Question

我正在将R代码转换为SQL。

我的R代码如下：

temp <- questions %>%
  select(UserId, ResultIndicator, ThemeId) %>%
  filter(UserId == 72) %>%
  group_by(ThemeId, ResultIndicator) %>%
  arrange(desc(ResultIndicator)) %>%
  summarise(Nominal = n()) %>% 
  mutate(Percent = Nominal/sum(Nominal)) %>%
  mutate(Percent = round(Percent, 3) * 100)  %>%  
  mutate(diff = Percent - lag(Percent, default = first(Percent)))

输出如下：

structure(list(ThemeId = c(11L, 11L, 12L, 12L, 13L, 19L), ResultIndicator = c("Correct", 
"Wrong", "Correct", "Wrong", "Correct", "Wrong"), Nominal = c(34L, 
4L, 25L, 2L, 10L, 1L), Percent = c(89.5, 10.5, 92.6, 7.4, 100, 
100), diff = c(0, -79, 0, -85.2, 0, 0)), class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"), row.names = c(NA, -6L), vars = "ThemeId", labels = structure(list(
    ThemeId = c(11L, 12L, 13L, 19L)), class = "data.frame", row.names = c(NA, 
-4L), vars = "ThemeId", labels = structure(list(ThemeId = c(11L, 
12L, 13L, 19L, 22L, 33L, 35L, 38L, 48L, 56L, 59L, 62L, 71L, 77L
)), row.names = c(NA, -14L), class = "data.frame", vars = "ThemeId", drop = TRUE), indices = list(
    0:1, 2:3, 4L, 5L, 6:7, 8:9, 10:11, 12L, 13:14, 15:16, 17:18, 
    19:20, 21:22, 23:24), drop = TRUE, group_sizes = c(2L, 2L, 
1L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), biggest_group_size = 2L), indices = list(
    0:1, 2:3, 4L, 5L), drop = TRUE, group_sizes = c(2L, 2L, 1L, 
1L), biggest_group_size = 2L)

对于非R用户，以上翻译为

  ThemeId ResultIndicator Nominal Percent  diff
1      11         Correct      34    89.5   0.0
2      11           Wrong       4    10.5 -79.0
3      12         Correct      25    92.6   0.0
4      12           Wrong       2     7.4 -85.2
5      13         Correct      10   100.0   0.0
6      19           Wrong       1   100.0   0.0

我在SQL中的尝试是：

SELECT Count(Id) as Nominal, ResultIndicator, ThemeId
FROM LogUserQuestions
WHERE UserId = 72
GROUP BY ThemeId, ResultIndicator
ORDER BY ThemeId

但是我不知道如何计算时滞。我尝试过：

(Nominal - lag(Nominal) over (partition by [not sure] order by [not sure])) as diff

但是我不能使用名义，因为它是后来创建的。

有任何提示吗？

Answer 1

我认为是这样的：

SELECT ResultIndicator, ThemeId, COUNT(*) as Nominal, 
       COUNT(*) * 1.0 / SUM(COUNT(*)) OVER (),
       COUNT(*) - LAG(COUNT(*)) OVER (ORDER BY ResultIndicator) as diff
FROM LogUserQuestions
WHERE UserId = 72
GROUP BY ThemeId, ResultIndicator
ORDER BY ResultIndicator DESC;

Answer 2

实际上，与您的假设相反，但我不能使用名义，因为它是后来创建的，您可以在汇总查询中使用带有CTE的字段。实际上，您可能想对计算的 Percent 使用多个CTE，该CTE是基于较早的汇总 Nominal 得出的。但是，正如@GordonLinoff所示，所有内容都可以在一个查询中运行，但是可读性可能成为问题。

WITH agg AS (    
   SELECT ThemeId, ResultIndicator, Count(Id) as Nominal, 
   FROM LogUserQuestions
   WHERE UserId = 72
   GROUP BY ThemeId, ResultIndicator    
), pct AS
   SELECT ResultIndicator, ThemeId, Nominal,
          Nominal / SUM(Nominal) OVER (PARTITION BY ThemeId, ResultIndicator) AS [Percent]
   FROM agg
)

SELECT ResultIndicator, ThemeId, Nominal, ROUND([Percent], 0) AS [Percent],
       -- LAG() maintains three arguments: expression, offset, default
       ([Percent] - LAG([Percent], 1, [Percent])
                        OVER (PARTITION BY ThemeId ORDER BY ResultIndicator DESC) as diff
FROM pct

Answer 3

我在SQL表中没有您的数据，所以这有点困难。通常，在这种情况下，我要做的是构建唯一键，将整个结果包装在子查询中，然后从中选择。像这样：

SELECT Nominal, 
  ResultIndicator, 
  ThemeId, 
  (Nominal - lag(Nominal) over (partition by myKey order by myKey)) as diff
FROM
(
  SELECT Count(Id) as Nominal, ResultIndicator, ThemeId, ResultIndicator + CAST(ThemeId as varchar(50)) as myKey
  FROM LogUserQuestions
  WHERE UserId = 72
  GROUP BY ThemeId, ResultIndicator
  ORDER BY ThemeId, ResultIndicator
) sub
order by Nominal

组中行之间的差异

3 个答案: