使用GROUP BY计算组的百分比

Question

我正在对数据集执行GROUP BY和COUNT(*),我想计算每个组在总数上的百分比.

例如,在这个查询中,我想知道每个状态的count()代表总数(从publicdata:samples.natality中选择count()):

SELECT state, count(*)
FROM [publicdata:samples.natality]
GROUP by state

在SQL中有几种方法可以做到这一点,但我还没有找到在Bigquery中做到这一点的方法,有人知道吗？

谢谢!

Answer 1

检查ratio_to_report,这是最近公布的窗口函数之一:

SELECT state, ratio * 100 AS percent FROM (
 SELECT state, count(*) AS total, RATIO_TO_REPORT(total) OVER() AS ratio
 FROM [publicdata:samples.natality]
 GROUP by state
)

state   percent
AL      1.4201828131159113   
AK      0.23521048665998198  
AZ      1.3332896746620975   
AR      0.7709591206172346   
CA      10.008298605982642

Answer 2

修改 Felipe 对标准 SQL BigQuery 方言而不是 Legacy SQL 方言的回答如下所示：

select state, 100*(state_count / total) as pct
from (
  SELECT state, count(*) AS state_count, sum(count(*)) OVER() AS total
  FROM `bigquery-public-data.samples.natality` 
  GROUP by state
) s

标准 SQL BigQuery 聚合分析函数（又名“窗口函数”）的文档在这里：https : //cloud.google.com/bigquery/docs/reference/standard-sql/analytic-function-concepts

Answer 3

您可以使用窗口函数按组获取总数的百分比，而不需要子查询（改进 evan_b 的解决方案）：

SELECT 
   state
   ,count(*) / (sum(count(*)) OVER()) as pct
FROM  
   `bigquery-public-data.samples.natality` 
GROUP BY 
   state

Answer 4

您可以使用虚拟值作为键对总数进行自连接。例如：

SELECT
  t1.state AS state,
  t1.cnt AS cnt,
  100 * t1.cnt / t2.total as percent
FROM (
  SELECT
    state,
    COUNT(*) AS cnt,
    1 AS key
  FROM
    [publicdata:samples.natality]
  WHERE state is not null
  GROUP BY
    state) AS t1
JOIN (
  SELECT
    COUNT(*) AS total,
    1 AS key
  FROM
    [publicdata:samples.natality]) AS t2
ON t1.key = t2.key
ORDER BY percent DESC