我的数据集: -
emp dataframe looks like this :-
emp.show()
+---+-----+------+----------+-------------+
| ID| NAME|salary|department| date|
+---+-----+------+----------+-------------+
| 1| sban| 100.0| IT| 2018-01-10|
| 2| abc| 200.0| HR| 2018-01-05|
| 3| Jack| 100.0| SALE| 2018-01-05|
| 4| Ram| 100.0| IT|2018-01-01-06|
| 5|Robin| 200.0| IT| 2018-01-07|
| 6| John| 200.0| SALE| 2018-01-08|
| 7| sban| 300.0| Director| 2018-01-01|
+---+-----+------+----------+-------------+
2-然后我通过使用名称进行分组并获取其最高薪水,说数据帧是grpEmpByName: -
val grpByName = emp.select(col("name")).groupBy(col("name")).agg(max(col("salary")).alias("max_salary"))
grpByName.select("*").show()
+-----+----------+
| name|max_salary|
+-----+----------+
| Jack| 100.0|
|Robin| 200.0|
| Ram| 100.0|
| John| 200.0|
| abc| 200.0|
| sban| 300.0|
+-----+----------+
3-然后尝试加入: -
val joinedBySalarywithMaxSal = emp.join(grpEmpByName, col("emp.salary") === col("grpEmpByName.max_salary") , "inner")
投掷
18/02/08 21:29:26 INFO CodeGenerator: Code generated in 13.667672 ms
Exception in thread "main" org.apache.spark.sql.AnalysisException: cannot resolve '`grpByName.max_salary`' given input columns: [NAME, department, date, ID, salary, max_salary, NAME];;
'Join Inner, (salary#2 = 'grpByName.max_salary)
:- Project [ID#0, NAME#1, salary#2, department#3, date#4]
: +- MetastoreRelation default, emp
+- Aggregate [NAME#44], [NAME#44, max(salary#45) AS max_salary#25]
+- Project [salary#45, NAME#44]
+- Project [ID#43, NAME#44, salary#45, department#46, date#47]
+- MetastoreRelation default, emp
我不知道为什么它不能像我检查时那样工作
grpByName.select(col("max_salary")).show()
+----------+
|max_salary|
+----------+
| 100.0|
| 200.0|
| 100.0|
| 200.0|
| 200.0|
| 300.0|
+----------+
提前致谢。
答案 0 :(得分:2)
点符号用于引用表内的嵌套结构,而不是引用表本身。
请调用col
上的DataFrame
方法定义,如下所示:
emp.join(grpEmpByName, emp.col("salary") === grpEmpByName.col("max_salary"), "inner")
您可以看到示例here。
此外,请注意join
s are inner
by default,因此您应该能够编写以下内容:
emp.join(grpEmpByName, emp.col("salary") === grpEmpByName.col("max_salary"))
答案 1 :(得分:1)
我不确定,希望可以提供帮助:
val joinedBySalarywithMaxSal = emp.join(grpEmpByName, emp.col("emp") === grpEmpByName.col("max_salary") , "inner")