如何根据 spark scala 中另一列的值创建新列？

Question

这是我的第一个 dataframe：

  val sample_df = Seq(("john","morning",1.5,0.0),("john","night",0.0,3.9),("bill","morning",0.4,0.0),("bill","night",0.0,2.3)).toDF("name","time_of_day","morning_min","night_min")

我想添加一个名为“row_min”的列，如果“time_of_day”行是“morning”，则取“morning_min”列的值，否则为“night_min”列。

以下是生成的 dataframe 应如下所示：

  val resulting_df = Seq(("john","morning",1.5,0.0,1.5),("john","night",0.0,3.9,3.9),("bill","morning",0.4,0.0,0.4),("bill","night",0.0,2.3,2.3)).toDF("name","time_of_day","morning_min","night_min","row_min")

任何帮助将不胜感激。 祝你有个愉快的一天。

Answer 1

这是我的第一个 dataframe：

  val sample_df = Seq(("john","morning",1.5,0.0),("john","night",0.0,3.9),("bill","morning",0.4,0.0),("bill","night",0.0,2.3)).toDF("name","time_of_day","morning_min","night_min")

我想添加一个名为“row_min”的列，如果“time_of_day”行是“morning”，则取“morning_min”列的值，否则为“night_min”列。

以下是生成的 dataframe 应如下所示：

  val resulting_df = Seq(("john","morning",1.5,0.0,1.5),("john","night",0.0,3.9,3.9),("bill","morning",0.4,0.0,0.4),("bill","night",0.0,2.3,2.3)).toDF("name","time_of_day","morning_min","night_min","row_min")

任何帮助将不胜感激。 祝你有个愉快的一天。