scala学习(3)-----wordcount【sparksession】

参考:

spark中文官方网址:http://spark.apachecn.org/#/

https://www.iteblog.com/archives/1674.html

一、知识点:

1、Dataframe新增一列:https://www.cnblogs.com/itboys/p/9762808.html

方法四和五是新增一列唯一ID

方法一:利用createDataFrame方法,新增列的过程包含在构建rdd和schema中
方法二:利用withColumn方法,新增列的过程包含在udf函数中
方法三:利用SQL代码,新增列的过程直接写入SQL代码中
方法四:以上三种是增加一个有判断的列,如果想要增加一列唯一序号,可以使用monotonically_increasing_id
方法五:使用zipWithUniqueId获取id 并重建 DataFrame.
// dataframe新增一列方法1,利用createDataFrame方法
val trdd = input.select(targetColumns).rdd.map(x=>{
  if (x.get(0).toString().toDouble > critValueR || x.get(0).toString().toDouble < critValueL) 
    Row(x.get(0).toString().toDouble,"F")
  else Row(x.get(0).toString().toDouble,"T")      
  })      
val schema = input.select(targetColumns).schema.add("flag", StringType, true)
val sample3 = ss.createDataFrame(trdd, schema).distinct().withColumnRenamed(targetColumns, "idx")

// dataframe新增一列方法2
val code :(Int => String) = (arg: Int) => {if (arg > critValueR || arg < critValueL) "F" else "T"}
val addCol = udf(code)
val sample3 = input.select(targetColumns).withColumn("flag", addCol(input(targetColumns)))
.withColumnRenamed(targetColumns, "idx")

// dataframe新增一列方法3
input.select(targetColumns).createOrReplaceTempView("tmp")
val sample3 = ss.sqlContext.sql("select distinct "+targetColname+
    " as idx,case when "+targetColname+">"+critValueR+" then 'F'"+
    " when "+targetColname+"<"+critValueL+" then 'F' else 'T' end as flag from tmp")

// 添加序号列新增一列方法4
import org.apache.spark.sql.functions.monotonically_increasing_id
val inputnew = input.withColumn("idx", monotonically_increasing_id)
// 这个id虽然是唯一的,但是不能从零开始,也不是顺序排列,可以简单理解为是随机产生的标识码

// 方法五:使用zipWithUniqueId获取id 并重建 DataFrame.

import spark.implicits._ 
import org.apache.spark.sql.Row
import org.apache.spark.sql.types.{StructType, StructField, LongType}
val df =Seq(("a", -1.0), ("b", -2.0), ("c", -3.0)).toDF("foo", "bar")
// 获取df 的表头
val s = df.schema
// 将原表转换成带有rdd, //再转换成带有id的rdd, //再展开成Seq方便转化成 Dataframe val rows = df.rdd.zipWithUniqueId.map{case (r: Row, id: Long) => Row.fromSeq(id +: r.toSeq)} // 再由 row 根据原表头进行转换 val dfWithPK = spark.createDataFrame( rows, StructType(StructField("id", LongType, false) +: s.fields))
 

 2、新增一列ID:https://blog.csdn.net/liaodaoluyun/article/details/86232639

 二、wordcount

package com.qihoo.spark.examles

import com.qihoo.spark.app.SparkAppJob
import org.apache.spark.SparkContext
import org.kohsuke.args4j.{Option => ArgOption}
import org.apache.spark.sql.functions.monotonically_increasing_id

class WordCount extends SparkAppJob {
  //input
  @ArgOption(name = "-i", required = true, aliases = Array("--input"), usage = "input")
  var input: String = _
  //output
  @ArgOption(name = "-o", required = true, aliases = Array("--output"), usage = "output")
  var output: String = _

  override protected def run(sc: SparkContext): Unit = {
    import sparkSession.implicits._
    val showDasouSegment = sparkSession.read.text(input).as[String].filter(_.trim.length() != 0)
    showDasouSegment.show()
    val words = showDasouSegment
      .map(line => line.split("	"))
      .flatMap(line => line(1).split(" "))
      .groupByKey(value=>value)
    // val counts = words.count() 这一句是才让wordcount有效。以下代码是增加一列word的ID。
  // counts.show() 打印结果 val res
= words.keys.withColumn("ID",monotonically_increasing_id) res.show() // res.write.text(output) 这句话应该会报错,因为要将dataframe所有列合并成一列才能采用text存储。
  //
val concatDf = res.select(concat_ws(" ", $"word", $"id")) 将res中的word和id列合并成一列。 } }
原文地址:https://www.cnblogs.com/Lee-yl/p/11050809.html