I can't find a way to convert a binary to a string representation without using a UDF. Is there a way with native PySpark functions and not a UDF?
from pyspark.sql import DataFrame, SparkSession
import pyspark.sql.functions as F
import uuid
from pyspark.sql.types import Row, StringType
spark_test_instance = (SparkSession
.builder
.master('local')
.getOrCreate())
df: DataFrame = spark_test_instance.createDataFrame([Row()])
df = df.withColumn("id", F.lit(uuid.uuid4().bytes))
df = df.withColumn("length", F.length(df["id"]))
uuidbytes_to_str = F.udf(lambda x: str(uuid.UUID(bytes=bytes(x), version=4)), StringType())
df = df.withColumn("id_str", uuidbytes_to_str(df["id"]))
df = df.withColumn("length_str", F.length(df["id_str"]))
df.printSchema()
df.show(1, truncate=False)
gives:
root
|-- id: binary (nullable = false)
|-- length: integer (nullable = false)
|-- id_str: string (nullable = true)
|-- length_str: integer (nullable = false)
+-------------------------------------------------+------+------------------------------------+----------+
|id |length|id_str |length_str|
+-------------------------------------------------+------+------------------------------------+----------+
|[0A 35 DC 67 13 C8 47 7E B0 80 9F AB 98 CA FA 89]|16 |0a35dc67-13c8-477e-b080-9fab98cafa89|36 |
+-------------------------------------------------+------+------------------------------------+----------+
You can use hex
for getting the id_str
:
from pyspark.sql import SparkSession
import pyspark.sql.functions as F
import uuid
spark = SparkSession.builder.getOrCreate()
data = [(uuid.uuid4().bytes,)]
df = spark.createDataFrame(data, ["id"])
df = df.withColumn("id_str", F.lower(F.hex("id")))
df.show(truncate=False)
# +-------------------------------------------------+--------------------------------+
# |id |id_str |
# +-------------------------------------------------+--------------------------------+
# |[8C 76 42 18 BD CA 47 A1 9C D7 9D 74 0C 3C A4 76]|8c764218bdca47a19cd79d740c3ca476|
# +-------------------------------------------------+--------------------------------+
Update:
You can also use regexp_replace
to get the exact id_str
, in the same format you have shared, as follows:
from pyspark.sql import SparkSession
import pyspark.sql.functions as F
import uuid
spark = SparkSession.builder.getOrCreate()
data = [(uuid.uuid4().bytes,)]
df = spark.createDataFrame(data, ["id"])
df = df.withColumn(
"id_str",
F.regexp_replace(
F.lower(F.hex("id")),
"(.{8})(.{4})(.{4})(.{4})(.{12})",
"$1-$2-$3-$4-$5"
)
)
df.show(truncate=False)
# +-------------------------------------------------+------------------------------------+
# |id |id_str |
# +-------------------------------------------------+------------------------------------+
# |[F8 25 99 3E 2D 7A 40 E4 A1 24 C0 28 B9 30 F6 03]|f825993e-2d7a-40e4-a124-c028b930f603|
# +-------------------------------------------------+------------------------------------+