scala - How can I change column types in Spark SQL's DataFrame?

Question

Suppose I'm doing something like:

val df = sqlContext.load("com.databricks.spark.csv", Map("path" -> "cars.csv", "header" -> "true"))
df.printSchema()

root
 |-- year: string (nullable = true)
 |-- make: string (nullable = true)
 |-- model: string (nullable = true)
 |-- comment: string (nullable = true)
 |-- blank: string (nullable = true)

df.show()
year make  model comment              blank
2012 Tesla S     No comment
1997 Ford  E350  Go get one now th...

But I really wanted the year as Int (and perhaps transform some other columns).

The best I could come up with was

df.withColumn("year2", 'year.cast("Int")).select('year2 as 'year, 'make, 'model, 'comment, 'blank)
org.apache.spark.sql.DataFrame = [year: int, make: string, model: string, comment: string, blank: string]

which is a bit convoluted.

I'm coming from R, and I'm used to being able to write, e.g.

df2 <- df %>%
   mutate(year = year %>% as.integer,
          make = make %>% toupper)

I'm likely missing something, since there should be a better way to do this in Spark/Scala...

Answer 1

Edit: Newest version

Since spark 2.x you can use .withColumn. Check the docs here:

https://spark.apache.org/docs/latest/api/scala/index.html#org.apache.spark.sql.Dataset@withColumn(colName:String,col:org.apache.spark.sql.Column):org.apache.spark.sql.DataFrame

Oldest answer

Since Spark version 1.4 you can apply the cast method with DataType on the column:

import org.apache.spark.sql.types.IntegerType
val df2 = df.withColumn("yearTmp", df.year.cast(IntegerType))
    .drop("year")
    .withColumnRenamed("yearTmp", "year")

If you are using sql expressions you can also do:

val df2 = df.selectExpr("cast(year as int) year", 
                        "make", 
                        "model", 
                        "comment", 
                        "blank")

For more info check the docs: http://spark.apache.org/docs/1.6.0/api/scala/#org.apache.spark.sql.DataFrame