问题
I got this query to solve for the first 20 but I don’t know how to extend that to the 5 rows
prepare dna_length(int) as
with t1 as (
select chr(65) as s
union select chr(67)
union select chr(71)
union select chr(84) )
, t2 as ( select s,
row_number() over() as rn from t1)
, t3 as ( select generate_series(1,$1) as i,
round(random() * 4 + 0.5) as rn )
, t4 as ( select t2.s
from t2 join t3 on (t2.rn=t3.rn))
select
array_to_string(array(select s from t4),'') as dna ;
enter image description here
回答1:
Does this work for you? You can put the $1 back to replace the 20 in t3 if it does.
with t1 as (
select 1 as rn, 'A' as s
union select 2, 'C'
union select 3, 'T'
union select 4, 'G'
), t2 as (
select generate_series(1, 5) as sample
), t3 as (
select t2.sample, generate_series(1,20) as i,
round(random() * 4 + 0.5) as rn
from t2
), t4 as (
select t3.sample, t3.i, t3.rn, t1.s
from t3
join t1 on t1.rn = t3.rn
)
select sample, string_agg(s, '' order by i)
from t4
group by sample
order by sample;
sample | string_agg
--------+----------------------
1 | AGCATCAAGCGGTAAAAAAG
2 | ATATCGCGCCGAGGGAAGAC
3 | GAAACCCCATCTTAACTGGA
4 | AGTGAGGCCGCCACTCTACC
5 | AGTCCCCACAACGATTAGAA
(5 rows)
来源：https://stackoverflow.com/questions/62584420/a-query-to-generate-5-random-dna-sequences-that-are-each-about-20-bases