Question

GBQ参考页面上的所有排名公式都假设有多行，一列。对于下面的内容，我试图找出Algo1到Algo5s的排名。

theTable：

cid algo_1  algo_2  algo_3  algo_4  algo_5   
1     4.31    4.15    4.33    4.35    4.35
2     1.31    4.15    4.33    4.34    3.35

所以，结果是

cid algo_1  algo_2  algo_3  algo_4  algo_5  algo_1_rank algo_2_rank algo_3_rank algo_4_rank algo_5_rank  
1     4.31     4.15   4.33    4.35    4.35            4           5           3           1           1  
2     1.31   4.15     4.33    4.34    3.35            5           3           2           1           4

P.S。实际上，我在SO之外被问到这个问题，所以决定在这里分享

Answer 1

选项1 - 标准SQL

WITH theTable AS (
  SELECT 1 AS cid, 4.31 AS algo_1, 4.15 AS algo_2, 4.33 AS algo_3, 4.35 AS algo_4, 4.35 AS algo_5 UNION ALL
  SELECT 2 AS cid, 1.31 AS algo_1, 4.15 AS algo_2, 4.33 AS algo_3, 4.35 AS algo_4, 3.35 AS algo_5
),
tempTable AS (
  SELECT 
    cid, 
    alg.k AS algo, 
    alg.v AS value, 
    RANK() OVER(PARTITION BY cid ORDER BY v DESC) AS rnk
  FROM theTable, UNNEST(
    ARRAY[STRUCT<k STRING, v FLOAT64>("algo_1", algo_1), STRUCT("algo_2", algo_2), 
          STRUCT("algo_3", algo_3), STRUCT("algo_4", algo_4), STRUCT("algo_5", algo_5)]
  ) AS alg
)
SELECT 
  cid,
  MAX(IF(algo = "algo_1", value, NULL)) AS algo_1,
  MAX(IF(algo = "algo_2", value, NULL)) AS algo_2,
  MAX(IF(algo = "algo_3", value, NULL)) AS algo_3,
  MAX(IF(algo = "algo_4", value, NULL)) AS algo_4,
  MAX(IF(algo = "algo_5", value, NULL)) AS algo_5,
  MAX(IF(algo = "algo_1", rnk, NULL)) AS algo_1_rank,
  MAX(IF(algo = "algo_2", rnk, NULL)) AS algo_2_rank,
  MAX(IF(algo = "algo_3", rnk, NULL)) AS algo_3_rank,
  MAX(IF(algo = "algo_4", rnk, NULL)) AS algo_4_rank,
  MAX(IF(algo = "algo_5", rnk, NULL)) AS algo_5_rank
FROM tempTable
GROUP BY cid

选项2 - 旧版SQL

SELECT
  cid,
  MAX(IF(num = 1, value, NULL)) AS algo_1,
  MAX(IF(num = 2, value, NULL)) AS algo_2,
  MAX(IF(num = 3, value, NULL)) AS algo_3,
  MAX(IF(num = 4, value, NULL)) AS algo_4,
  MAX(IF(num = 5, value, NULL)) AS algo_5,
  MAX(IF(num = 1, rnk, NULL)) AS algo_1_rank,
  MAX(IF(num = 2, rnk, NULL)) AS algo_2_rank,
  MAX(IF(num = 3, rnk, NULL)) AS algo_3_rank,
  MAX(IF(num = 4, rnk, NULL)) AS algo_4_rank,
  MAX(IF(num = 5, rnk, NULL)) AS algo_5_rank  
FROM (
  SELECT 
    cid, num, 
    CASE 
      WHEN num = 1 THEN algo_1
      WHEN num = 2 THEN algo_2
      WHEN num = 3 THEN algo_3
      WHEN num = 4 THEN algo_4
     WHEN num = 5 THEN algo_5
    END AS value,
    RANK() OVER(PARTITION BY cid ORDER BY value DESC) AS rnk
  FROM (
    SELECT * FROM 
      (SELECT 1 AS cid, 4.31 AS algo_1, 4.15 AS algo_2, 4.33 AS algo_3, 4.35 AS algo_4, 4.35 AS algo_5),
      (SELECT 2 AS cid, 1.31 AS algo_1, 4.15 AS algo_2, 4.33 AS algo_3, 4.35 AS algo_4, 3.35 AS algo_5)
  ) AS theTable
  CROSS JOIN (
    SELECT INTEGER(SPLIT("1,2,3,4,5")) AS num FROM (SELECT 1)
  ) AS nums
)
GROUP BY cid

