Store string in a column as nested JSON to a JSON file - Pyspark

user_13161656 0
Viewed 830

I have a pyspark dataframe, this is what it looks like

+------------------------------------+-------------------+-------------+--------------------------------+---------+
|member_uuid                         |Timestamp          |updated      |member_id                       |easy_id  |
+------------------------------------+-------------------+-------------+--------------------------------+---------+
|027130fe-584d-4d8e-9fb0-b87c984a0c20|2020-02-11 19:15:32|password_hash|ajuypjtnlzmk4na047cgav27jma6_STG|993269700|

I transformed the above dataframe to this,

 +---------+---------+-------------------------------------------------------------------------------------------------------------------------------------------------+-------------------+
|attribute|operation|params                                                                                                                                           |timestamp          |
+---------+---------+-------------------------------------------------------------------------------------------------------------------------------------------------+-------------------+
|profile  |UPDATE   |{"member_uuid":"027130fe-584d-4d8e-9fb0-b87c984a0c20","member_id":"ajuypjtnlzmk4na047cgav27jma6_STG","easy_id":993269700,"field":"password_hash"}|2020-02-11 19:15:32|

Using the following code,

ll = ['member_uuid', 'member_id', 'easy_id', 'field']
df = df.withColumn('timestamp', col('Timestamp')).withColumn('attribute', lit('profile')).withColumn('operation', lit(col_name)) \
                    .withColumn('field', col('updated')).withColumn('params', F.to_json(struct([x for x in ll])))
    df = df.select('attribute', 'operation', 'params', 'timestamp')

I have save this dataframe df to a text file after converting it to JSON. I tried using the following code to do the same,

df_final.toJSON().coalesce(1).saveAsTextFile('file')

The file contains,

{"attribute":"profile","operation":"UPDATE","params":"{\"member_uuid\":\"027130fe-584d-4d8e-9fb0-b87c984a0c20\",\"member_id\":\"ajuypjtnlzmk4na047cgav27jma6_STG\",\"easy_id\":993269700,\"field\":\"password_hash\"}","timestamp":"2020-02-11T19:15:32.000Z"}

I want it to save in this format,

{"attribute":"profile","operation":"UPDATE","params":{"member_uuid":"027130fe-584d-4d8e-9fb0-b87c984a0c20","member_id":"ajuypjtnlzmk4na047cgav27jma6_STG","easy_id":993269700,"field":"password_hash"},"timestamp":"2020-02-11T19:15:32.000Z"}

to_json saves the value in the params columns as a string, is there a way to keep the json context here so I can save it as the desired output?

pyspark-sql pyspark-dataframes pyspark python json
2 Answers

Don't use to_json to create params column in dataframe.

  • The trick here is just create struct and write to the file (using .saveAsTextFile (or) .write.json()) Spark will create JSON for the Struct field.

  • if we already created json object and writing in json format Spark will add \ to escape the quotes already exists in Json string.

Example:

from pyspark.sql.functions import *

#sample data
df=spark.createDataFrame([("027130fe-584d-4d8e-9fb0-b87c984a0c20","2020-02-11 19:15:32","password_hash","ajuypjtnlzmk4na047cgav27jma6_STG","993269700")],["member_uuid","Timestamp","updated","member_id","easy_id"])

df1=df.withColumn("attribute",lit("profile")).withColumn("operation",lit("UPDATE"))

df1.selectExpr("struct(member_uuid,member_id,easy_id) as params","attribute","operation","timestamp").write.format("json").mode("overwrite").save("<path>")

#{"params":{"member_uuid":"027130fe-584d-4d8e-9fb0-b87c984a0c20","member_id":"ajuypjtnlzmk4na047cgav27jma6_STG","easy_id":"993269700"},"attribute":"profile","operation":"UPDATE","timestamp":"2020-02-11 19:15:32"}

df1.selectExpr("struct(member_uuid,member_id,easy_id) as params","attribute","operation","timestamp").toJSON().saveAsTextFile("<path>")

#{"params":{"member_uuid":"027130fe-584d-4d8e-9fb0-b87c984a0c20","member_id":"ajuypjtnlzmk4na047cgav27jma6_STG","easy_id":"993269700"},"attribute":"profile","operation":"UPDATE","timestamp":"2020-02-11 19:15:32"}
user_76326950

A simple way to handle it is to just do a replace operation on the file

sourceData=open('file').read().replace('"{','{').replace('}"','}').replace('\\','')
with open('file','w') as final:
    final.write(sourceData)

This might not be what you are looking for, but will achieve the end result.

user_28478570
Related