Spark dataframe get column value into a string variable - scala

Question

I am trying extract column value into a variable so that I can use the value somewhere else in the code. I am trying like the following
val name= test.filter(test("id").equalTo("200")).select("name").col("name")
It returns
name org.apache.spark.sql.Column = name
how to get the value?

Answer 1

The col("name") gives you a column expression. If you want to extract data from column "name" just do the same thing without col("name"):
val names = test.filter(test("id").equalTo("200"))
.select("name")
.collectAsList() // returns a List[Row]
Then for a row you could get name in String by:
val name = row.getString(0)

Answer 2

val maxDate = spark.sql("select max(export_time) as export_time from tier1_spend.cost_gcp_raw").first()
val rowValue = maxDate.get(0)

Answer 3

By this snippet, you can extract all the values in a column into a string.
Modify the snippet with where clauses to get your desired value.
val df = Seq((5, 2), (10, 1)).toDF("A", "B")
val col_val_df = df.select($"A").collect()
val col_val_str = col_val_df.map(x => x.get(0)).mkString(",")
/*
df: org.apache.spark.sql.DataFrame = [A: int, B: int]
col_val_row: Array[org.apache.spark.sql.Row] = Array([5], [10])
col_val_str: String = 5,10
*/
The value of entire column is stored in col_val_str
col_val_str: String = 5,10

Answer 4

Let us assume you need to pick the name from the below table for a particular Id and store that value in a variable.
+-----+-------+
| id | name |
+-----+-------+
| 100 | Alex |
| 200 | Bidan |
| 300 | Cary |
+-----+-------+
SCALA
-----------
Irrelevant data is filtered out first and then the name column is selected and finally stored into name variable
var name = df.filter($"id" === "100").select("name").collect().map(_.getString(0)).mkString("")
PYTHON (PYSPARK)
-----------------------------
For simpler usage, I have created a function that returns the value by passing the dataframe and the desired column name to this (this is spark Dataframe and not Pandas Dataframe). Before passing the dataframe to this function, filter is applied to filter out other records.
def GetValueFromDataframe(_df,columnName):
for row in _df.rdd.collect():
return row[columnName].strip()
name = GetValueFromDataframe(df.filter(df.id == "100"),"name")
There might be more simpler approach than this using 3x version of Python. The code which I showed above was tested for 2.7 version.
Note :
It is most likely to encounter out of memory error (Driver memory) since we use the collect function. Hence it is always recommended to apply transformations (like filter,where etc) before you call the collect function. If you
still encounter with driver out of memory issue, you could pass --conf spark.driver.maxResultSize=0 as command line argument to make use of unlimited driver memory.

Answer 5

For anyone interested below is an way to turn a column into an Array, for the below case we are just taking the first value.
val names= test.filter(test("id").equalTo("200")).selectExpr("name").rdd.map(x=>x.mkString).collect
val name = names(0)

Answer 6

s is the string of column values
.collect() converts columns/rows to an array of lists, in this case, all rows will be converted to a tuple, temp is basically an array of such tuples/row.
x(n-1) retrieves the n-th column value for x-th row, which is by default of type "Any", so needs to be converted to String so as to append to the existing strig.
s =""
// say the n-th column is the target column
val temp = test.collect() // converts Rows to array of list
temp.foreach{x =>
s += (x(n-1).asInstanceOf[String])
}
println(s)