Number of palindromes in character strings - sql

Question

I'm trying to gather a list of 6 letter palindromes and the number of times they occur using Postgres 9.3.5.
This is the query I've tried:
SELECT word, count(*)
FROM ( SELECT regexp_split_to_table(read_sequence, '([ATCG])([ATCG])([ATCG])(\3)(\2)(\1)') as word
FROM reads ) t
GROUP BY word;
However this brings up results that a) aren't palindromic and b) greater or less than 6 letters long.
\d reads
Table "public.reads"
Column | Type | Modifiers
--------------+---------+-----------
read_header | text | not null
read_sequence | text |
option | text |
quality_score | text |
pair_end | text | not null
species_id | integer |
Indexes:
"reads_pkey" PRIMARY KEY, btree (read_header, pair_end)
read_sequence contains DNA sequences, 'ATGCTGATGCGGCGTAGCTGGATCGA' for example.
I'd like to see the number of palindromes in each sequence so the example would contain 1 another sequence could have 4 another 3 and so on.

Answer 1

Count per row:
SELECT read_header, pair_end, substr(read_sequence, i, 6) AS word, count(*) AS ct
FROM reads r
, generate_series(1, length(r.read_sequence) - 5 ) i
WHERE substr(read_sequence, i, 6) ~ '([ATCG])([ATCG])([ATCG])\3\2\1'
GROUP BY 1,2,3
ORDER BY 1,2,3,4 DESC;
Count per read_header and palindrome:
SELECT read_header, substr(read_sequence, i, 6) AS word, count(*) AS ct
FROM
...
GROUP BY 1,2
ORDER BY 1,2,3 DESC;
Count per read_header:
SELECT read_header, count(*) AS ct
FROM
...
GROUP BY 1
ORDER BY 1,2 DESC;
Count per palindrome:
SELECT substr(read_sequence, i, 6) AS word, count(*) AS ct
FROM
...
GROUP BY 1
ORDER BY 1,2 DESC;
SQL Fiddle.
Explain
A palindrome could start at any position 5 characters shy of the end to allow a length of 6. And palindromes can overlap. So:
Generate a list of possible starting positions with generate_series() in a LATERAL join, and based on this all possible 6-character strings.
Test for palindrome with regular expression with back references, similar to what you had, but regexp_split_to_table() is not the right function here. Use a regular expression match (~).
Aggregate, depending on what you actually want.