如何识别Hive中重复出现的字符串列？

Question

我在Hive中有这样的观点：

id        sequencenumber          appname
242539622              1          A
242539622              2          A
242539622              3          A
242539622              4          B
242539622              5          B
242539622              6          C
242539622              7          D
242539622              8          D
242539622              9          D
242539622             10          B
242539622             11          B
242539622             12          D
242539622             13          D
242539622             14          F 

我希望每个人都有以下观点：

id        sequencenumber          appname    appname_c
242539622              1          A             A
242539622              2          A             A
242539622              3          A             A
242539622              4          B             B_1
242539622              5          B             B_1
242539622              6          C             C
242539622              7          D             D_1
242539622              8          D             D_1
242539622              9          D             D_1
242539622             10          B             B_2
242539622             11          B             B_2
242539622             12          D             D_2
242539622             13          D             D_2
242539622             14          F             F 

或者与此接近的任何东西，可以识别序列中给定事件的重新出现。

我的最终目标是计算在每组事件中花费的时间（或者如果您希望在马尔可夫建模的上下文中说明），考虑是否存在任何环回。例如，上述示例中在B_1中花费的时间可以与B_2进行非常比较。

已经在Hive（link）中搜索了窗口函数，但我认为他们不能像R / Python那样进行行式比较。

Answer 1

使用Hive窗口函数的解决方案。我用你的数据测试它，删除your_table CTE并改用你的表。结果如预期。

with your_table as (--remove this CTE, use your table instead
select stack(14,
'242539622', 1,'A',
'242539622', 2,'A',
'242539622', 3,'A',
'242539622', 4,'B',
'242539622', 5,'B',
'242539622', 6,'C',
'242539622', 7,'D',
'242539622', 8,'D',
'242539622', 9,'D',
'242539622',10,'B',
'242539622',11,'B',
'242539622',12,'D',
'242539622',13,'D',
'242539622',14,'F'
) as (id,sequencenumber,appname)
) --remove this CTE, use your table instead

select id,sequencenumber,appname, 
       case when sum(new_grp_flag) over(partition by id, group_name) = 1 then appname --only one group of consequent runs exists (like A)
            else        
            nvl(concat(group_name, '_', 
                       sum(new_grp_flag) over(partition by id, group_name order by sequencenumber) --rolling sum of new_group_flag
                       ),appname) 
        end appname_c       
from
(       

select id,sequencenumber,appname,
       case when appname=prev_appname or appname=next_appname then appname end group_name, --identify group of the same app
       case when appname<>prev_appname or prev_appname is null then 1 end new_grp_flag     --one 1 per each group
from       
(
select id,sequencenumber,appname,
       lag(appname)  over(partition by id order by sequencenumber) prev_appname, --need these columns
       lead(appname) over(partition by id order by sequencenumber) next_appname  --to identify groups of records w same app
from your_table --replace with your table
)s
)s
order by id,sequencenumber
;

结果：

OK
id        sequencenumber     appname    appname_c
242539622       1       A       A
242539622       2       A       A
242539622       3       A       A
242539622       4       B       B_1
242539622       5       B       B_1
242539622       6       C       C
242539622       7       D       D_1
242539622       8       D       D_1
242539622       9       D       D_1
242539622       10      B       B_2
242539622       11      B       B_2
242539622       12      D       D_2
242539622       13      D       D_2
242539622       14      F       F
Time taken: 232.319 seconds, Fetched: 14 row(s)

Answer 2

您需要执行2个窗口函数才能实现该结果。

使用pyspark并假设df是您的数据帧：

from pyspark.sql import functions as F, Window

df.withColumn(
    "fg",
    F.lag("appname").over(Window.partitionBy("id").orderBy("sequencenumber)
).withColumn(
    "fg",
    F.when(
        F.col("fg")==F.col("id"),
        0
    ).otherwise(1)
).withColumn(
    "fg",
    F.sum("fg").over(Window.partitionBy("id", "appname"))
).show()