如何在BigQuery中的标准SQL中实现RATIO

本文介绍了如何在BigQuery中的标准SQL中实现RATIO_TO_REPORT（）？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个使用RATIO_TO_REPORT（）的遗留SQL查询 - 它不使用开放式访问表，但这是它的样子：

我试图迁移从传统SQL到标准SQL（在使用BigQuery之前从未使用SQL），所以提示将非常感谢！只需直接计算比例即可：
 
解决方案
  SELECT Mutation_AA， Gene_name， CaseCount，（CaseCount / SUM（CaseCount）OVER（PARTITION BY Gene_name））AS比率。 。 。   
您不需要子查询：  
  SELECT Mutation_AA，Gene_name， COUNT（DISTINCT ID_tumour，50000）AS CaseCount， COUNT（DISTINCT ID_tumour，50000）/ SUM（COUNT（DISTINCT ID_tumour，50000） ））OVER（PARTITION BY Gene_Name）作为比例 FROM [isb-cgc：COSMIC.grch38_v79]  GROUP BY Mutation_AA，Gene_name;   
 
I have a legacy SQL query that uses RATIO_TO_REPORT() -- it doesn't use open-access tables, but this is what it looks like:
SELECT
  Mutation_AA,
  Gene_name,
  CaseCount,
  RATIO_TO_REPORT(CaseCount) OVER (PARTITION BY Gene_name) AS ratio
FROM (
  SELECT
    COUNT(DISTINCT ID_tumour, 50000) AS CaseCount,
    Mutation_AA,
    Gene_name
  FROM
    [isb-cgc:COSMIC.grch38_v79]
  GROUP BY
    Mutation_AA,
    Gene_name )
I'm trying to migrate from legacy SQL to standard SQL (never having used SQL prior to using BigQuery), so tips would be much appreciated!  thx
 解决方案 
Just directly calculate the ratio:
SELECT Mutation_AA,
       Gene_name,
       CaseCount,
       (CaseCount / SUM(CaseCount) OVER (PARTITION BY Gene_name)) AS ratio
. . .
You don't need the subquery:
SELECT Mutation_AA, Gene_name,
       COUNT(DISTINCT ID_tumour, 50000) AS CaseCount,
       COUNT(DISTINCT ID_tumour, 50000) / SUM(COUNT(DISTINCT ID_tumour, 50000)) OVER (PARTITION BY Gene_Name) as ratio
FROM [isb-cgc:COSMIC.grch38_v79]
GROUP BY Mutation_AA, Gene_name ;