Question

如何将spark数据帧中的所有列名称转换为Seq变量。

输入数据＆amp;模式

val dataset1 = Seq(("66", "a", "4"), ("67", "a", "0"), ("70", "b", "4"), ("71", "d", "4")).toDF("KEY1", "KEY2", "ID")

dataset1.printSchema()
root
|-- KEY1: string (nullable = true)
|-- KEY2: string (nullable = true)
|-- ID: string (nullable = true)

我需要使用scala编程将所有列名存储在变量中。我试过如下，但它没有用。

val selectColumns = dataset1.schema.fields.toSeq

selectColumns: Seq[org.apache.spark.sql.types.StructField] = WrappedArray(StructField(KEY1,StringType,true),StructField(KEY2,StringType,true),StructField(ID,StringType,true))

预期产出：

val selectColumns = Seq(
  col("KEY1"),
  col("KEY2"),
  col("ID")
)

selectColumns: Seq[org.apache.spark.sql.Column] = List(KEY1, KEY2, ID)

Answer 1

您可以使用以下命令：

val selectColumns = dataset1.columns.toSeq

scala> val dataset1 = Seq(("66", "a", "4"), ("67", "a", "0"), ("70", "b", "4"), ("71", "d", "4")).toDF("KEY1", "KEY2", "ID")
dataset1: org.apache.spark.sql.DataFrame = [KEY1: string, KEY2: string ... 1 more field]

scala> val selectColumns = dataset1.columns.toSeq
selectColumns: Seq[String] = WrappedArray(KEY1, KEY2, ID)

Answer 2

val selectColumns = dataset1.columns.toList.map(col(_))

Answer 3

我像这样使用columns属性

val cols = dataset1.columns.toSeq

然后，如果您稍后按顺序从头到尾选择所有列，则可以使用

val orderedDF = dataset1.select(cols.head, cols.tail:_ *)

Answer 4

也可以从架构中获取列。

val dataset1 = Seq(("66", "a", "4"), ("67", "a", "0"), ("70", "b", "4"), ("71", "d", "4")).toDF("KEY1", "KEY2", "ID")
dataset1.printSchema()
root
 |-- KEY1: string (nullable = true)
 |-- KEY2: string (nullable = true)
 |-- ID: string (nullable = true)

val selectColumns = dataset1.schema.fieldNames
selectColumns: Array[String] = Array(KEY1, KEY2, ID)

val selectColumns2 = dataset1.schema.fieldNames.toSeq 
selectColumns2: Seq[String] = WrappedArray(KEY1, KEY2, ID)

Answer 5

我们可以通过以下方式将数据集/表的列名放入Sequence变量中。

来自数据集，

val col_seq:Seq[String] = dataset.columns.toSeq

从表中

val col_seq:Seq[String] = spark.table("tablename").columns.toSeq
                           or
val col_seq:Seq[String] = spark.catalog.listColumns("tablename").select('name).collect.map(col=>col.toString).toSeq

获取Spark数据帧列列表

5 个答案: