如何在 pyspark 数据框上应用分组依据并对结果对象进行转换
How to apply group by on pyspark dataframe and a transformation on the resulting object
我有一个 spark 数据框
| item_id | attribute_key| attribute_value
____________________________________________________________________________
| id_1 brand Samsung
| id_1 ram 6GB
| id_2 brand Apple
| id_2 ram 4GB
_____________________________________________________________________________
我想按 item_id
对这个数据框进行分组并输出为一个文件,每一行都是一个 json
对象
{id_1: "properties":[{"brand":['Samsung']},{"ram":['6GB']} ]}
{id_2: "properties":[{"brand":['Apple']},{"ram":['4GB']} ]}
这是一个很大的分布式数据框架,因此不能转换为 pandas。
在 pyspark
中甚至可以进行这种转换吗?
在 scala 中,但 python 版本将非常相似 (sql.functions):
val df = Seq((1,"brand","Samsung"),(1,"ram","6GB"),(1,"ram","8GB"),(2,"brand","Apple"),(2,"ram","6GB")).toDF("item_id","attribute_key","attribute_value")
+-------+-------------+---------------+
|item_id|attribute_key|attribute_value|
+-------+-------------+---------------+
| 1| brand| Samsung|
| 1| ram| 6GB|
| 1| ram| 8GB|
| 2| brand| Apple|
| 2| ram| 6GB|
+-------+-------------+---------------+
df.groupBy('item_id,'attribute_key)
.agg(collect_list('attribute_value).as("list2"))
.groupBy('item_id)
.agg(map(lit("properties"),collect_list(map('attribute_key,'list2))).as("prop"))
.select(to_json(map('item_id,'prop)).as("json"))
.show(false)
输出:
+------------------------------------------------------------------+
|json |
+------------------------------------------------------------------+
|{"1":{"properties":[{"ram":["6GB","8GB"]},{"brand":["Samsung"]}]}}|
|{"2":{"properties":[{"brand":["Apple"]},{"ram":["6GB"]}]}} |
+------------------------------------------------------------------+
我有一个 spark 数据框
| item_id | attribute_key| attribute_value
____________________________________________________________________________
| id_1 brand Samsung
| id_1 ram 6GB
| id_2 brand Apple
| id_2 ram 4GB
_____________________________________________________________________________
我想按 item_id
对这个数据框进行分组并输出为一个文件,每一行都是一个 json
对象
{id_1: "properties":[{"brand":['Samsung']},{"ram":['6GB']} ]}
{id_2: "properties":[{"brand":['Apple']},{"ram":['4GB']} ]}
这是一个很大的分布式数据框架,因此不能转换为 pandas。 在 pyspark
中甚至可以进行这种转换吗?在 scala 中,但 python 版本将非常相似 (sql.functions):
val df = Seq((1,"brand","Samsung"),(1,"ram","6GB"),(1,"ram","8GB"),(2,"brand","Apple"),(2,"ram","6GB")).toDF("item_id","attribute_key","attribute_value")
+-------+-------------+---------------+
|item_id|attribute_key|attribute_value|
+-------+-------------+---------------+
| 1| brand| Samsung|
| 1| ram| 6GB|
| 1| ram| 8GB|
| 2| brand| Apple|
| 2| ram| 6GB|
+-------+-------------+---------------+
df.groupBy('item_id,'attribute_key)
.agg(collect_list('attribute_value).as("list2"))
.groupBy('item_id)
.agg(map(lit("properties"),collect_list(map('attribute_key,'list2))).as("prop"))
.select(to_json(map('item_id,'prop)).as("json"))
.show(false)
输出:
+------------------------------------------------------------------+
|json |
+------------------------------------------------------------------+
|{"1":{"properties":[{"ram":["6GB","8GB"]},{"brand":["Samsung"]}]}}|
|{"2":{"properties":[{"brand":["Apple"]},{"ram":["6GB"]}]}} |
+------------------------------------------------------------------+