迭代 Spark 数据帧中的行和列

问题描述

我有以下动态创建的 Spark 数据框:

I have the following Spark dataframe that is created dynamically:

val sf1 = StructField("name", StringType, nullable = true)
val sf2 = StructField("sector", StringType, nullable = true)
val sf3 = StructField("age", IntegerType, nullable = true)

val fields = List(sf1,sf2,sf3)
val schema = StructType(fields)

val row1 = Row("Andy","aaa",20)
val row2 = Row("Berta","bbb",30)
val row3 = Row("Joe","ccc",40)

val data = Seq(row1,row2,row3)

val df = spark.createDataFrame(spark.sparkContext.parallelize(data), schema)

df.createOrReplaceTempView("people")
val sqlDF = spark.sql("SELECT * FROM people")

现在，我需要迭代sqlDF中的每一行和每一列来打印每一列，这是我的尝试:

Now, I need to iterate each row and column in sqlDF to print each column, this is my attempt:

sqlDF.foreach { row =>
  row.foreach { col => println(col) }
}

row 是 Row 类型，但不可迭代，这就是为什么这段代码会在 row.foreach 中引发编译错误.如何迭代Row中的每一列?

row is type Row, but is not iterable that's why this code throws a compilation error in row.foreach. How to iterate each column in Row?

推荐答案

您可以使用 toSeq 将 Row 转换为 Seq.一旦转向 Seq，你可以像往常一样使用 foreach、map 或任何你需要的东西

You can convert Row to Seq with toSeq. Once turned to Seq you can iterate over it as usual with foreach, map or whatever you need

    sqlDF.foreach { row =>
           row.toSeq.foreach{col => println(col) }
    }

输出:

Berta
bbb
30
Joe
Andy
aaa
20
ccc
40

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