Question

我的桌子如下：

文档：

 *= require_tree .
 *= require_self
 */

@import "bootstrap-sprockets";
@import "bootstrap";

(And some custom styling here)

我想获得最近30天的得分中位数。

我当前的查询是：

+-----+-------------+-------------------------+
| dId | score       | datetime                |
+-----+-------------+-------------------------+
| A   | 100.0       | 2019-03-08 16:17:34.043 |
| B   | 80.5        | 2019-02-15 16:17:34.043 |
| C   | 70.1        | 2019-03-08 16:17:34.043 |
+-----+-------------+-------------------------+

我该怎么做？

Answer 1

以下是用于BigQuery标准SQL

#standardSQL
CREATE TEMP FUNCTION Median(arr ARRAY<INT64>) AS (
  IF(MOD(ARRAY_LENGTH(arr), 2) = 1, arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2))],
      (arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2) - 1)] + arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2))]) / 2)
);
SELECT Median(ARRAY_AGG(score ORDER BY score)) Median
FROM `project.dataset.document`
WHERE DATE(dt) >= DATE_SUB(CURRENT_DATE(), INTERVAL 30 DAY)

您可以使用问题中的示例数据来进行推文和上面的游戏，如下例所示

#standardSQL
CREATE TEMP FUNCTION Median(arr ARRAY<INT64>) AS (
  IF(MOD(ARRAY_LENGTH(arr), 2) = 1, arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2))],
      (arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2) - 1)] + arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2))]) / 2)
);
WITH `project.dataset.document` AS (
  SELECT 'A' dId, 100 score, DATETIME '2019-03-08 16:17:34.043' dt UNION ALL 
  SELECT 'B', 80, '2019-02-15 16:17:34.043' UNION ALL 
  SELECT 'C', 70, '2019-03-08 16:17:34.043'     
)
SELECT Median(ARRAY_AGG(score ORDER BY score)) Median
FROM `project.dataset.document`
WHERE DATE(dt) >= DATE_SUB(CURRENT_DATE(), INTERVAL 30 DAY)

有结果

Row Median   
1   85.0

请注意，您可以使用CREATE TEMP FUNCTION Median(arr ANY TYPE) AS (...使其更通用，并接受任何类型的序列

更新

以下示例应适用于NUMERIC

#standardSQL
CREATE TEMP FUNCTION Median(arr ANY TYPE) AS (
  IF(MOD(ARRAY_LENGTH(arr), 2) = 1, arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2))],
      (arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2) - 1)] + arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2))]) / 2)
);
WITH `project.dataset.document` AS (
  SELECT 'A' dId, CAST(100.0 AS numeric) score, DATETIME '2019-03-08 16:17:34.043' datetime UNION ALL 
  SELECT 'B', 80.5, '2019-02-15 16:17:34.043' UNION ALL 
  SELECT 'C', 70.1, '2019-03-08 16:17:34.043'     
)
SELECT Median(ARRAY_AGG(CAST(score AS FLOAT64) ORDER BY score)) Median
FROM `project.dataset.document`
WHERE DATE(datetime) >= DATE_SUB(CURRENT_DATE(), INTERVAL 30 DAY)

更新

好的。发现内部错误的原因-这是由于按数字值排序
因此，最终版本是：

#standardSQL
CREATE TEMP FUNCTION Median(arr ANY TYPE) AS (
  IF(MOD(ARRAY_LENGTH(arr), 2) = 1, arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2))],
      (arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2) - 1)] + arr[OFFSET(DIV(ARRAY_LENGTH(arr), 2))]) / 2)
);
WITH `project.dataset.document` AS (
  SELECT 'A' dId, CAST(100.0 AS numeric) score, DATETIME '2019-03-08 16:17:34.043' datetime UNION ALL 
  SELECT 'B', 80.5, '2019-02-15 16:17:34.043' UNION ALL 
  SELECT 'C', 70.1, '2019-03-08 16:17:34.043'     
)
SELECT Median(ARRAY_AGG(score ORDER BY CAST(score AS FLOAT64))) Median
FROM `project.dataset.document`
WHERE DATE(datetime) >= DATE_SUB(CURRENT_DATE(), INTERVAL 30 DAY)

Answer 2

您可以使用PERCENTILE_CONT进行操作。只需使用0.5 PERCENTILE_CONT子句在上个月筛选的所有分数中找到WHERE。如果您想以一种原始方式使用它。这是查询...

SELECT
  PERCENTILE_CONT(score, 0.5) OVER() AS Median
FROM
  `document` d
WHERE
   d.datetime >= TIMESTAMP_SUB(CURRENT_TIMESTAMP(), INTERVAL 30 day)

Bigquery Standard SQL-最近30天获得中位数

2 个答案: