在Spark Scala中重命名DataFrame的列名

Question

我试图转换DataFrameSpark-Scala中的所有标题/列名称.截至目前,我想出了以下代码,它只替换了一个列名.

for( i <- 0 to origCols.length - 1) {
  df.withColumnRenamed(
    df.columns(i), 
    df.columns(i).toLowerCase
  );
}

Answer 1

如果结构是平的:

val df = Seq((1L, "a", "foo", 3.0)).toDF
df.printSchema
// root
//  |-- _1: long (nullable = false)
//  |-- _2: string (nullable = true)
//  |-- _3: string (nullable = true)
//  |-- _4: double (nullable = false)

你能做的最简单的事就是使用toDF方法:

val newNames = Seq("id", "x1", "x2", "x3")
val dfRenamed = df.toDF(newNames: _*)

dfRenamed.printSchema
// root
// |-- id: long (nullable = false)
// |-- x1: string (nullable = true)
// |-- x2: string (nullable = true)
// |-- x3: double (nullable = false)

如果要重命名单个列,可以使用以下任select一项alias:

df.select($"_1".alias("x1"))

可以很容易地推广到多列:

val lookup = Map("_1" -> "foo", "_3" -> "bar")

df.select(df.columns.map(c => col(c).as(lookup.getOrElse(c, c))): _*)

或者withColumnRenamed:

df.withColumnRenamed("_1", "x1")

这与使用foldLeft重命名多个列:

lookup.foldLeft(df)((acc, ca) => acc.withColumnRenamed(ca._1, ca._2))

使用嵌套结构(structs),一种可能的选择是通过选择整个结构来重命名:

val nested = spark.read.json(sc.parallelize(Seq(
    """{"foobar": {"foo": {"bar": {"first": 1.0, "second": 2.0}}}, "id": 1}"""
)))

nested.printSchema
// root
//  |-- foobar: struct (nullable = true)
//  |    |-- foo: struct (nullable = true)
//  |    |    |-- bar: struct (nullable = true)
//  |    |    |    |-- first: double (nullable = true)
//  |    |    |    |-- second: double (nullable = true)
//  |-- id: long (nullable = true)

@transient val foobarRenamed = struct(
  struct(
    struct(
      $"foobar.foo.bar.first".as("x"), $"foobar.foo.bar.first".as("y")
    ).alias("point")
  ).alias("location")
).alias("record")

nested.select(foobarRenamed, $"id").printSchema
// root
//  |-- record: struct (nullable = false)
//  |    |-- location: struct (nullable = false)
//  |    |    |-- point: struct (nullable = false)
//  |    |    |    |-- x: double (nullable = true)
//  |    |    |    |-- y: double (nullable = true)
//  |-- id: long (nullable = true)

请注意,它可能会影响nullability元数据.另一种可能性是通过强制重命名:

nested.select($"foobar".cast(
  "struct<location:struct<point:struct<x:double,y:double>>>"
).alias("record")).printSchema

// root
//  |-- record: struct (nullable = true)
//  |    |-- location: struct (nullable = true)
//  |    |    |-- point: struct (nullable = true)
//  |    |    |    |-- x: double (nullable = true)
//  |    |    |    |-- y: double (nullable = true)

要么:

import org.apache.spark.sql.types._

nested.select($"foobar".cast(
  StructType(Seq(
    StructField("location", StructType(Seq(
      StructField("point", StructType(Seq(
        StructField("x", DoubleType), StructField("y", DoubleType)))))))))
).alias("record")).printSchema

// root
//  |-- record: struct (nullable = true)
//  |    |-- location: struct (nullable = true)
//  |    |    |-- point: struct (nullable = true)
//  |    |    |    |-- x: double (nullable = true)
//  |    |    |    |-- y: double (nullable = true)

Answer 2

对于那些对PySpark版本感兴趣的人(实际上它在Scala中是相同的 - 请参阅下面的评论):

merchants_df_renamed = merchants_df.toDF(
    'merchant_id', 'category', 'subcategory', 'merchant')

merchants_df_renamed.printSchema()

结果:

root
| - merchant_id:integer(nullable = true)
| - category:string(nullable = true)
| - subcategory:string(nullable = true)
| - merchant:string(nullable = true)

Answer 3

def aliasAllColumns(t: DataFrame, p: String = "", s: String = ""): DataFrame =
{
  t.select( t.columns.map { c => t.col(c).as( p + c + s) } : _* )
}

如果不明显，这会为每个当前列名称添加一个前缀和一个后缀。当您有两个表具有一个或多个具有相同名称的列，并且您希望连接它们但仍然能够消除结果表中的列的歧义时，这会很有用。如果在“普通”SQL 中有类似的方法来执行此操作，那肯定会很好。