Pyspark - 根据一行中的条件过滤掉多行

Question

I have a table like so:

--------------------------------------------
| Id  |  Value   | Some Other Columns Here
| 0   |  5       |
| 0   |  4       |
| 0   |  0       |
| 1   |  3       |
| 2   |  1       |
| 2   |  8       |
| 3   |  -4      |
--------------------------------------------

I would like to remove all IDs which have any Value <= 0, so the result would be:

--------------------------------------------
| Id  |  Value   | Some Other Columns Here
| 1   |  3       |
| 2   |  1       |
| 2   |  8       |
--------------------------------------------

I tried doing this by filtering to only rows with Value<=0, selecting the distinct IDs from this, converting that to a list, and then removing any rows in the original table that have an ID in that list using df.filter(~df.Id.isin(mylist))

However, I have a huge amount of data, and this ran out of memory making the list, so I need to come up with a pure pyspark solution.

Answer 1

You can use window functions:

select t.*
from (select t.*, min(value) over (partition by id) as min_value
      from t
     ) t
where min_value > 0

Answer 2

As Gordon mentions, you may need a window for this, here is a pyspark version:

import pyspark.sql.functions as F
from pyspark.sql.window import Window

w = Window.partitionBy("Id")
(df.withColumn("flag",F.when(F.col("Value")<=0,0).otherwise(1))
   .withColumn("Min",F.min("flag").over(w)).filter(F.col("Min")!=0)
   .drop("flag","Min")).show()

---

+---+-----+
| Id|Value|
+---+-----+
|  1|    3|
|  2|    1|
|  2|    8|
+---+-----+

Brief summary of approach taken:

• Set a flag when Value<=0 then 0 else `1
• get min over a partition of id (will return 0 if any of prev cond is met)
• filter only when this Min value is not 0

`