Split 1 column into 3 columns in spark scala

问题:

I have a dataframe in Spark using scala that has a column that I need split.

scala> test.show
+-------------+
|columnToSplit|
+-------------+
|        a.b.c|
|        d.e.f|
+-------------+

I need this column split out to look like this:

+--------------+
|col1|col2|col3|
|   a|   b|   c|
|   d|   e|   f|
+--------------+

I'm using Spark 2.0.0

Thanks

回答1:

Try:

df.withColumn("_tmp", split($"columnToSplit", "\\.")).select(
  $"_tmp".getItem(0).as("col1"),
  $"_tmp".getItem(1).as("col2"),
  $"_tmp".getItem(2).as("col3")
).drop("_tmp")

回答2:

To do this programmatically, you can create a sequence of expressions with (0 until 3).map(i => col("temp").getItem(i).as(s"col$i")) (assume you need 3 columns as result) and then apply it to select with : _* syntax:

df.withColumn("temp", split(col("columnToSplit"), "\\.")).select(
    (0 until 3).map(i => col("temp").getItem(i).as(s"col$i")): _*
).show
+----+----+----+
|col0|col1|col2|
+----+----+----+
|   a|   b|   c|
|   d|   e|   f|
+----+----+----+

---

To keep all columns:

df.withColumn("temp", split(col("columnToSplit"), "\\.")).select(
    col("*") +: (0 until 3).map(i => col("temp").getItem(i).as(s"col$i")): _*
).show
+-------------+---------+----+----+----+
|columnToSplit|     temp|col0|col1|col2|
+-------------+---------+----+----+----+
|        a.b.c|[a, b, c]|   a|   b|   c|
|        d.e.f|[d, e, f]|   d|   e|   f|
+-------------+---------+----+----+----+

---

If you are using pyspark, use a list comprehension to replace the map in scala:

df = spark.createDataFrame([['a.b.c'], ['d.e.f']], ['columnToSplit'])
from pyspark.sql.functions import col, split

(df.withColumn('temp', split('columnToSplit', '\\.'))
   .select(*(col('temp').getItem(i).alias(f'col{i}') for i in range(3))
).show()
+----+----+----+
|col0|col1|col2|
+----+----+----+
|   a|   b|   c|
|   d|   e|   f|
+----+----+----+

回答3:

A solution which avoids the select part. This is helpful when you just want to append the new columns:

case class Message(others: String, text: String)

val r1 = Message("foo1", "a.b.c")
val r2 = Message("foo2", "d.e.f")

val records = Seq(r1, r2)
val df = spark.createDataFrame(records)

df.withColumn("col1", split(col("text"), "\\.").getItem(0))
  .withColumn("col2", split(col("text"), "\\.").getItem(1))
  .withColumn("col3", split(col("text"), "\\.").getItem(2))
  .show(false)

+------+-----+----+----+----+
|others|text |col1|col2|col3|
+------+-----+----+----+----+
|foo1  |a.b.c|a   |b   |c   |
|foo2  |d.e.f|d   |e   |f   |
+------+-----+----+----+----+

Update: I highly recommend to use Psidom's implementation to avoid splitting three times.

回答4:

This appends columns to the original DataFrame and doesn't use select, and only splits once using a temporary column:

import spark.implicits._

df.withColumn("_tmp", split($"columnToSplit", "\\."))
  .withColumn("col1", $"_tmp".getItem(0))
  .withColumn("col2", $"_tmp".getItem(1))
  .withColumn("col3", $"_tmp".getItem(2))
  .drop("_tmp")

回答5:

This expands on Psidom's answer and shows how to do the split dynamically, without hardcoding the number of columns. This answer runs a query to calculate the number of columns.

val df = Seq(
  "a.b.c",
  "d.e.f"
).toDF("my_str")
.withColumn("letters", split(col("my_str"), "\\."))

val numCols = df
  .withColumn("letters_size", size($"letters"))
  .agg(max($"letters_size"))
  .head()
  .getInt(0)

df
  .select(
    (0 until numCols).map(i => $"letters".getItem(i).as(s"col$i")): _*
  )
  .show()