选项3 - 带标量UDF的标准SQL

CREATE TEMPORARY FUNCTION myRank(a float64, b float64, c float64, d float64, e float64)
RETURNS ARRAY<int64>
LANGUAGE js AS """
  var arr = [a, b, c, d, e];
  var sorted = arr.slice().sort(function(a,b){return b-a})
  var ranks = arr.slice().map(function(v){ return sorted.indexOf(v)+1 });
  return ranks
  """;

WITH theTable AS (
  SELECT 1 AS cid, 4.31 AS algo_1, 4.15 AS algo_2, 4.33 AS algo_3, 4.35 AS algo_4, 4.35 AS algo_5 UNION ALL
  SELECT 2 AS cid, 1.31 AS algo_1, 4.15 AS algo_2, 4.33 AS algo_3, 4.35 AS algo_4, 3.35 AS algo_5
), 
tempTable AS (
  SELECT *, myRank(algo_1, algo_2, algo_3, algo_4, algo_5) AS ranks
  FROM theTable
)
SELECT 
  cid, algo_1, algo_2, algo_3, algo_4, algo_5,
  ranks[ORDINAL(1)] AS algo_1_rank, 
  ranks[ORDINAL(2)] AS algo_2_rank, 
  ranks[ORDINAL(3)] AS algo_3_rank, 
  ranks[ORDINAL(4)] AS algo_4_rank, 
  ranks[ORDINAL(5)] AS algo_5_rank
FROM tempTable

选项4 - 提前处理排名

通常，作为theTable的架构表是从多行生成的，每行只有一个测试条目（cid，算法，值如下面的theOriginalData）
在这一点上实际排名是最有意义的

SELECT 
  cid,
  MAX(IF(algo = "algo_1", value, NULL)) AS algo_1,
  MAX(IF(algo = "algo_2", value, NULL)) AS algo_2,
  MAX(IF(algo = "algo_3", value, NULL)) AS algo_3,
  MAX(IF(algo = "algo_4", value, NULL)) AS algo_4,
  MAX(IF(algo = "algo_5", value, NULL)) AS algo_5,

  MAX(IF(algo = "algo_1", rnk, NULL)) AS algo_1_rank,
  MAX(IF(algo = "algo_2", rnk, NULL)) AS algo_2_rank,
  MAX(IF(algo = "algo_3", rnk, NULL)) AS algo_3_rank,
  MAX(IF(algo = "algo_4", rnk, NULL)) AS algo_4_rank,
  MAX(IF(algo = "algo_5", rnk, NULL)) AS algo_5_rank,

FROM (
  SELECT
    cid, algo, value,
    RANK() OVER(PARTITION BY cid ORDER BY value DESC) AS rnk
  FROM (
    SELECT * FROM 
      (SELECT 1 AS cid, "algo_1" AS algo, 4.31 AS value),
      (SELECT 1 AS cid, "algo_2" AS algo, 4.15 AS value),
      (SELECT 1 AS cid, "algo_3" AS algo, 4.33 AS value),
      (SELECT 1 AS cid, "algo_4" AS algo, 4.35 AS value),
      (SELECT 1 AS cid, "algo_5" AS algo, 4.35 AS value),
      (SELECT 2 AS cid, "algo_1" AS algo, 1.31 AS value),
      (SELECT 2 AS cid, "algo_2" AS algo, 4.15 AS value),
      (SELECT 2 AS cid, "algo_3" AS algo, 4.33 AS value),
      (SELECT 2 AS cid, "algo_4" AS algo, 4.34 AS value),
      (SELECT 2 AS cid, "algo_5" AS algo, 3.35 AS value)
  ) AS theOriginalData
)
GROUP BY cid

注意：如果由我决定 - 我会选择选项＃4 以防万一由于某种原因（例如表格已经存在））我会选择选项＃3 ，因为它看起来最优雅

Answer 2

米哈伊尔为你的问题提供了一些很好的解决方案 - 如果其中一个人为你服务，请将他的回答标记为已被接受。如果你最终使用它，我想给出＃3的替代形式;你可以用SQL UDF来表达转型：

batches

如何在一行中排列多个列？

2 个答案: