获取组内特定值之前的所有行和之后的 n 行

Question

我正在寻找执行以下操作的简单 R/SQL 代码。在特定公司中，它会在 VALUE 列中查找 1 的第一个实例，然后

1. 提取之前的所有行（VALUE 列中的所有行都为 0）和
2. 只要这两行的 VALUE 为 0，则正好是两行。

代码为所有公司执行此操作。

所以这张桌子....

----------------------
| FIRM | YEAR | VALUE |
----------------------
|  A   | 2007 |  0   |
----------------------
|  A   | 2008 |  0   |
----------------------
|  A   | 2009 |  0   |
----------------------
|  A   | 2010 |  1   |
----------------------
|  A   | 2011 |  0   |
----------------------
|  A   | 2012 |  0   |
----------------------
|  B   | 2009 |  0   |
----------------------
|  B   | 2010 |  1   |
----------------------
|  B   | 2011 |  0   |
----------------------
|  C   | 2010 |  0   |
----------------------
|  C   | 2011 |  1   |
----------------------
|  C   | 2012 |  1   |
----------------------

看起来像这样......

--------------------------
| FIRM  | YEAR  | VALUE  |
--------------------------
|  A   | 2007 |  0   |
----------------------
|  A   | 2008 |  0   |
----------------------
|  A   | 2009 |  0   |
----------------------
|  A   | 2010 |  1   |
----------------------
|  A   | 2011 |  0   |
----------------------
|  A   | 2012 |  0   |
----------------------

非常感谢您的帮助。谢谢。

Answer 1

使用 R，您可以创建一个函数，该函数将返回要选择的行号。

get_rows <- function(VALUE) {
   ind <- which(VALUE == 1)[1]
   if ((ind + 2) <= length(VALUE) && all(VALUE[c(ind + 1,ind + 2)] == 0))
     sort(c(which(VALUE[seq_len(ind + 2)] == 0), ind))
   else 0
}

并将其应用于每个FIRM。

library(dplyr)
df %>% group_by(FIRM) %>% slice(get_rows(VALUE))

#  FIRM   YEAR VALUE
#  <fct> <int> <int>
#1 A      2007     0
#2 A      2008     0
#3 A      2009     0
#4 A      2010     1
#5 A      2011     0
#6 A      2012     0

数据

df <- structure(list(FIRM = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 
2L, 2L, 3L, 3L, 3L), .Label = c("A", "B", "C"), class = "factor"), 
YEAR = c(2007L, 2008L, 2009L, 2010L, 2011L, 2012L, 2009L, 
2010L, 2011L, 2010L, 2011L, 2012L), VALUE = c(0L, 0L, 0L, 
1L, 0L, 0L, 0L, 1L, 0L, 0L, 1L, 1L)), class = "data.frame",row.names = c(NA, -12L))

Answer 2

您可以计算最小年份，然后使用此信息：

with t as (
      select firm, min(year) as min_year_1
      from tab t
      where value = 1
      group by firm
     )
select t.*
from (select t.*,
             lag(value) over (partition by firm order by year) as prev_value,
             lead(value) over (partition by firm order by year) as next_value
      from tab t
     ) t join
     tt
     on tt.firm = t.firm
where t.year <= tt.min_year or
      (t.year = tt.min_year + 1 and
       t.value = 0 and
       t.next_value = 0
      ) or
      (t.year = tt.min_year + 2 and
       t.value = 0 and
       t.prev_value = 0
      );

0 之后两行的最后一个条件相当棘手。

这假设年份是连续的，没有间隔，这与您问题中的数据一致。

编辑：

您可以只使用窗口函数来做到这一点：

select t.*
from (select t.*,
             count(*) over (partition by firm, running_value) as cnt,
             row_number() over (partition by firm, running_value) as seqnum
      from (select t.*,
                   sum(value) over (partition by firm order by year) as running_value
            from tab t
           ) t
      ) t
where running_value = 0 or
      (running_value = 1 and seqnum = 1) or  -- first "1"
      (running_value = 1 and seqnum <= 3 and
       cnt >= 3);