Pyspark将多列合并成一个json列

DBA*_*642 4 python dataframe apache-spark pyspark

不久前我为 python 提出了这个问题,但现在我需要在 PySpark 中做同样的事情。

我有一个像这样的数据框(df):

|cust_id|address    |store_id|email        |sales_channel|category|
-------------------------------------------------------------------
|1234567|123 Main St|10SjtT  |idk@gmail.com|ecom         |direct  |
|4567345|345 Main St|10SjtT  |101@gmail.com|instore      |direct  |
|1569457|876 Main St|51FstT  |404@gmail.com|ecom         |direct  |
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我想将最后 4 个字段组合成一个元数据字段,它是一个 json 格式,如下所示:

|cust_id|address    |metadata                                                                                     |
-------------------------------------------------------------------------------------------------------------------
|1234567|123 Main St|{'store_id':'10SjtT', 'email':'idk@gmail.com','sales_channel':'ecom', 'category':'direct'}   |
|4567345|345 Main St|{'store_id':'10SjtT', 'email':'101@gmail.com','sales_channel':'instore', 'category':'direct'}|
|1569457|876 Main St|{'store_id':'51FstT', 'email':'404@gmail.com','sales_channel':'ecom', 'category':'direct'}   |
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这是我用来在 python 中执行此操作的代码:

cols = [
    'store_id',
    'store_category',
    'sales_channel',
    'email'
]

df1 = df.copy()
df1['metadata'] = df1[cols].to_dict(orient='records')
df1 = df1.drop(columns=cols)
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但我想将其转换为 PySpark 代码以使用 spark 数据框;我不想在 Spark 中使用熊猫。

Shu*_*Shu 12

使用to_json函数创建json对象!

Example:

from pyspark.sql.functions import *

#sample data
df=spark.createDataFrame([('1234567','123 Main St','10SjtT','idk@gmail.com','ecom','direct')],['cust_id','address','store_id','email','sales_channel','category'])

df.select("cust_id","address",to_json(struct("store_id","category","sales_channel","email")).alias("metadata")).show(10,False)

#result
+-------+-----------+----------------------------------------------------------------------------------------+
|cust_id|address    |metadata                                                                                |
+-------+-----------+----------------------------------------------------------------------------------------+
|1234567|123 Main St|{"store_id":"10SjtT","category":"direct","sales_channel":"ecom","email":"idk@gmail.com"}|
+-------+-----------+----------------------------------------------------------------------------------------+
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to_json by passing list of columns:

ll=['store_id','email','sales_channel','category']

df.withColumn("metadata", to_json(struct([x for x in ll]))).drop(*ll).show()

#result
+-------+-----------+----------------------------------------------------------------------------------------+
|cust_id|address    |metadata                                                                                |
+-------+-----------+----------------------------------------------------------------------------------------+
|1234567|123 Main St|{"store_id":"10SjtT","email":"idk@gmail.com","sales_channel":"ecom","category":"direct"}|
+-------+-----------+----------------------------------------------------------------------------------------+
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  • 有史以来最好的答案!我将从 Kafka -> Spark -> Kafka 出发,这个答案正是我所需要的。 (3认同)