问题描述
我正在尝试读取一个架构文件(它是一个文本文件)并将其应用到我的 CSV 文件中,而没有标题.由于我已经有一个架构文件,我不想使用 InferSchema
选项,这是一个开销.
I am trying to read a Schema file (which is a text file) and apply it to my CSV file without a header. Since I already have a schema file I don't want to use InferSchema
option which is an overhead.
我的输入架构文件如下所示,
My input schema file looks like below,
"num IntegerType","letter StringType"
我正在尝试使用以下代码来创建架构文件,
I am trying the below code to create a schema file,
val schema_file = spark.read.textFile("D:\\Users\\Documents\\schemaFile.txt")
val struct_type = schema_file.flatMap(x => x.split(",")).map(b => (b.split(" ")(0).stripPrefix("\"").asInstanceOf[String],b.split(" ")(1).stripSuffix("\"").asInstanceOf[org.apache.spark.sql.types.DataType])).foreach(x=>println(x))
我收到如下错误
Exception in thread "main" java.lang.UnsupportedOperationException: No Encoder found for org.apache.spark.sql.types.DataType
- 字段(类:org.apache.spark.sql.types.DataType",名称:_2")- 根类:scala.Tuple2"
- field (class: "org.apache.spark.sql.types.DataType", name: "_2")- root class: "scala.Tuple2"
并尝试将其用作架构文件,同时使用 spark.read.csv
如下所示并将其编写为 ORC 文件
and trying to use this as a schema file while using spark.read.csv
like below and write it as an ORC file
val df=spark.read
.format("org.apache.spark.csv")
.option("header", false)
.option("inferSchema", true)
.option("samplingRatio",0.01)
.option("nullValue", "NULL")
.option("delimiter","|")
.schema(schema_file)
.csv("D:\\Users\\sampleFile.txt")
.toDF().write.format("orc").save("D:\\Users\\ORC")
需要帮助将文本文件转换为架构文件并将我的输入 CSV 文件转换为 ORC.
Need help to convert a text file into a schema file and convert my input CSV file to ORC.
推荐答案
要从 text
文件创建模式,请创建一个函数以 match
type
并返回 DataType
作为
To create a schema from a text
file create a function to match
the type
and return DataType
as
def getType(raw: String): DataType = {
raw match {
case "ByteType" => ByteType
case "ShortType" => ShortType
case "IntegerType" => IntegerType
case "LongType" => LongType
case "FloatType" => FloatType
case "DoubleType" => DoubleType
case "BooleanType" => BooleanType
case "TimestampType" => TimestampType
case _ => StringType
}
}
现在通过读取架构文件来创建架构
Now create a schema by reading a schema file as
val schema = Source.fromFile("schema.txt").getLines().toList
.flatMap(_.split(",")).map(_.replaceAll("\"", "").split(" "))
.map(x => StructField(x(0), getType(x(1)), true))
现在读取csv文件为
spark.read
.option("samplingRatio", "0.01")
.option("delimiter", "|")
.option("nullValue", "NULL")
.schema(StructType(schema))
.csv("data.csv")
希望这会有所帮助!
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