这是解决方案。根据您的数据集,我提出了问题 -
下面的数据框有不正确的条目。我想删除所有不正确的记录,只保留正确的记录 -
val Friends = Seq(
("Rahul", "99", "AA"),
("Rahul", "20", "BB"),
("Rahul", "30", "BB"),
("Mahesh", "55", "CC"),
("Mahesh", "88", "DD"),
("Mahesh", "44", "FF"),
("Ramu", "30", "FF"),
("Gaurav", "99", "PP"),
("Gaurav", "20", "HH")).toDF("Name", "Age", "City")
过滤数组 -
val Name = List("Rahul", "Mahesh", "Gaurav")
val IncorrectAge = List(20, 55)
数据操作-
Friends.filter(!(col("Name").isin(Name: _*) && col("Age").isin(IncorrectAge: _*))).show
这是输出 -
+------+---+----+
| Name|Age|City|
+------+---+----+
| Rahul| 99| AA|
| Rahul| 30| BB|
|Mahesh| 88| DD|
|Mahesh| 44| FF|
| Ramu| 30| FF|
|Gaurav| 99| PP|
+------+---+----+
你也可以在joins的帮助下做到这一点..
创建不良记录 df -
val badrecords = Friends.filter(col("Name").isin(Name: _*) && col("Age").isin(IncorrectAge: _*))
用户 left_anti 加入选择好友减去不良记录 -
Friends.alias("left").join(badrecords.alias("right"), Seq("Name", "Age"), "left_anti").show
这是输出 -
+------+---+----+
| Name|Age|City|
+------+---+----+
| Rahul| 99| AA|
| Rahul| 30| BB|
|Mahesh| 88| DD|
|Mahesh| 44| FF|
| Ramu| 30| FF|
|Gaurav| 99| PP|
+------+---+----+