计算每一行的频率

Question

我正在尝试计算每行中我的元素的频率，我将解释：我 select 来自 table 包含一些元素，例如 "pos,chr,ref,alt,id_disease".

从这些我必须提取我的 ref 的频率，alt 即：

num_occurrencies_of(ref='A' and alt='C')/total number of rows.

通过这个查询，我几乎没有接近我的 objective，事实上 id 没有正确计算频率它 returns 总是 a constant

SELECT pos, chr, upper(ref||' '||alt) AS refalt, id_disease AS lvl15, t1.tot_var, t1.freq
FROM varianti 
JOIN ( SELECT count(*) AS tot_var,(count(*)::numeric / sum(count(*)) over ()) as freq
       FROM varianti)t1 ON TRUE
WHERE length(ref)=1 AND length(alt)=1 AND chr similar to 'chr[\d X Y]*'

我只想像这样检索数据：

chr pos refalt lvl15 freq tot_var
1   120  AT     15    0.3  1000
1   150  CG     30    0.01 1000

tot_var = 计算我需要的行总数（它不能是 1，我计算每一行！）

ref 和 alt 都可以在每个可能的排列中具有这些值 (A、T、C、G)，AA、AT、TA、TC、CT 等。

我的代码中缺少什么？

如果您想了解更多信息，请告诉我

变量示例：

chr pos ref alt id_disease
chr1 152 A   C    15
chr3 487 T   T    74

这是我的查询的输出：

pos          chr    refalt  lvl15   tot_var freq
124338543   chr11   G A      69      1     0.000000677833751782702767
124338595   chr11   C T      28      1      0.000000677833751782702767
124361862   chr11   C .      53      1     0.000000677833751782702767
124361899   chr11   T A      20      1     0.000000677833751782702767

Answer 1

根据您提供的信息

SELECT DISTINCT chr, pos, 
upper(ref||' '||alt) AS refalt, id_disease AS lvl15, 
SUM(CASE WHEN (ref == 'A' AND alt == 'C')THEN 1 ELSE 0 END)/COUNT(*) AS 'freq', 
COUNT(*) AS 'tot_var' 
FROM varianti

我仍然不确定 'tot_var' 是什么。获取实际数据样本以及该数据样本本身的预期输出会很有用。

编辑 1：获取数据集中每对夫妇的频率

SELECT DISTINCT upper(ref||' '||alt) AS refalt, 
COUNT(chr)/COUNT(*) AS 'freq' 
FROM varianti 
GROUP BY refalt

编辑 2：根据要求更新了查询

SELECT varianti.chr, varianti.pos, 
upper(varianti.ref||' '||varianti.alt) AS refalt, varianti.id_disease AS lvl15, COUNT(*) AS 'tot_var', 
FROM varianti
JOIN 
( SELECT DISTINCT upper(ref||' '||alt) AS refalt, 
  COUNT(chr)/COUNT(*) AS 'freq' 
  FROM varianti 
  GROUP BY refalt 
) refalt_table ON refalt_table.refalt = varianti.refalt

编辑 3：根据错误更新了查询

SELECT chr, pos, upper(ref||' '||alt) as refalt, id_disease AS lvl15, refalt_table.freq as 'freq', (SELECT COUNT(*) FROM varianti tot where tot.pos = v.pos) as 'tot_var'
FROM varianti v
LEFT JOIN 
( SELECT DISTINCT UPPER(ref) as 'ref',UPPER(alt) as 'alt', 
  COUNT(pos)/(SELECT COUNT(*) FROM varianti vcount) AS 'freq' 
  FROM varianti
  GROUP BY ref,alt
) refalt_table ON refalt_table.ref = v.ref and refalt_table.alt = v.alt

计算每一行的频率

Calculate frequency for each row

sql

postgresql-9.5