How to get the last row from DataFrame? - scala

Question

I hava a DataFrame,the DataFrame hava two column 'value' and 'timestamp',,the 'timestmp' is ordered,I want to get the last row of the DataFrame,what should I do?
this is my input:
+-----+---------+
|value|timestamp|
+-----+---------+
| 1| 1|
| 4| 2|
| 3| 3|
| 2| 4|
| 5| 5|
| 7| 6|
| 3| 7|
| 5| 8|
| 4| 9|
| 18| 10|
+-----+---------+
this is my code:
val arr = Array((1,1),(4,2),(3,3),(2,4),(5,5),(7,6),(3,7),(5,8),(4,9),(18,10))
var df=m_sparkCtx.parallelize(arr).toDF("value","timestamp")
this is my expected result:
+-----+---------+
|value|timestamp|
+-----+---------+
| 18| 10|
+-----+---------+

Answer 1

Try this, it works for me.
df.orderBy($"value".desc).show(1)

Answer 2

I would use simply the query that
- orders your table by descending order
- takes 1st value from this order
df.createOrReplaceTempView("table_df")
query_latest_rec = """SELECT * FROM table_df ORDER BY value DESC limit 1"""
latest_rec = self.sqlContext.sql(query_latest_rec)
latest_rec.show()

Answer 3

I'd simply reduce:
df.reduce { (x, y) =>
if (x.getAs[Int]("timestamp") > y.getAs[Int]("timestamp")) x else y
}

Answer 4

The most efficient way is to reduce your DataFrame. This gives you a single row which you can convert back to a DataFrame, but as it contains only 1 record, this does not make much sense.
sparkContext.parallelize(
Seq(
df.reduce {
(a, b) => if (a.getAs[Int]("timestamp") > b.getAs[Int]("timestamp")) a else b
} match {case Row(value:Int,timestamp:Int) => (value,timestamp)}
)
)
.toDF("value","timestamp")
.show
+-----+---------+
|value|timestamp|
+-----+---------+
| 18| 10|
+-----+---------+
Less efficient (as it needs shuffling) although shorter is this solution:
df
.where($"timestamp" === df.groupBy().agg(max($"timestamp")).map(_.getInt(0)).collect.head)

Answer 5

If your timestamp column is unique and is in increasing order then there are following ways to get the last row
println(df.sort($"timestamp", $"timestamp".desc).first())
// Output [1,1]
df.sort($"timestamp", $"timestamp".desc).take(1).foreach(println)
// Output [1,1]
df.where($"timestamp" === df.count()).show
Output:
+-----+---------+
|value|timestamp|
+-----+---------+
| 18| 10|
+-----+---------+
If not create a new column with the index and select the last index as below
val df1 = spark.sqlContext.createDataFrame(
df.rdd.zipWithIndex.map {
case (row, index) => Row.fromSeq(row.toSeq :+ index)
},
StructType(df.schema.fields :+ StructField("index", LongType, false)))
df1.where($"timestamp" === df.count()).drop("index").show
Output:
+-----+---------+
|value|timestamp|
+-----+---------+
| 18| 10|
+-----+---------+

Answer 6

Java:
Dataset<Row> sortDF = inputDF.orderBy(org.apache.spark.sql.functions.col(config.getIncrementingColumn()).desc());
Row row = sortDF.first()

Answer 7

You can also use this function desc: Column desc(String columnName)
df.orderBy(desc("value")).show(1)
which gives same result as
df.orderBy($"value".desc).show(1)