check data size spark dataframes - scala

I have the following question :
Actually I am working with the following csv file:
""job"";""marital"""
""management"";""married"""
""technician"";""single"""
I loaded it into a spark dataframe as follows:
My aim is to check the length and type of each field in the dataframe following the set od rules below :
col type
job char10
marital char7
I started implementing the check of the length of each field but I am getting a compilation error :
val data = spark.read.option("inferSchema", "true").option("header", "true").csv("file:////home/user/Desktop/user/file.csv")
data.map(line => {
val fields = line.toString.split(";")
fields(0).size
fields(1).size
})
The expected output should be:
List(10,10)
As for the check of the types I don't have any idea about how to implement it as we are using dataframes. Any idea about a function verifying the data format ?
Thanks a lot in advance for your replies.
ata

I see you are trying to use Dataframe, But if there are multiple double quotes then you can read as a textFile and remove them and convert to Dataframe as below
import org.apache.spark.sql.functions._
import spark.implicits._
val raw = spark.read.textFile("path to file ")
.map(_.replaceAll("\"", ""))
val header = raw.first
val data = raw.filter(row => row != header)
.map { r => val x = r.split(";"); (x(0), x(1)) }
.toDF(header.split(";"): _ *)
You get with data.show(false)
+----------+-------+
|job |marital|
+----------+-------+
|management|married|
|technician|single |
+----------+-------+
To calculate the size you can use withColumn and length function and play around as you need.
data.withColumn("jobSize", length($"job"))
.withColumn("martialSize", length($"marital"))
.show(false)
Output:
+----------+-------+-------+-----------+
|job |marital|jobSize|martialSize|
+----------+-------+-------+-----------+
|management|married|10 |7 |
|technician|single |10 |6 |
+----------+-------+-------+-----------+
All the column type are String.
Hope this helps!

You are using a dataframe. So when you use the map method, you are processing Row in your lambda.
so line is a Row.
Row.toString will return a string representing the Row, so in your case 2 structfields typed as String.
If you want to use map and process your Row, you have to get the vlaue inside the fields manually. with getAsString and getAsString.
Usually when you use Dataframes, you have to work in column's logic as in SQL using select, where... or directly the SQL syntax.

Related

Create SOAP XML REQUEST from selected dataframe columns in Scala

Is there a way to create an XML SOAP REQUEST by extracting a few columns from each row of a dataframe ? 10 records in a dataframe means 10 separate SOAP XML REQUESTs.
How would you make the function call using map now?
You can do that by applying a map function to the dataframe.
val df = your dataframe
df.map(x => convertToSOAP(x))
// convertToSOAP is your function.
Putting up an example based on your comment, hope you find this useful.
case class emp(id:String,name:String,city:String)
val list = List(emp("1","user1","NY"),emp("2","user2","SFO"))
val rdd = sc.parallelize(list)
val df = rdd.toDF
df.map(x => "<root><name>" + x.getString(1) + "</name><city>"+ x.getString(2) +"</city></root>").show(false)
// Note: x is a type of org.apache.spark.sql.Row
Output will be as follows :
+--------------------------------------------------+
|value |
+--------------------------------------------------+
|<root><name>user1</name><city>NY</city></root> |
|<root><name>user2</name><city>SFO</city></root> |
+--------------------------------------------------+

