计算每一行的频率

Question

我试图计算每行中我的元素的频率，我将解释： 我从包含诸如“pos，chr，ref，alt，id_disease”等元素的表中进行选择。 。

从这些我将不得不提取我REF频率，ALT即：

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

与此查询我得到几乎接近我的目标，其实id不计算正确的频率，它返回总是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]*' 

所有我想要的是获取这样的数据：

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

tot_var =计数我所需要的行总数（它不能是1我计算每一行！）

ref和alt都可以有这些值（A， T，C，G）在每一个可能的排列，AA，AT，TA，TC，CT等。

我在代码中丢失了什么？

告诉我，如果你想了解更多的相关信息varianti的

---

例子：

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

这里是我的查询的输出：

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

根据这些信息，您所提供

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 

我还不确定'tot_var'是什么。获取实际的数据样本以及该数据样本本身的预期输出将非常有用。

编辑1：要获得每对夫妇的频率数据集中

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

编辑2：基于需求

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更新查询：更新查询基于错误

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