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