Process all columns / the entire row in a Spark UDF

For a dataframe containing a mix of string and numeric datatypes, the goal is to create a new features column that is a minhash of all of them.
While this could be done by performing a dataframe.toRDD it is expensive to do that when the next step will be to simply convert the RDD back to a dataframe.
So is there a way to do a udf along the following lines:
val wholeRowUdf = udf( (row: Row) => computeHash(row))
Row is not a spark sql datatype of course - so this would not work as shown.
Update/clarifiction I realize it is easy to create a full-row UDF that runs inside withColumn. What is not so clear is what can be used inside a spark sql statement:
val featurizedDf = spark.sql("select wholeRowUdf( what goes here? ) as features
from mytable")
Row is not a spark sql datatype of course - so this would not work as shown.
I am going to show that you can use Row to pass all the columns or selected columns to a udf function using struct inbuilt function
First I define a dataframe
val df = Seq(
("a", "b", "c"),
("a1", "b1", "c1")
).toDF("col1", "col2", "col3")
// +----+----+----+
// |col1|col2|col3|
// +----+----+----+
// |a |b |c |
// |a1 |b1 |c1 |
// +----+----+----+
Then I define a function to make all the elements in a row as one string separated by , (as you have computeHash function)
import org.apache.spark.sql.Row
def concatFunc(row: Row) = row.mkString(", ")
Then I use it in udf function
import org.apache.spark.sql.functions._
def combineUdf = udf((row: Row) => concatFunc(row))
Finally I call the udf function using withColumn function and struct inbuilt function combining selected columns as one column and pass to the udf function
df.withColumn("contcatenated", combineUdf(struct(col("col1"), col("col2"), col("col3")))).show(false)
// +----+----+----+-------------+
// |col1|col2|col3|contcatenated|
// +----+----+----+-------------+
// |a |b |c |a, b, c |
// |a1 |b1 |c1 |a1, b1, c1 |
// +----+----+----+-------------+
So you can see that Row can be used to pass whole row as an argument
You can even pass all columns in a row at once
val columns = df.columns
df.withColumn("contcatenated", combineUdf(struct(columns.map(col): _*)))
Updated
You can achieve the same with sql queries too, you just need to register the udf function as
df.createOrReplaceTempView("tempview")
sqlContext.udf.register("combineUdf", combineUdf)
sqlContext.sql("select *, combineUdf(struct(`col1`, `col2`, `col3`)) as concatenated from tempview")
It will give you the same result as above
Now if you don't want to hardcode the names of columns then you can select the column names according to your desire and make it a string
val columns = df.columns.map(x => "`"+x+"`").mkString(",")
sqlContext.sql(s"select *, combineUdf(struct(${columns})) as concatenated from tempview")
I hope the answer is helpful
I came up with a workaround: drop the column names into any existing spark sql function to generate a new output column:
concat(${df.columns.tail.mkString(",'-',")}) as Features
In this case the first column in the dataframe is a target and was excluded. That is another advantage of this approach: the actual list of columns many be manipulated.
This approach avoids unnecessary restructuring of the RDD/dataframes.

How to concatenate multiple columns into single column (with no prior knowledge on their number)?

Let say I have the following dataframe:
agentName|original_dt|parsed_dt| user|text|
+----------+-----------+---------+-------+----+
|qwertyuiop| 0| 0|16102.0| 0|
I wish to create a new dataframe with one more column that has the concatenation of all the elements of the row:
agentName|original_dt|parsed_dt| user|text| newCol
+----------+-----------+---------+-------+----+
|qwertyuiop| 0| 0|16102.0| 0| [qwertyuiop, 0,0, 16102, 0]
Note: This is a just an example. The number of columns and names of them is not known. It is dynamic.
TL;DR Use struct function with Dataset.columns operator.
Quoting the scaladoc of struct function:
struct(colName: String, colNames: String*): Column Creates a new struct column that composes multiple input columns.
There are two variants: string-based for column names or using Column expressions (that gives you more flexibility on the calculation you want to apply on the concatenated columns).
From Dataset.columns:
columns: Array[String] Returns all column names as an array.
Your case would then look as follows:
scala> df.withColumn("newCol",
struct(df.columns.head, df.columns.tail: _*)).
show(false)
+----------+-----------+---------+-------+----+--------------------------+
|agentName |original_dt|parsed_dt|user |text|newCol |
+----------+-----------+---------+-------+----+--------------------------+
|qwertyuiop|0 |0 |16102.0|0 |[qwertyuiop,0,0,16102.0,0]|
+----------+-----------+---------+-------+----+--------------------------+
I think this works perfect for your case
here is with an example
val spark =
SparkSession.builder().master("local").appName("test").getOrCreate()
import spark.implicits._
val data = spark.sparkContext.parallelize(
Seq(
("qwertyuiop", 0, 0, 16102.0, 0)
)
).toDF("agentName","original_dt","parsed_dt","user","text")
val result = data.withColumn("newCol", split(concat_ws(";", data.schema.fieldNames.map(c=> col(c)):_*), ";"))
result.show()
+----------+-----------+---------+-------+----+------------------------------+
|agentName |original_dt|parsed_dt|user |text|newCol |
+----------+-----------+---------+-------+----+------------------------------+
|qwertyuiop|0 |0 |16102.0|0 |[qwertyuiop, 0, 0, 16102.0, 0]|
+----------+-----------+---------+-------+----+------------------------------+
Hope this helped!
In general, you can merge multiple dataframe columns into one using array.
df.select($"*",array($"col1",$"col2").as("newCol")) \\$"*" will capture all existing columns
Here is the one line solution for your case:
df.select($"*",array($"agentName",$"original_dt",$"parsed_dt",$"user", $"text").as("newCol"))
You can use udf function to concat all the columns into one. All you have to do is define a udf function and pass all the columns you want to concat to the udf function and call the udf function using .withColumn function of dataframe
Or
You can use concat_ws(java.lang.String sep, Column... exprs) function available for dataframe.
var df = Seq(("qwertyuiop",0,0,16102.0,0))
.toDF("agentName","original_dt","parsed_dt","user","text")
df.withColumn("newCol", concat_ws(",",$"agentName",$"original_dt",$"parsed_dt",$"user",$"text"))
df.show(false)
Will give you output as
+----------+-----------+---------+-------+----+------------------------+
|agentName |original_dt|parsed_dt|user |text|newCol |
+----------+-----------+---------+-------+----+------------------------+
|qwertyuiop|0 |0 |16102.0|0 |qwertyuiop,0,0,16102.0,0|
+----------+-----------+---------+-------+----+------------------------+
That will get you the result you want
There may be syntax errors in my answer. This is useful if you are using java<8 and spark<2.
String columns=null
For ( String columnName : dataframe.columns())
{
Columns = columns == null ? columnName : columns+"," + columnName;
}
SqlContext.sql(" select *, concat_ws('|', " +columns+ ") as complete_record " +
"from data frame ").show();

