscription:
http://www.postgresql.org/mailpref/pgsql-sql

Reply via email to