问题
I have a spark frame with two columns which looks like:
+-------------------------------------------------------------+------------------------------------+
|docId |id |
+-------------------------------------------------------------+------------------------------------+
|DYSDG6-RTB-91d663dd-949e-45da-94dd-e604b6050cb5-1537142434000|91d663dd-949e-45da-94dd-e604b6050cb5|
|VAVLS7-RTB-8e2c1917-0d6b-419b-a59e-cd4acc255bb7-1537142445000|8e2c1917-0d6b-419b-a59e-cd4acc255bb7|
|VAVLS7-RTB-c818dcde-7a68-4c1e-9cc4-c841660732d2-1537146854000|c818dcde-7a68-4c1e-9cc4-c841660732d2|
|IW2BYL-RTB-E9727F7D-D1BA-479C-9D3A-931F87E78B0A-1537146572000|E9727F7D-D1BA-479C-9D3A-931F87E78B0A|
|DYSDG6-RTB-f50f79e9-3ec3-4bd8-8e53-f62c3f80bcb0-1537146220000|f50f79e9-3ec3-4bd8-8e53-f62c3f80bcb0|
+-------------------------------------------------------------+------------------------------------+
I have a function that convert the id column into an 85 bit encoded string :
def convert_id(id):
import base64 as bs
id_str = str(id).replace("-", "")
return str(bs.a85encode(bytearray.fromhex(id_str)))[2:-1]
I want to transform this using pandas udf which is reported to be faster than the normal udf's.
How can I achieve this ? TIA.
回答1:
Done. Simple function can help to achieve this:
@pandas_udf(returnType=StringType())
def convert_id(id):
converted = id.map(lambda x : str(bs.a85encode(bytearray.fromhex(str(x).replace("-", ""))))[2:-1])
return converted
来源：https://stackoverflow.com/questions/52401542/how-to-use-pandas-udf-functionality-in-pyspark