Spark dataframe get column value into a string variable

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?
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)
val maxDate = spark.sql("select max(export_time) as export_time from tier1_spend.cost_gcp_raw").first()
val rowValue = maxDate.get(0)
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($"A").collect()
val col_val_str = col_val_df.map(x => x.get(0)).mkString(",")
/*
df: org.apache.spark.sql.DataFrame = [A: int, B: int]
col_val_row: Array[org.apache.spark.sql.Row] = Array([5], [10])
col_val_str: String = 5,10
*/
The value of entire column is stored in col_val_str
col_val_str: String = 5,10
Let us assume you need to pick the name from the below table for a particular Id and store that value in a variable.
+-----+-------+
| id | name |
+-----+-------+
| 100 | Alex |
| 200 | Bidan |
| 300 | Cary |
+-----+-------+
SCALA
-----------
Irrelevant data is filtered out first and then the name column is selected and finally stored into name variable
var name = df.filter($"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.
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)
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)

Scala spark Select as not working as expected

Hope someone can help. Fairly certain this is something I'm doing wrong.
I have a dataframe called uuidvar with 1 column called 'uuid' and another dataframe, df1, with a number of columns, one of which is also 'uuid'. I would like to select from from df1 all of the rows which have a uuid that appear in uuidvar. Now, having the same column names is not ideal so I tried to do it with
val uuidselection=df1.join(uuidvar, df1("uuid") === uuidvar("uuid").as("another_uuid"), "right_outer").select("*")
However when I show uuidselection I have 2 columns called "uuid". Furthermore, if I try and select the specific columns I want, I am told
cannot resolve 'uuidvar' given input columns
or similar depending on what I try and select.
I have tried to make it simpler and just do
val uuidvar2=uuidvar.select("uuid").as("uuidvar")
and this doesn't rename the column in uuidvar.
Does 'as' not operate as I am expecting it to, am I making some other fundamental error or is it broken?
I'm using spark 1.5.1 and scala 1.10.
Answer
You can't use as when specifying the join-criterion.
Use withColumnRenamed to modify the column before the join.
Seccnd, use generic col function for accessing columns via name (instead of using the dataframe's apply method, e.g. df1(<columnname>)
case class UUID1 (uuid: String)
case class UUID2 (uuid: String, b:Int)
class UnsortedTestSuite2 extends SparkFunSuite {
configuredUnitTest("SO - uuid") { sc =>
val sqlContext = new SQLContext(sc)
import sqlContext.implicits._
val uuidvar = sc.parallelize( Seq(
UUID1("cafe-babe-001"),
UUID1("cafe-babe-002"),
UUID1("cafe-babe-003"),
UUID1("cafe-babe-004")
)).toDF()
val df1 = sc.parallelize( Seq(
UUID2("cafe-babe-001", 1),
UUID2("cafe-babe-002", 2),
UUID2("cafe-babe-003", 3)
)).toDF()
val uuidselection=df1.join(uuidvar.withColumnRenamed("uuid", "another_uuid"), col("uuid") === col("another_uuid"), "right_outer")
uuidselection.show()
}
}
delivers
+-------------+----+-------------+
| uuid| b| another_uuid|
+-------------+----+-------------+
|cafe-babe-001| 1|cafe-babe-001|
|cafe-babe-002| 2|cafe-babe-002|
|cafe-babe-003| 3|cafe-babe-003|
| null|null|cafe-babe-004|
+-------------+----+-------------+
Comment
.select("*") does not have any effect. So
df.select("*") =^= df
I've always used the withColumnRenamed api to rename columns:
Take this table as an example:
| Name | Age |
df.withColumnRenamed('Age', 'newAge').show()
| Name | newAge |
So to make it work with your code, something like this should work:
val uuidvar_another = uuidvar.withColumnRenamed("uuid", "another_uuid")
val uuidselection=df1.join(uuidvar, df1("uuid") === uuidvar("another_uuid"), "right_outer").select("*")