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我的 Redshift 表中有 BIGINT 列，我想要一个查询：

  1. 计算该列所有行中二进制值的每个位位置出现值“1”的次数
  2. 将以一种我能够获取x最高位位置的方式显示它。

例如（我已经将整数值写为二进制以简化示例）：

column
--------
11011110  = 222
00000000  = 0
11111100  = 252
00011000  = 24
11111100  = 252
00011000  = 24
11000010  = 194

76543210 <- bit_position

将返回如下表：

bit_position   count
0              0
1              2
2              3
3              5
4              5
5              2
6              4
7              4

在这种情况下，我将能够获得前五位 bit_position：（3,4,6,7,2）

注意：我的列可能最多有 64 个位位置。

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1 回答 1

3

您可以使用按位 AND&来检查每个位置。

这是一个跨行的示例：

SELECT SUM(CASE WHEN bit_col & 64 > 0 THEN 1 ELSE 0 END) "1000000"
     , SUM(CASE WHEN bit_col & 32 > 0 THEN 1 ELSE 0 END) "0100000"
     , SUM(CASE WHEN bit_col & 16 > 0 THEN 1 ELSE 0 END) "0010000"
     , SUM(CASE WHEN bit_col & 8 > 0 THEN 1 ELSE 0 END)  "0001000"
     , SUM(CASE WHEN bit_col & 4 > 0 THEN 1 ELSE 0 END)  "0000100"
     , SUM(CASE WHEN bit_col & 2 > 0 THEN 1 ELSE 0 END)  "0000010"
     , SUM(CASE WHEN bit_col & 1 > 0 THEN 1 ELSE 0 END)  "0000001"
FROM my_table
;
 1000000 | 0100000 | 0010000 | 0001000 | 0000100 | 0000010 | 0000001
---------+---------+---------+---------+---------+---------+---------
      11 |       8 |      11 |      13 |      11 |       9 |       8

要将结果放在单个列中，您需要使用联合：

          SELECT 1 AS "col", SUM(CASE WHEN bit_col & 64 > 0 THEN 1 ELSE 0 END) AS bit_count FROM my_table
UNION ALL SELECT 2 AS "col", SUM(CASE WHEN bit_col & 32 > 0 THEN 1 ELSE 0 END) AS bit_count FROM my_table
UNION ALL SELECT 3 AS "col", SUM(CASE WHEN bit_col & 16 > 0 THEN 1 ELSE 0 END) AS bit_count FROM my_table
UNION ALL SELECT 4 AS "col", SUM(CASE WHEN bit_col &  8 > 0 THEN 1 ELSE 0 END) AS bit_count FROM my_table
UNION ALL SELECT 5 AS "col", SUM(CASE WHEN bit_col &  4 > 0 THEN 1 ELSE 0 END) AS bit_count FROM my_table
UNION ALL SELECT 6 AS "col", SUM(CASE WHEN bit_col &  2 > 0 THEN 1 ELSE 0 END) AS bit_count FROM my_table
UNION ALL SELECT 7 AS "col", SUM(CASE WHEN bit_col &  1 > 0 THEN 1 ELSE 0 END) AS bit_count FROM my_table
ORDER BY bit_count DESC
;
 position | bit_count
----------+-----------
        6 |         6
        7 |         6
        4 |         4
        5 |         4
        2 |         0
        3 |         0
        1 |         0

http://docs.aws.amazon.com/redshift/latest/dg/r_OPERATOR_SYMBOLS.html

编辑：如果你想要更动态的东西，你需要考虑使用 UDF。您可以从我的f_bitwise_to_stringUDF 作为模板开始，然后从那里添加您需要的内容。https://github.com/awslabs/amazon-redshift-udfs/blob/master/scalar-udfs/f_bitwise_to_string.sql

于 2017-07-26T20:36:11.227 回答