Spark DataFrame如何实现日期加1并自动跳过周末取下一工作日

方案1：使用内置日期函数（无UDF，性能更优） #

核心逻辑利用Spark内置dayofweek函数判断星期，规则为：dayofweek返回值1对应周日、2对应周一...7对应周六，根据加1后的日期星期数做顺延：

• 加1后为周六：额外再加2天到周一
• 加1后为周日：额外再加1天到周一
• 其余情况直接使用加1后的日期

实现代码：

import org.apache.spark.sql.functions._

val df = Seq(
  ("50312", "2021-12-01", "0.9992019"),
  ("50312", "2021-12-02", "0.20171201"),
  ("50312", "2021-12-03", "2.9992019")
).toDF("id","some_date","item_value")
.withColumn("next_day_temp", date_add(col("some_date"),1))
.withColumn("nextworking_day", 
  when(dayofweek(col("next_day_temp")) === 7, date_add(col("next_day_temp"),2))
  .when(dayofweek(col("next_day_temp")) === 1, date_add(col("next_day_temp"),1))
  .otherwise(col("next_day_temp"))
)
.drop("next_day_temp")

方案2：自定义UDF（方便后续扩展节假日规则） #

如果后续需要叠加法定节假日排除逻辑，用UDF实现灵活性更高：

import java.time.LocalDate
import java.time.DayOfWeek
import org.apache.spark.sql.functions.udf

val getNextWorkingDay = udf { dateStr: String =>
  var currentDate = LocalDate.parse(dateStr).plusDays(1)
  // 跳过周末，需要扩展节假日的话可在此处新增判断逻辑
  while (currentDate.getDayOfWeek == DayOfWeek.SATURDAY || currentDate.getDayOfWeek == DayOfWeek.SUNDAY) {
    currentDate = currentDate.plusDays(1)
  }
  currentDate.toString
}

// 调用UDF实现
val df = Seq(
  ("50312", "2021-12-01", "0.9992019"),
  ("50312", "2021-12-02", "0.20171201"),
  ("50312", "2021-12-03", "2.9992019")
).toDF("id","some_date","item_value")
.withColumn("nextworking_day", getNextWorkingDay(col("some_date")))

内容的提问来源于stack exchange，提问作者Shasu