在Spark中使用Scala基于条件获取row_number()可以通过以下步骤实现: 1. 导入必要的Spark相关库和函数: ```scala import org.apache.spark.s...
row_number降序 按照id分组,根据age字段进行组内排序,排序方式为降序 valwindowSpec1=Window.partitionBy("id").orderBy(col("age").desc) df.withColumn("rw",row_number.over(windowSpec1)).show() 1. 2. +---+---+---+---+---+---+ | id|age|label|pro0|pro1| rw| +---+---+---...
下面的代码中Window.partitionBy()的参数是cols:Column*,支持传入一个可变长度的Column序列 这个时候如果只是传入rowKey必须是报错的,因为rowKey是一个Array类型,通过:_*就把一个Array拆开成了一个变长序列 val rowKey= primaryKey.split(",").map(x=>col(x)) newDF = dataDF.withColumn("row_num", row_nu...
import org.apache.spark.sql.expressions.Window // 假设我们根据某个字段来更新数据 val windowSpec = Window.partitionBy("partition_column").orderBy("order_column") val updateDF = updatedDF.withColumn("row_number", row_number().over(windowSpec)) 执行批量更新操作: 最后,使用Spark的DataFrame API将...
、n进行编号,当数量n为奇数时,取编号(n + 1)/2位置的数即可,当n为偶数时,取(int)(n + 1)/2位置和(int)(n + 1)/2 + 1位置的数取平均即可。...首先使用row_number()给数据进行编号: val windowFun = Window.orderBy(col("feature3").asc) df.withColumn("rank",row_number...使用lit方法...
To create a row in Scala, we can use theRowclass constructor and pass the values as parameters. The number and type of parameters should match the schema or structure of the data. importorg.apache.spark.sql.Rowvalrow=Row("John",30,"USA") ...
SELECT ta.colName, ta.value, ta.num, ROW_NUMBER() OVER (PARTITION BY ta.colName ORDER BY ta.num DESC) AS row FROM (""")var vergin= 0for(col <-strColArr){if(vergin == 1){ strHistogramSql.append(" UNION ALL ") } vergin= 1strHistogramSql.append(s"""SELECT 'StrHistogram_$col...
2)))valdf2=spark.createDataFrame(rowRDD,schema)df2.show()+---+---+---+|number|word|index...
1.需要导入一个window,还需要一个sql.row_numbere 2.over里面的orderby 降序,scala是desc,Pythondesc() 3。Python版本可以直接在select里面追加新列,scala只能用withColumn python-pyspark frompyspark.sql.sessionimportSparkSessionfrompyspark.sql.typesimport*frompyspark.sql.functionsimport*frompyspark.sqlimportRowfrom...
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