通过检查值是否在列表中以及其他条件来过滤 Spark DataFrame
Filter Spark DataFrame by checking if value is in a list, with other criteria
作为一个简化示例,我尝试使用以下代码过滤 Spark DataFrame:
val xdf = sqlContext.createDataFrame(Seq(
("A", 1), ("B", 2), ("C", 3)
)).toDF("name", "cnt")
xdf.filter($"cnt" >1 || $"name" isin ("A","B")).show()
然后报错:
org.apache.spark.sql.AnalysisException: cannot resolve '((cnt > 1) || name)' due to data type mismatch: differing types in '((cnt > 1) || name)' (boolean and string).;
正确的做法是什么?在我看来,它在 name
列之后停止阅读。它是解析器中的错误吗?我正在使用 Spark 1.5.1
您必须将各个表达式括起来:
xdf.filter(($"cnt" > 1) || ($"name" isin ("A","B"))).show()
val list = List("x","y","t")
xdf.filter($"column".isin(list: _*))
作为一个简化示例,我尝试使用以下代码过滤 Spark DataFrame:
val xdf = sqlContext.createDataFrame(Seq(
("A", 1), ("B", 2), ("C", 3)
)).toDF("name", "cnt")
xdf.filter($"cnt" >1 || $"name" isin ("A","B")).show()
然后报错:
org.apache.spark.sql.AnalysisException: cannot resolve '((cnt > 1) || name)' due to data type mismatch: differing types in '((cnt > 1) || name)' (boolean and string).;
正确的做法是什么?在我看来,它在 name
列之后停止阅读。它是解析器中的错误吗?我正在使用 Spark 1.5.1
您必须将各个表达式括起来:
xdf.filter(($"cnt" > 1) || ($"name" isin ("A","B"))).show()
val list = List("x","y","t")
xdf.filter($"column".isin(list: _*))