上下文
sqlContext.sql(s"""
SELECT
school_name,
name,
age
FROM my_table
""")
询问
鉴于上表,我想按学校名称分组并将名称、年龄收集到 Map[String, Int]
例如 - 伪代码
val df = sqlContext.sql(s"""
SELECT
school_name,
age
FROM my_table
GROUP BY school_name
""")
------------------------
school_name | name | age
------------------------
school A | "michael"| 7
school A | "emily" | 5
school B | "cathy" | 10
school B | "shaun" | 5
df.groupBy("school_name").agg(make_map)
------------------------------------
school_name | map
------------------------------------
school A | {"michael": 7, "emily": 5}
school B | {"cathy": 10, "shaun": 5}
最佳答案
以下内容适用于 Spark 2.0。您可以使用map自 2.0 版本以来可用的函数可将列获取为 Map。
val df1 = df.groupBy(col("school_name")).agg(collect_list(map($"name",$"age")) as "map")
df1.show(false)
这将为您提供以下输出。
+-----------+------------------------------------+
|school_name|map |
+-----------+------------------------------------+
|school B |[Map(cathy -> 10), Map(shaun -> 5)] |
|school A |[Map(michael -> 7), Map(emily -> 5)]|
+-----------+------------------------------------+
现在您可以使用 UDF
将各个 Map 连接成单个 Map,如下所示。
import org.apache.spark.sql.functions.udf
val joinMap = udf { values: Seq[Map[String,Int]] => values.flatten.toMap }
val df2 = df1.withColumn("map", joinMap(col("map")))
df2.show(false)
这将为 Map[String,Int]
提供所需的输出。
+-----------+-----------------------------+
|school_name|map |
+-----------+-----------------------------+
|school B |Map(cathy -> 10, shaun -> 5) |
|school A |Map(michael -> 7, emily -> 5)|
+-----------+-----------------------------+
如果您想将列值转换为 JSON 字符串,则 Spark 2.1.0 引入了 to_json功能。
val df3 = df2.withColumn("map",to_json(struct($"map")))
df3.show(false)
to_json
函数将返回以下输出。
+-----------+-------------------------------+
|school_name|map |
+-----------+-------------------------------+
|school B |{"map":{"cathy":10,"shaun":5}} |
|school A |{"map":{"michael":7,"emily":5}}|
+-----------+-------------------------------+
关于apache-spark - 如何使用 groupBy 将行收集到 map 中?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41819275/