SQL 用于对对话数据进行排序和聚合的排名函数

Question

我正在寻找 rank/aggregate SQL 中的对话数据（特别是 BigQuery）。数据是对话数据，每一行代表一个句子。在下图中，我添加了说话人、句子和 sequence_start 的示例数据。 desired_rank 是目标结果（或类似的东西）。

我相信应该有一个像 rank/lag/first 这样的 window 函数，它应该以编程方式达到所需的排名。

我得到的最接近的最初是：

WITH DATA AS (
SELECT 'Speaker A' as speaker, 'Sentence 1' as sentence, 1 as sentence_start, 1 as desired_rank
UNION ALL SELECT 'Speaker A' as speaker, 'Sentence 2' as sentence, 9 as sentence_start, 1 as desired_rank
UNION ALL SELECT 'Speaker B' as speaker, 'Sentence 3' as sentence, 27 as sentence_start, 2 as desired_rank
UNION ALL SELECT 'Speaker C' as speaker, 'Sentence 4' as sentence, 46 as sentence_start, 3 as desired_rank
UNION ALL SELECT 'Speaker A' as speaker, 'Sentence 5' as sentence, 78 as sentence_start, 4 as desired_rank
)
SELECT speaker, sentence, sentence_start, desired_rank,

FIRST_VALUE(sentence_start)
  OVER (
    PARTITION BY speaker
    ORDER BY sentence_start
    RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW)

FROM DATA
ORDER BY sentence_start

结果的问题是说话者 A 总是排名为 1，而它应该是 4（或类似的东西）。

感谢您的帮助。谢谢！

Answer 1

也许 RANK, DENSE_RANK and ROW_NUMBER functions 之一会有所帮助。用 sentence_start = 9 向 DATA 添加了一行以指出不同之处：

WITH DATA AS (
  SELECT 'Speaker A' as speaker, 'Sentence 1' as sentence, 1 as sentence_start, 1 as desired_rank
  UNION ALL SELECT 'Speaker A' as speaker, 'Sentence 2' as sentence, 9 as sentence_start, 1 as desired_rank
  UNION ALL SELECT 'Speaker A' as speaker, 'Sentence 2' as sentence, 9 as sentence_start, 1 as desired_rank
  UNION ALL SELECT 'Speaker B' as speaker, 'Sentence 3' as sentence, 27 as sentence_start, 2 as desired_rank
  UNION ALL SELECT 'Speaker C' as speaker, 'Sentence 4' as sentence, 46 as sentence_start, 3 as desired_rank
  UNION ALL SELECT 'Speaker A' as speaker, 'Sentence 5' as sentence, 78 as sentence_start, 4 as desired_rank
)
SELECT 
  speaker,
  sentence,
  sentence_start,
  desired_rank,
  RANK() OVER (ORDER BY sentence_start) AS rank,
  DENSE_RANK() OVER (ORDER BY sentence_start) AS dense_rank,
  ROW_NUMBER() OVER (ORDER BY sentence_start) AS row_number,
FROM DATA

Answer 2

想通了。需要加入到下一行以确定更改。添加了说话人 A 说第 5 句和第 6 句的复杂情况。

WITH data AS (
SELECT          'Speaker A' as speaker, 'Sentence 1' as sentence, 1 as sentence_start, 1 as desired_rank
UNION ALL SELECT 'Speaker A' as speaker, 'Sentence 2' as sentence, 9 as sentence_start, 1 as desired_rank
UNION ALL SELECT 'Speaker B' as speaker, 'Sentence 3' as sentence, 27 as sentence_start, 2 as desired_rank
UNION ALL SELECT 'Speaker C' as speaker, 'Sentence 4' as sentence, 46 as sentence_start, 3 as desired_rank
UNION ALL SELECT 'Speaker A' as speaker, 'Sentence 5' as sentence, 78 as sentence_start, 4 as desired_rank
UNION ALL SELECT 'Speaker A' as speaker, 'Sentence 6' as sentence, 90 as sentence_start, 4 as desired_rank
),
data_ranked AS (
SELECT speaker, sentence, sentence_start, desired_rank,
COALESCE(LEAD(sentence_start) OVER (ORDER BY sentence_start asc),9999999999999) AS next_sentence_start
FROM DATA
ORDER BY sentence_start
),
sentence_information AS (
SELECT sentence_information.speaker, sentence_information.sentence, sentence_information.sentence_start, sentence_information.next_sentence_start
  , CASE WHEN sentence_information.speaker <> next_sentence_information.speaker THEN TRUE ELSE FALSE END as next_sentence_speaker_change_indicator
FROM DATA_RANKED as sentence_information
  LEFT OUTER JOIN DATA AS next_sentence_information ON sentence_information.next_sentence_start = next_sentence_information.sentence_start
),
compiled_sentence_information AS (SELECT sentence_information.speaker, sentence_information.sentence, sentence_information.sentence_start, sentence_information.next_sentence_start
, COALESCE(next_sentence_information.next_sentence_speaker_change_indicator, FALSE) as speaker_change_indicator
, CASE WHEN COALESCE(next_sentence_information.next_sentence_speaker_change_indicator, FALSE) THEN 1 ELSE 0 END as speaker_change_number
, SUM(CASE WHEN COALESCE(next_sentence_information.next_sentence_speaker_change_indicator, FALSE) THEN 1 ELSE 0 END) OVER (ORDER BY sentence_information.sentence_start ASC) AS speaker_sentence_rank
, CASE WHEN sentence_information.next_sentence_start = 9999999999999 THEN TRUE ELSE sentence_information.next_sentence_speaker_change_indicator END as final_sentence_in_paragraph
FROM sentence_information 
  LEFT OUTER JOIN sentence_information as next_sentence_information on sentence_information.sentence_start = next_sentence_information.next_sentence_start
),
paragraphs as (
SELECT *, STRING_AGG(sentence, " ") OVER (PARTITION BY speaker_sentence_rank ORDER BY sentence_start) as paragraph
FROM compiled_sentence_information
)
SELECT speaker, paragraph
FROM paragraphs
WHERE final_sentence_in_paragraph = TRUE
ORDER BY sentence_start

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