CreateDataFrame takes 2 arguments , an rdd and schema.
my schema is like this
val schemas= StructType(
Seq(
StructField("number",IntegerType,false),
StructField("notation", StringType,false)
)
)
in one case i am able to create dataframe from RDD like below:
`val data1=Seq(Row(1,"one"),Row(2,"two"))
val rdd=spark.sparkContext.parallelize(data1)
val final_df= spark.createDataFrame(rdd,schemas)`
In other case like below .. i am not able to
`val data2=Seq((1,"one"),(2,"two"))
val rdd=spark.sparkContext.parallelize(data2)
val final_df= spark.createDataFrame(rdd,schemas)`
Whats wrong with data2 for not able to become a valid rdd for Dataframe?
but we can able to create dataframe using toDF() with data2 but not CreateDataFrame.
val data2_DF=Seq((1,"one"),(2,"two")).toDF("number", "notation")
Please help me understand this behaviour.
Is Row mandatory while creating dataframe?
In the second case, just do :
val final_df = spark.createDataFrame(rdd)
Because your RDD is an RDD of Tuple2 (which is a Product), the schema is known at compile time, so you don't need to specify a schema
Related
I have got probably easy and quick question regarding the DataFrames in the Scala in Spark.
I have an existing Spark DataFrame (operate with Scala 2.10.5 and Spark 1.6.3) and would like to add a new column with the ArrayType or MapType, but don't know how to achieve that. But don't know how to deal with that. I would not like to create multiple columns with the 'single' values, but store them in one column. It would shorten my code and make it more changes prone.
import org.apache.spark.sql.types.MapType
...
// DataFrame initial creation
val df = ...
// adding new columns
val df_new = df
.withColumn("new_col1", lit("something_to_add") // add a literal
.withColumn("new_col2"), MapType("key1" -> "val1", "key2" -> "val2")) // ???
You could try something like
val df_new = df
.withColumn("new_col1", lit("something_to_add") // add a literal
.withColumn("new_col2"), typedLit[Map[String, String]](Map("key1" -> "val1"), ("key2" -> "val2")))
I have a dataframe with headers for example outputDF. I now want to take outputDF.columns and create a new dataframe with just one row which contains column names.
I then want to union both these dataframes with option("head=false") which spark can then write to a HDFS.
How do i do that?
below is an example
Val df = spark.read.csv("path")
val newDf = df.columns.toSeq.toDF
val unoindf= df.union(newDf);
I'm trying to do a count in scala with dataframe. My data has 3 columns and I've already loaded the data and split by tab. So I want to do something like this:
val file = file.map(line=>line.split("\t"))
val x = file1.map(line=>(line(0), line(2).toInt)).reduceByKey(_+_,1)
I want to put the data in dataframe, and having some trouble on the syntax
val file = file.map(line=>line.split("\t")).toDF
val file.groupby(line(0))
.count()
Can someone help check if this is correct?
spark needs to know the schema of the df
there are many ways to specify the schema, here is one option:
val df = file
.map(line=>line.split("\t"))
.map(l => (l(0), l(1).toInt)) //at this point spark knows the number of columns and their types
.toDF("a", "b") //give the columns names for ease of use
df
.groupby('a)
.count()
I loaded a csv as dataframe. I would like to cast all columns to float, knowing that the file is to big to write all columns names:
val spark = SparkSession.builder.master("local").appName("my-spark-app").getOrCreate()
val df = spark.read.option("header",true).option("inferSchema", "true").csv("C:/Users/mhattabi/Desktop/dataTest2.csv")
Given this DataFrame as example:
val df = sqlContext.createDataFrame(Seq(("0", 0),("1", 1),("2", 0))).toDF("id", "c0")
with schema:
StructType(
StructField(id,StringType,true),
StructField(c0,IntegerType,false))
You can loop over DF columns by .columns functions:
val castedDF = df.columns.foldLeft(df)((current, c) => current.withColumn(c, col(c).cast("float")))
So the new DF schema looks like:
StructType(
StructField(id,FloatType,true),
StructField(c0,FloatType,false))
EDIT:
If you wanna exclude some columns from casting, you could do something like (supposing we want to exclude the column id):
val exclude = Array("id")
val someCastedDF = (df.columns.toBuffer --= exclude).foldLeft(df)((current, c) =>
current.withColumn(c, col(c).cast("float")))
where exclude is an Array of all columns we want to exclude from casting.
So the schema of this new DF is:
StructType(
StructField(id,StringType,true),
StructField(c0,FloatType,false))
Please notice that maybe this is not the best solution to do it but it can be a starting point.
This article claims that a DataFrame in Spark is equivalent to a Dataset[Row], but this blog post shows that a DataFrame has a schema.
Take the example in the blog post of converting an RDD to a DataFrame: if DataFrame were the same thing as Dataset[Row], then converting an RDD to a DataFrameshould be as simple
val rddToDF = rdd.map(value => Row(value))
But instead it shows that it's this
val rddStringToRowRDD = rdd.map(value => Row(value))
val dfschema = StructType(Array(StructField("value",StringType)))
val rddToDF = sparkSession.createDataFrame(rddStringToRowRDD,dfschema)
val rDDToDataSet = rddToDF.as[String]
Clearly a dataframe is actually a dataset of rows and a schema.
In Spark 2.0, in code there is:
type DataFrame = Dataset[Row]
It is Dataset[Row], just because of definition.
Dataset has also schema, you can print it using printSchema() function. Normally Spark infers schema, so you don't have to write it by yourself - however it's still there ;)
You can also do createTempView(name) and use it in SQL queries, just like DataFrames.
In other words, Dataset = DataFrame from Spark 1.5 + encoder, that converts rows to your classes. After merging types in Spark 2.0, DataFrame becomes just an alias for Dataset[Row], so without specified encoder.
About conversions: rdd.map() also returns RDD, it never returns DataFrame. You can do:
// Dataset[Row]=DataFrame, without encoder
val rddToDF = sparkSession.createDataFrame(rdd)
// And now it has information, that encoder for String should be used - so it becomes Dataset[String]
val rDDToDataSet = rddToDF.as[String]
// however, it can be shortened to:
val dataset = sparkSession.createDataset(rdd)
Note (in addition to the answer of T Gaweda) that there is a schema associated to each Row (Row.schema). However, this schema is not set until it is integrated in a DataFrame (or Dataset[Row])
scala> Row(1).schema
res12: org.apache.spark.sql.types.StructType = null
scala> val rdd = sc.parallelize(List(Row(1)))
rdd: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] = ParallelCollectionRDD[5] at parallelize at <console>:28
scala> spark.createDataFrame(rdd,schema).first
res15: org.apache.spark.sql.Row = [1]
scala> spark.createDataFrame(rdd,schema).first.schema
res16: org.apache.spark.sql.types.StructType = StructType(StructField(a,IntegerType,true))