Filtering rows based on column values in spark dataframe scala

Question

I have a dataframe(spark):

id  value 
3     0
3     1
3     0
4     1
4     0
4     0

I want to create a new dataframe:

3 0         


        

Answer 1

One way is to use monotonically_increasing_id() and a self-join:

val data = Seq((3,0),(3,1),(3,0),(4,1),(4,0),(4,0)).toDF("id", "value")
data.show
+---+-----+
| id|value|
+---+-----+
|  3|    0|
|  3|    1|
|  3|    0|
|  4|    1|
|  4|    0|
|  4|    0|
+---+-----+

Now we generate a column named idx with an increasing Long:

val dataWithIndex = data.withColumn("idx", monotonically_increasing_id())
// dataWithIndex.cache()

Now we get the min(idx) for each id where value = 1:

val minIdx = dataWithIndex
               .filter($"value" === 1)
               .groupBy($"id")
               .agg(min($"idx"))
               .toDF("r_id", "min_idx")

Now we join the min(idx) back to the original DataFrame:

dataWithIndex.join(
  minIdx,
  ($"r_id" === $"id") && ($"idx" <= $"min_idx")
).select($"id", $"value").show
+---+-----+
| id|value|
+---+-----+
|  3|    0|
|  3|    1|
|  4|    1|
+---+-----+

Note: monotonically_increasing_id() generates its value based on the partition of the row. This value may change each time dataWithIndex is re-evaluated. In my code above, because of lazy evaluation, it's only when I call the final show that monotonically_increasing_id() is evaluated.

If you want to force the value to stay the same, for example so you can use show to evaluate the above step-by-step, uncomment this line above:

//  dataWithIndex.cache()