Question

我有这段代码，我想把它写成SQL。有谁知道等效的SQL代码会是什么样子？

lags = range(1, 5)
df = df.assign(**{
    '{}{}'.format('lag', t): df.groupby('article_id').num_views.shift(t) for t in lags
})

更新：

我正在寻找SQL标准方言。这是一个数据集示例（部分前10行）：

  article_id section time   num_views   comments
0   abc111b     A   00:00   15            0
1   abc111b     A   01:00   36            0
2   abc111b     A   02:00   36            0
3   bbbddd222hf A   03:00   41            0
4   bbbddd222hf B   04:00   44            0
5   nnn678www   B   05:00   39            0
6   nnn678www   B   06:00   38            0
7   nnn678www   B   07:00   66            0
8   nnn678www   C   08:00   65            0
9   nnn678www   C   09:00   87            1

Answer 1

您可以使用LAG() function，属于SQL-99 ANSI标准"windowing functions"：

select
  article_id, section, time, num_views, comments,
  lag(num_views, 1, 0) over(partition by article_id order by article_id, time) as lag1,
  lag(num_views, 2, 0) over(partition by article_id order by article_id, time) as lag2,
  lag(num_views, 3, 0) over(partition by article_id order by article_id, time) as lag3,
  lag(num_views, 4, 0) over(partition by article_id order by article_id, time) as lag4
from tab;

Complete and working SQLFiddle example...

PS please be aware that not all RDBMS systems implement "windowing functions"

Pandas的等效SQL [df.groupby（...）[＆＃39; col_name＆＃39;] .shift（1）]

1 个答案: